Superregular Matrices over Finite Fields

Abstract

A trivially zero minor of a matrix is a minor having all its terms in the Leibniz formula equal to zero. A matrix is superregular if all of its minors that are not trivially zero are nonzero. In the area of Coding Theory, superregular matrices over finite fields are connected with codes with optimum error correcting capabilities. There are two types of superregular matrices that yield two different types of codes. One has in all of its entries a nonzero element, and these are called full superregular matrices. The second interesting class of superregular matrices is formed by lower triangular Toeplitz matrices. In contrast to full superregular matrices, all general constructions of these matrices require very large field sizes. In this work, we investigate the construction of lower triangular Toeplitz superregular matrices over small finite prime fields. Instead of computing all possible minors, we study the structure of finite fields in order to reduce the possible nonzero minors. This allows us to restrict the huge number of possibilities that one needs to check and come up with novel constructions of superregular matrices over relatively small fields. Finally, we present concrete examples of lower triangular Toeplitz superregular matrices of sizes up to 10.

1. Introduction and Preliminaries

Let $\mathbb{F}$ denote a finite field, $F={\left({\mu }_{i,j}\right)}_{1\le i,j\le m}\in {\mathbb{F}}^{m\times m}$, and let ${\mathcal{S}}_m$ represent the symmetric group of order m. Recall that the determinant of F is given by

$$\left|F\right|=\sum _{\sigma \in {\mathcal{S}}_m}\mathrm{sgn}\left(\sigma \right){\mu }_{1\sigma \left(1\right)}\cdots {\mu }_{m\sigma \left(m\right)},$$

(1)

where the sign of the permutation $\sigma$, denoted by $\mathrm{sgn}\left(\sigma \right)$, is 1 (resp., $-1$) if $\sigma$ can be written as product of an even (resp., odd) number of transpositions. A trivial term of the determinant is a term of (1), ${\mu }_{1\sigma \left(1\right)}\cdots {\mu }_{m\sigma \left(m\right)}$, equal to zero. If F is a square submatrix of a matrix B, with entries in $\mathbb{F}$, and all the terms of the determinant of F are trivial, we say that $\left|F\right|$ is a trivial minor of B. We say that B is superregular if all its nontrivial minors are different from zero. Notice that B can be any rectangular matrix. Hence, these matrices can be considered as the correspondents of totally positive (non-negative) matrices for finite fields since they are finite totally positive matrices with elements over a finite field $\mathbb{F}$ whose minors with the possibility to be nonzero, are nonzero; see [1,2,3].

Several notions of superregular matrices have appeared in different areas of Mathematics and Engineering having in common the specification of some properties regarding their minors [4,5,6,7]. In the context of Coding Theory, these matrices have entries in a finite field $\mathbb{F}$ and can be used to generate linear codes with good distance properties. A class of these matrices, which we will call full superregular, were first introduced in the context of block codes. A full superregular matrix is a matrix with all of its minors different from zero and therefore, all of its entries nonzero. For instance, Cauchy matrices are full superregular and can be used to build the so-called Reed–Solomon block codes. Also, circulant Cauchy matrices can be used to construct MDS codes; see [8]. It is well known that a systematic generator matrix $G=\left[I\right|B]$ generates a maximum distance separable (MDS) block code if and only if B is full superregular [8]. The MDS conjecture ([9] Conjecture 6.13) states that if a matrix $B\in {\mathbb{F}}^{k\times (n-k)}$ is superregular with $4\le k\le \left|\mathbb{F}\right|-2$, then $\left|\mathbb{F}\right|\ge n-1$. This conjecture was verified for prime finite fields and remains open for general finite fields. The q-analog of Cauchy superregular matrices has been recently studied in depth (see [10,11]). The problem of building these matrices is related but different to the one studied in [12,13].

Convolutional codes are more involved than block codes and, for this reason, a more general class of superregular matrices had to be introduced.

Definition 1

([14], Definition 3.3). A Toeplitz lower triangular matrix B is defined to be LT-superregular if all of its square submatrices, with the property that all the entries in their diagonal come from the lower triangular part of B, are nonsingular.

Note that due to such a lower triangular configuration, the remaining minors are necessarily zero. Roughly speaking, superregularity asks for all minors that are possibly nonzero, to be nonzero. In [14], it was shown that Toeplitz LT-superregular matrices can be used to construct convolutional codes of rate $k/n$ and degree $\delta$ that are strongly MDS or MDP; see [15,16,17] for recent results on the construction of these codes over small fields. For the rank analog of Toeplitz LT-superregular matrices in the context of the rank metric, see [10,11]. Again, this is due to the fact that the combination of columns of superregular matrices ensures the largest number of possible nonzero entries for any $\mathbb{F}$-linear combination (for this particular lower triangular structure). In other words, it can be deduced from [14] that a lower triangular matrix $B=[b_0{b}_1\dots b_{n-1}]\in {\mathbb{F}}^{n\times n}$, $b_i$ the columns of B, is LT-superregular if and only if for any $\mathbb{F}$-linear combination b of columns $b_{i_1},b_{i_2},\dots ,b_{i_N}$ of B, with $i_j<i_{j+1}$, then $wt\left(b\right)\ge wt\left(b_{i_1}\right)-N+1=(n-i_1)-N+1$, where $wt\left(v\right)$ is the Hamming weight of a vector v, i.e., its number or nonzero coordinates. For a similar result but for more general classes of superregular matrices, not necessarily lower triangular, see [18] (Theorem 3.1).

It is important to note that in this case, due to this triangular configuration, it is hard to come up with an algebraic construction of LT-superregular matrices. There exist, however, two general constructions of these matrices [14,19] although they need very large field sizes. In this paper, we will be interested in finding Toeplitz LT-superregular matrices over small finite prime fields. So, our matrices will be of this form

$$A_{\gamma }=\left[\begin{array}{cccc}{a}_1& 0& \cdots & 0\\ a_2& a_1& \ddots & 0\\ ⋮& \ddots & \ddots & ⋮\\ a_{\gamma }& \cdots & a_2& a_1\end{array}\right].$$

(2)

One important question is how large a finite field must be such that a superregular matrix of a given order can exist over that field. For example, there exists no LT-superregular matrix of order 3 over the field ${\mathbb{F}}_2$ because all the entries in the lower triangular part of a superregular matrix must be nonzero, which means that in this case all such entries would have to be 1; clearly, this does not result in a superregular matrix, since the lower left submatrix of order 2 is singular. The size of the smallest finite field for which exists an LT-superregular matrix of order $\gamma \le 9$ can be seen in

Table 1. Comparison of actual required field sizes and $N_{\gamma }+1$.

$\mathsf{\gamma }$	Minimum Field Size	Upper Bound (${\mathit{N}}_{\mathsf{\gamma }}+1$)
3	3	3
4	5	5
5	7	11
6	11	27
7	17	77
8	31	233
9	59	751
10	≤127	2495

. For $\gamma \ge 10$, the smallest finite field for which exists a Toeplitz LT-superregular matrix of order $\gamma$ is still unknown, but in [20], Hutchinson et al. obtained an upper bound for its size and in [21] the authors showed the existence of LT-superregular matrices of order $10\times 10$ over the field ${\mathbb{F}}_{2^8}$. In [20] (Conjecture 3.5) and in [14] it was conjectured, based on several examples, that an LT-superregular matrix of order $\gamma$ exists over ${\mathbb{F}}_{2^{\gamma }}$ for $\gamma \ge 5$. Recently, new upper bounds on the necessary field size for the existence of these matrices and other superregular matrices with different structure were presented in [22].

Next, we will give a brief description of a method to derive an upper bound on the minimum size a finite field must have such that a superregular matrix of a given order can exist over that field. This idea was first presented in [20] and it will be the starting point of our investigation. Consider

$$X_{\gamma }=\left[\begin{array}{cccc}{x}_1& 0& \cdots & 0\\ x_2& x_1& \ddots & 0\\ ⋮& \ddots & \ddots & ⋮\\ x_{\gamma }& \cdots & x_2& x_1\end{array}\right].$$

a lower triangular Toeplitz matrix with indeterminate entries $x_1,x_2,\dots ,x_{\gamma }$. The determinants of the proper square submatrices of such a matrix are given by nonzero polynomials in these indeterminates. Notice that in any of these polynomials at most the first power of $x_{\gamma }$ can appear; i.e., for each of these polynomials either $x_{\gamma }$ is linear or $x_{\gamma }$ does not appear in any of its terms. We study now those proper square submatrices of $X_{\gamma }$ whose determinants are linear in $x_{\gamma }$. Denote by $L_{\gamma }$ the set of such submatrices and by $L_{\gamma }^{\prime }$, the subset of $L_{\gamma }$ formed by the submatrices of $X_{\gamma }$ which are symmetric over the antidiagonal. Hutchinson et al. proved in [20] that $N_{\gamma }:={\displaystyle \frac{1}{2}}\left(\left|L_{\gamma }\right|+\left|L_{\gamma }^{\prime }\right|\right)$ is an upper bound for the number of different polynomials that can appear as the determinants of elements of $L_{\gamma }$. By computer search, we found that $N_{\gamma }$ is actually the exact number of such polynomials, for $\gamma \le 7$ and $\gamma =9$, but for $\gamma =8$ we have 231 different polynomials and for $\gamma =10$ we have 2489 different polynomials, whereas $N_8=232$ and $N_{10}=2494$. In [20], it is also proved that

$$N_{\gamma }={\displaystyle \frac{\frac{1}{\gamma }\left(\begin{array}{c}2\gamma -2\\ \gamma -1\end{array}\right)+\left(\begin{array}{c}\gamma -1\\ ⌊\frac{\gamma -1}{2}⌋\end{array}\right)}{2}}.$$

(3)

Therefore, given $\gamma \ge 1$, a field $\mathbb{F}$ and a lower triangular Toeplitz matrix $A_{\gamma }\in {\mathbb{F}}^{\gamma \times \gamma }$ (as in (2)), then $A_{\gamma }$ has at most $N_{\gamma }$ different minors that depend on the entry $a_{\gamma }$, all of them being linear on $a_{\gamma }$.

Remark 1.

Notice that ${\left(N_i\right)}_{i=2}^{\infty }$ is an increasing sequence (and $N_1=N_2=1$). If we choose a finite field $\mathbb{F}$, such that $\left|\mathbb{F}\right|>N_{\gamma }$, then we may choose $a_1\in \mathbb{F}$ such that $a_1\ne 0$, then select $a_2\in \mathbb{F}$ such that all the minors involving $a_2$ in the matrix

$$A_2=\left[\begin{array}{cc}{a}_1& 0\\ a_2& a_1\end{array}\right]$$

are nonzero (i.e., any $a_2\ne 0$), then again we can choose $a_3\in \mathbb{F}$ such that all the minors involving $a_3$ in the matrix

$$A_3=\left[\begin{array}{ccc}{a}_1& 0& 0\\ a_2& a_1& 0\\ a_3& a_2& a_1\end{array}\right]$$

are nonzero, and continuing in this way, we may eventually choose $a_{\gamma }\in \mathbb{F}$ such that all of the minors involving $a_{\gamma }$ in the matrix $A_{\gamma }$ (as in (2)) are nonzero. Therefore, all the nontrivial minors of this matrix $A_{\gamma }$ just constructed, are nonzero and so, $A_{\gamma }$ is LT-superregular. This is the idea of the proof of Theorem 1.

Theorem 1

([20]). Let $\mathbb{F}$ be a finite field such that $\left|\mathbb{F}\right|>N_{\gamma }$, then there exists a $\gamma \times \gamma$ LT-superregular matrix over $\mathbb{F}$.

Unfortunately, this upper bound for the minimum field size is not very sharp, as Table 1 (obtained in [20]) demonstrates. The actual minimum field sizes displayed in the table were obtained by exhaustive computer search.

In this paper, we continue the study of lower triangular Toeplitz superregular matrices and focus on finite fields $\mathbb{F}={\mathbb{F}}_p$, where p is an odd prime number. In particular, we investigate the number of different nonzero minors linear on $a_{\gamma }$ of the matrices $A_{\gamma }$, for each odd prime p. In the cases considered, we show that this number is significantly smaller than the $N_{\gamma }$ derived in [20].

2. Smallest Number of Different Nonzero Minors of an LT-Superregular Toeplitz Matrix

Since the multiplication by a constant does not change the superregularity of a matrix, we may assume that $a_1=1$. The following lemma implies that we can also assume that $a_2=1$.

Lemma 1

([23], Theorem 5.8). Suppose that the matrix $A_{\gamma }$ in (2) is LT-superregular and let $\alpha \in {\mathbb{F}}^{\ast }$. Then, the matrix

$$\alpha \otimes A_{\gamma }=\left[\begin{array}{cccc}{a}_1& 0& \cdots & 0\\ \alpha a_2& a_1& \ddots & 0\\ ⋮& \ddots & \ddots & ⋮\\ {\alpha }^{\gamma -1}{a}_{\gamma }& \cdots & \alpha a_2& a_1\end{array}\right]$$

is also superregular.

From now on, we will consider

$$A_{\gamma }=A_{\gamma }(1,1,a_3,\cdots ,a_{\gamma })=\left[\begin{array}{ccccc}1& 0& \cdots & \cdots & 0\\ 1& 1& \ddots & & ⋮\\ a_3& 1& \ddots & \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & 0\\ a_{\gamma }& \cdots & a_3& 1& 1\end{array}\right]\in {\mathbb{F}}^{\gamma \times \gamma }.$$

(4)

In this section, we are interested in studying the smallest possible number of different nonzero minors of $A_{\gamma }$ linear on $a_{\gamma }$, for each $\gamma$ with $3\le \gamma \le 9$. For $\gamma \le 6$, we are able to compute the smallest number of different nonzero minors for every finite prime field. We will also exhibit plenty of superregular matrices for each $3\le \gamma \le 10$.

Throughout this section, the following result about quadratic residues will be helpful.

Lemma 2.

Let $p>5$ be an odd prime number. Then, the following hold:

1. 

$x^2\equiv -1modp$ is solvable if and only if $p\equiv 1mod4$;

2. 

$x^2\equiv 2modp$ is solvable if and only if $p\equiv \pm 1mod8$;

3. 

$x^2\equiv 3modp$ is solvable if and only if $p\equiv \pm 1mod12$;

4. 

$x^2\equiv -3modp$ is solvable if and only if $p\equiv 1mod3$;

5. 

$x^2\equiv 5modp$ is solvable if and only if $p\equiv \pm 1mod5$.

Proof. 

(1) and (2) are well-known results [24]. Using the quadratic reciprocity law, we easily obtain the remaining statements. □

Throughout this work, we will denote by $\sqrt{u}$ the smallest solution of $x^2\equiv umodp$, when it exists, belonging to the complete system of residues $\{0,1,\dots ,p-1\}$.

2.1. Cases $\gamma =3$ and $\gamma =4$

If $\gamma =3$, then there are two minors with the entry $a_3$, namely $\left|\right[a_3\left]\right|$ and $\left|\begin{array}{cc}1& 1\\ a_3& 1\end{array}\right|$, so

$$a_3\notin S_3=\{0,1\}.$$

(5)

Hence $\mathbb{F}$ must have at least 3 elements. Therefore $p\ge 3$. For example, if we take $a_3\equiv -1modp$, or $a_3\equiv {\displaystyle \frac{1}{2}}modp$, then $A_3(1,1,a_3)$ is LT-superregular.

If $\gamma =4$, then $N_4=4$, i.e., there are four different nonzero minors with the entry $a_4$, on the variables $a_3$ and $a_4$, namely,

$$|\left[a_4\right]|,\left|\begin{array}{cc}1& 1\\ a_4& a_3\end{array}\right|,\left|\begin{array}{cc}{a}_3& 1\\ a_4& a_3\end{array}\right|\mathrm{and}\left|\begin{array}{ccc}1& 1& 0\\ a_3& 1& 1\\ a_4& a_3& 1\end{array}\right|.$$

So, we must have

$$a_4\notin S_4=\left\{0,a_3,a_3^2,2a_3-1\right\}modp.$$

(6)

Since $a_3\notin S_3$, we only have $\left|S_4\right|=3$ when $a_3={\displaystyle \frac{1}{2}}modp$; also, we cannot have less than three different minors. Hence, in this case we can always choose, for example, $a_4=1$. Therefore, we just proved the following result.

Theorem 2.

For any $p\ge 5$, the minimum number of different minors of $A_4$ involving $a_4$ is 3. Moreover, if we take $a_3\equiv {\displaystyle \frac{1}{2}}modp$, and if $a_4\notin \{0,{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{4}}\}modp$, then $A_4(1,1,{\displaystyle \frac{1}{2}},a_4)$ is LT-superregular over $\mathbb{F}$.

2.2. Case $\gamma =5$

Although $N_5=10$, in this subsection we will prove that it is possible to construct LT-superregular matrices $A_5$ over $\mathbb{F}$, with $p\ge 7$ as the number of different minors involving $a_5$ can be reduced to 6 for $p=7$, and it can be reduced to 7 for all $p\ge 11$, by selecting properly $a_3$ and $a_4$.

The ten different minors in the variables $a_3,a_4$, and $a_5$ are

$$|\left[a_5\right]|,\left|\begin{array}{cc}1& 1\\ a_5& a_4\end{array}\right|,\left|\begin{array}{cc}{a}_3& 1\\ a_5& a_4\end{array}\right|,\left|\begin{array}{cc}{a}_3& 1\\ a_5& a_3\end{array}\right|,\left|\begin{array}{cc}{a}_4& a_3\\ a_5& a_4\end{array}\right|,$$

$$\left|\begin{array}{ccc}1& 1& 0\\ a_3& 1& 1\\ a_5& a_4& a_3\end{array}\right|,\left|\begin{array}{ccc}1& 1& 0\\ a_4& a_3& 1\\ a_5& a_4& a_3\end{array}\right|,\left|\begin{array}{ccc}1& 1& 0\\ a_4& a_3& 1\\ a_5& a_4& 1\end{array}\right|,\left|\begin{array}{ccc}{a}_3& 1& 1\\ a_4& a_3& 1\\ a_5& a_4& a_3\end{array}\right|,$$

$$\mathrm{and}\left|\begin{array}{cccc}1& 1& 0& 0\\ a_3& 1& 1& 0\\ a_4& a_3& 1& 1\\ a_5& a_4& a_3& 1\end{array}\right|.$$

Now, if p is sufficiently large such that $a_3$ satisfies (5), $a_4$ satisfies (6), and we also have

$$\begin{array}{cc}\hfill a_5\notin S_5=\left\{0,a_4,a_3{a}_4,a_3^2,{\displaystyle \frac{a_4^2}{a_3}},\right.& a_3^2-a_3+a_4,-a_3^2+a_3{a}_4+a_4,\hfill \\ \hfill & \left.2a_4-a_3,{\displaystyle \frac{a_3^3-2a_3{a}_4+a_4^2}{a_3-1}},a_3^2-3a_3+2a_4+1\right\}modp,\hfill \end{array}$$

(7)

then $A_5$ is LT-superregular over $\mathbb{F}$.

Next, we will find out what is the minimum number of nontrivial minors involving $a_5$, of $A_5$, for any prime $p\ge 5$.

Since $a_3$ satisfies (5) and $a_4$ satisfies (6), any two of the expressions $0,a_4,a_3{a}_4$, and $a_3^2$ cannot be equal. Therefore, $|S_5|\ge 4$. Also, we cannot have ${\displaystyle \frac{a_4^2}{a_3}}\in \{0,a_4,a_3{a}_4\}$, but we can have equality of other members of $S_5$. We will analyze each of the pertinent cases, but only explaining all the steps in the first case, since the other cases are similar.

2.2.1. If $a_4^2=a_3^3$

Let $p\ge 5$ be a prime number. If $a_4^2=a_3^3$ then the Legendre symbol $\left({\displaystyle \frac{a_3}{p}}\right)=1$. Let $b\in \mathbb{F}$ such that $a_3=b^2\notin S_3$ and $a_4=b^3\notin S_4$. Then,

$$\begin{array}{cc}\hfill S_5=\left\{0,b^3,b^5,b^4,b^4,b^4+b^3-b^2,b^5-b^4+b^3,2b^3-b^2,\right.& {\displaystyle \frac{2b^6-2b^5}{b^2-1}},\hfill \\ \hfill & \left.b^4+2b^3-3b^2+1\right\}modp.\hfill \end{array}$$

(8)

If $p=5$, $A_5$ is a LT-superregular matrix only if $|S_5|=4$, but this never happens, because b cannot be 0, 1, or $-1$, and if $b=\pm 2$, $S_{5}mod5$ will have 5 elements.

Let $p\ge 7$. Although we are only interested in finding the values of b that reduce the size of $S_5$ the most, using the help of a computer program, we found out all the possibilities of having at least two elements of $S_5$ equal, with the conditions (5) and (6) being satisfied. This happens when b satisfies at least one of the conditions below:

(a) 

$b^4+b^3-b^2\equiv 0modp$, which implies $b\equiv {\displaystyle \frac{-1\pm \sqrt{5}}{2}}modp$ and, using property $\left(5\right)$ of Lemma 2, $p\equiv \pm 1mod5$;

(b) 

$b^5-b^4+b^3\equiv 0modp$, which implies $b\equiv {\displaystyle \frac{1\pm \sqrt{-3}}{2}}modp$ and, using property $\left(4\right)$ of Lemma 2, $p\equiv 1mod3$;

(c) 

$2b^3-b^2\equiv 0modp$, which implies $b\equiv {\displaystyle \frac{1}{2}}modp$;

(d) 

$2b^3-b^2\equiv b^{5}modp$, which implies $b\equiv {\displaystyle \frac{-1\pm \sqrt{5}}{2}}modp$ and, using property $\left(5\right)$ of Lemma 2, $p\equiv \pm 1mod5$;

(e) 

${\displaystyle \frac{2b^6-2b^5}{b^2-1}}\equiv b^{3}modp$, which implies $b\equiv -{\displaystyle \frac{1}{2}}modp$;

(f) 

${\displaystyle \frac{2b^6-2b^5}{b^2-1}}\equiv 2b^3-b^{2}modp$, which implies $b\equiv \pm {\displaystyle \frac{\sqrt{2}}{2}}modp$ and, using property $\left(2\right)$ of Lemma 2, $p\equiv \pm 1mod8$;

(g) 

$b^4+2b^3-3b^2+1\equiv 0modp$;

(h) 

$b^4+2b^3-3b^2+1\equiv b^{3}modp$, which implies that b is a solution of the congruence $x^3+2x^2-x-1\equiv 0modp$;

(i) 

$b^4+2b^3-3b^2+1\equiv b^{5}modp$, which implies that b is a solution of the congruence $x^4-2x^2+x+1\equiv 0modp$;

(j) 

$b^4+2b^3-3b^2+1\equiv b^{4}modp$, which implies $b\equiv -{\displaystyle \frac{1}{2}}modp$.

Remark 2.

The careful reader may have noticed that we obtain $b\equiv {\displaystyle \frac{1\pm \sqrt{5}}{2}}modp$ when any two of the elements of the set $\{b^4+b^3-b^2,b^5-b^4+b^3,{\displaystyle \frac{2b^6-2b^5}{b^2-1}},b^4+2b^3-3b^2+1\}$ are equal, but this would imply $a_4=2a_3-1$, which violates condition (6).

Next, we analyze when two conditions above can be simultaneously satisfied:

(i) 

If ${\displaystyle \frac{-1\pm \sqrt{5}}{2}}\equiv {\displaystyle \frac{1\pm \sqrt{-3}}{2}}modp$, then $\pm 2\sqrt{-15}\equiv 2modp$, which never happens for any prime $p\ge 7$;

(ii) 

Clearly, ${\displaystyle \frac{-1\pm \sqrt{5}}{2}}\notin \{\pm {\displaystyle \frac{1}{2}}\}modp$, for any prime $p\ge 7$;

(iii) 

If ${\displaystyle \frac{-1\pm \sqrt{5}}{2}}\equiv \pm {\displaystyle \frac{\sqrt{2}}{2}}modp$, then $\pm 2\sqrt{10}\equiv -6modp$, which never happens for any prime $p\ge 7$;

(iv) 

It can be seen that $b\equiv {\displaystyle \frac{-1\pm \sqrt{5}}{2}}modp$ is not a solution of the congruences in conditions g), h), and i) for any prime $p\ge 7$;

(v) 

We never have ${\displaystyle \frac{1\pm \sqrt{-3}}{2}}\equiv {\displaystyle \frac{1}{2}}modp$, if $p\ge 7$;

(vi) 

If ${\displaystyle \frac{1\pm \sqrt{-3}}{2}}\equiv -{\displaystyle \frac{1}{2}}modp$ then $p=7$, and $b=3$;

(vii) 

If ${\displaystyle \frac{1\pm \sqrt{-3}}{2}}\equiv \pm {\displaystyle \frac{\sqrt{2}}{2}}modp$, then $\pm 2\sqrt{-6}\equiv 2modp$, which implies $p=7$ and $b=5$;

(viii) 

$b\equiv {\displaystyle \frac{1\pm \sqrt{-3}}{2}}modp$ does not satisfy the condition g) for any prime $p\ge 7$, but it satisfies the condition h), when $p=13$ and $b=4$, and satisfies the condition i), when $p=7$ and $b=5$;

(ix) 

If $b\equiv {\displaystyle \frac{1}{2}}modp$ then $b\notin \left\{-{\displaystyle \frac{1}{2}},\pm {\displaystyle \frac{\sqrt{2}}{2}}\right\}modp$;

(x) 

$b\equiv {\displaystyle \frac{1}{2}}modp$ does not satisfy the condition g) for any prime $p\ge 7$, but it satisfies the condition h), when $p=7$ (i. e. $b=4$), and satisfies the condition i), when $p=17$ (i.e., $b=9$);

(xi) 

Clearly, $-{\displaystyle \frac{1}{2}}\neg \equiv \pm {\displaystyle \frac{\sqrt{2}}{2}}modp$, for any prime $p\ge 7$. Also, $b\equiv -{\displaystyle \frac{1}{2}}modp$ is not a solution of the congruences in conditions g), h), and i) for any prime $p\ge 7$;

(xii) 

$b\equiv \pm {\displaystyle \frac{\sqrt{2}}{2}}modp$ does not satisfy the condition h) for any prime $p\ge 7$, but it satisfies the conditions g) when $p=7$ and $b=2$, and i) when $p=7$ and $b=5$.

From the above, when $a_4^2=a_3^3$, we conclude that the minimum number of different minors of $A_5$ is as follows:

• 6 if $p=7$, obtained when

  $$b\equiv -{\displaystyle \frac{1}{2}}\equiv {\displaystyle \frac{1-\sqrt{-3}}{2}}\equiv 3mod7$$

  and when

  $$b\equiv {\displaystyle \frac{1\pm \sqrt{-3}}{2}}\equiv \pm {\displaystyle \frac{\sqrt{2}}{2}}\equiv 5mod7;$$
• 7 if $p=13$, obtained when $b\equiv -{\displaystyle \frac{1}{2}}\equiv 6mod13$ and when

  $$b\equiv {\displaystyle \frac{1-\sqrt{-3}}{2}}\equiv 4mod13;$$
• 7 if $p=17$, obtained when $b\equiv -{\displaystyle \frac{1}{2}}\equiv 8mod17$ and when $b\equiv {\displaystyle \frac{1}{2}}\equiv 9mod17$;
• 7 if $p\ge 11$ and $p\equiv \pm 1mod5$, obtained when $b\equiv {\displaystyle \frac{-1\pm \sqrt{5}}{2}}modp$ or when $b\equiv -{\displaystyle \frac{1}{2}}modp$;
• 7 if $p\ge 23$ and $p\equiv \pm 2mod5$, obtained only when $b\equiv -{\displaystyle \frac{1}{2}}modp$.

From now on, we assume $a_4^2\ne a_3^3$, so $|S_5|\ge 5$. Therefore, we only consider $p\ge 7$.

Unlike the previous case, where we analyzed when each pair of elements of $S_5$ could be equal, from now on we will only describe when we obtain the minimum number of elements of $S_5$, for each prime $p\ge 7$.

2.2.2. If $a_4={\displaystyle \frac{a_3}{2}}$

It can be seen that if $a_4={\displaystyle \frac{a_3}{2}}$, then the eighth element of $S_5$ is zero, and the only way to have $S_5$ with at most 7 elements is when the following hold:

• $a_3={\displaystyle \frac{3}{4}}$. In this case, $S_5=\left\{0,{\displaystyle \frac{3}{8}},{\displaystyle \frac{9}{32}},{\displaystyle \frac{9}{16}},{\displaystyle \frac{3}{16}},{\displaystyle \frac{3}{16}},{\displaystyle \frac{3}{32}},0,0,{\displaystyle \frac{1}{16}}\right\}$, which has 6 elements when $p=7$ and 7 elements when $p>7$.
• $p=13$ and $a_3\equiv {\displaystyle \frac{1}{3}}\equiv 9mod13$, having the fourth element of $S_5$ equal to the seventh and the fifth equal to the tenth.

2.2.3. If $a_4=3a_3-a_3^2-1$

Another pertinent case is when $a_4=3a_3-a_3^2-1$, making the second element of $S_5$ equal to the tenth, and it can be seen that $S_5$ only has at most 7 elements when the following hold:

• $p\equiv \pm 1mod5$ and $a_3\equiv {\displaystyle \frac{-1\pm \sqrt{5}}{2}}modp$, having the fourth element of $S_5$ equal to the seventh and the eighth equal to the ninth;
• $p=11$ and $a_3\equiv {\displaystyle \frac{2}{3}}\equiv 8mod11$, having the fourth element of $S_5$ equal to the eighth and the ninth equal to zero;
• $p=13$ and $a_3\equiv {\displaystyle \frac{2}{3}}\equiv 5mod13$, having the fourth element of $S_5$ equal to the eighth and the seventh equal to zero.

The following cases only give the minimum number of different minors for particular small values of p.

2.2.4. If $a_4={\displaystyle \frac{a_3+a_3^2}{2}}$

In this case, we have the fourth element of $S_5$ equal to the eighth element, and the only possibilities for $S_5$ to have at most 7 elements are when the following hold:

• $p=13$ and $a_3\equiv {\displaystyle \frac{1\pm \sqrt{-1}}{2}}\equiv 3,11mod13$, having the tenth element of $S_5$ equal to zero and the fifth equal to the sixth, or the second equal to the second equal to the ninth, depending on the value of $a_3$;
• $p=17$ and $a_3\equiv -3mod17$ or $a_3\equiv \sqrt{-1}\equiv 4mod17$, having the seventh element of $S_5$ equal to the zero and the fifth equal to the tenth.

2.2.5. If $a_4={\displaystyle \frac{a_3^2}{a_3+1}}$

In this case, we have the seventh element of $S_5$ equal to zero, and the only possibilities for $S_5$ to have at most 7 elements are when the following hold:

• $p=7$ and $a_3\equiv {\displaystyle \frac{1-\sqrt{-3}}{2}}\equiv 3mod7$, having the third element of $S_5$ equal to the eighth, the fourth equal to the tenth and the fifth equal to the sixth;
• $p=13$ and $a_3\equiv {\displaystyle \frac{1-\sqrt{-3}}{2}}\equiv 4mod13$, having the third element of $S_5$ equal to the eighth and the fifth equal to the tenth;
• $p=17$ and $a_3\equiv \sqrt{2}-1\equiv 5mod17$, having the fourth element of $S_5$ equal to the tenth and the eighth equal to the ninth.

2.2.6. If $a_4=a_3-a_3^2$

In this case, we have the sixth element of $S_5$ equal to zero, and the only possibilities for $S_5$ to have at most 7 elements are when the following hold:

• $p=11$ and $a_3\equiv {\displaystyle \frac{1}{3}}\equiv 4mod11$, having the fourth element of $S_5$ equal to the eighth and the fifth equal to the tenth;
• $p=13$ and $a_3\equiv {\displaystyle \frac{1}{3}}\equiv 9mod13$, having the fourth element of $S_5$ equal to the eighth and the third equal to the tenth;

2.2.7. If $a_4={\displaystyle \frac{a_3^2-3a_3+1}{a_3-2}}$

In this case, we have the third element of $S_5$ equal to the tenth, and the only possibilities for $S_5$ to have at most 7 elements are when the following hold:

• $p=7$ and $a_3\equiv {\displaystyle \frac{1+\sqrt{-3}}{2}}\equiv 10mod7$, having the fourth element of $S_5$ equal to the seventh, the eighth equal to zero, and the second equal to the ninth;
• $p=13$ and $a_3\equiv {\displaystyle \frac{3+\sqrt{-3}}{2}}\equiv 11mod13$, having the eighth element of $S_5$ equal to the ninth and the sixth equal to zero.
• $p=13$ and $a_3\equiv {\displaystyle \frac{1+\sqrt{-3}}{2}}\equiv 10mod13$, having the fourth element of $S_5$ equal to the seventh and the ninth equal to zero.

Remark 3.

Notice that in the second possibility above, we also have $a_4=a_3-a_3^2$, and so it could be included in the previous subsection, but it is here, because we obtained this expression for $a_3$ starting with $a_4={\displaystyle \frac{a_3^2-3a_3+1}{a_3-2}}$ and making the eighth element equal to the ninth, obtaining the sixth element equal to $-{\displaystyle \frac{\sqrt{-3}+7}{\sqrt{-3}-1}}$, which is zero for $p=13$ and $\sqrt{-3}=6$, making $\left|S_5\right|=7$.

2.2.8. If $a_4={\displaystyle \frac{a_3}{2-a_3}}$

In this case, we have the third element of $S_5$ equal to the eighth, and the only possibilities for $S_5$ to have at most 7 elements are when the following hold:

• $p=7$ and $a_3\equiv {\displaystyle \frac{3-\sqrt{-3}}{2}}\equiv 4mod7$, having the ninth element of $S_5$ equal to zero, the fourth equal to the seventh, and the fifth equal to the tenth;
• $p=13$ and $a_3\equiv {\displaystyle \frac{3-\sqrt{-3}}{2}}\equiv 5mod13$, having the ninth element of $S_5$ equal to zero and the fourth equal to the tenth.

Define

$$P_{5,7}=\left\{(a_3,a4)\right|\left|S_5\right|=6, \mathrm{and} \left(5\right) \mathrm{and} \left(6\right) \mathrm{are} \mathrm{satisfied}\}$$

and, for $p\ge 7$ prime,

$$P_{5,p}=\left\{(a_3,a4)\right|\left|S_5\right|=7, \mathrm{and} \left(5\right) \mathrm{and} \left(6\right) \mathrm{are} \mathrm{satisfied}\}.$$

Combining all the subsections above, we obtain the following result:

Theorem 3.

1. $\left|P_{5,p}\right|=12$ elements if $p=13$;

2. 

$\left|P_{5,p}\right|=8$ elements if $p=11$;

3. 

$\left|P_{5,p}\right|=6$ if $p=7$, $p=17$ or $11\ne p\equiv \pm 1mod5$;

4. 

$\left|P_{5,p}\right|=2$ if $p\ne 7$, $p\ne 13$, $p\ne 17$ and $p\equiv \pm 2mod5$.

We finish this section with three particular pairs $(a_3,a_4)$, where two of them give the minimum number of elements of $S_5$ for every prime $p\ge 7$.

If $a_3\equiv {\displaystyle \frac{1}{4}}modp$ and $a_4\equiv -{\displaystyle \frac{1}{8}}modp$, then

$$S_5=\left\{0,-{\displaystyle \frac{1}{8}},-{\displaystyle \frac{1}{32}},{\displaystyle \frac{1}{16}},{\displaystyle \frac{1}{16}},-{\displaystyle \frac{5}{16}},-{\displaystyle \frac{7}{32}},-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{8}},{\displaystyle \frac{1}{16}}\right\},$$

i.e.,

$$a_3^2={\displaystyle \frac{a_4^2}{a_3}}=a_3^2-3a_3+2a_4+1\equiv {\displaystyle \frac{1}{16}}modp$$

and

$${\displaystyle \frac{a_3^3-2a_3{a}_4+a_4^2}{a_3-1}}=a_4\equiv -{\displaystyle \frac{1}{8}}modp.$$

In the case $p=7$, we even have $-a_3^2+a_3{a}_4+a_4\equiv 0mod7$. Also, if $a_3\equiv {\displaystyle \frac{3}{4}}modp$ and $a_4\equiv {\displaystyle \frac{3}{8}}$, then

$$S_5=\left\{0,{\displaystyle \frac{3}{8}},{\displaystyle \frac{9}{32}},{\displaystyle \frac{9}{16}},{\displaystyle \frac{3}{16}},{\displaystyle \frac{3}{16}},{\displaystyle \frac{3}{32}},0,0,{\displaystyle \frac{1}{16}}\right\},$$

i.e.,

$$2a_4-a_3={\displaystyle \frac{a_3^3-2a_3{a}_4+a_4^2}{a_3-1}}=0$$

and

$${\displaystyle \frac{a_4^2}{a_3}}=a_3^2-a_3+a_4\equiv {\displaystyle \frac{3}{16}}modp.$$

In this case, we also have $a_3{a}_4\equiv a_3^2-3a_3+2a_4+1\equiv 4mod7$. Moreover, it is easy to see that if $a_5\equiv {\displaystyle \frac{1}{4}}modp$, then $a_5\notin S_5$ in both cases.

If we follow the previous subsections and consider $a_3={\displaystyle \frac{1}{2}}$ and $a_4=1$, we obtain

$$S_5=\left\{0,1,{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{4}},2,{\displaystyle \frac{3}{4}},{\displaystyle \frac{5}{4}},{\displaystyle \frac{3}{2}},-{\displaystyle \frac{1}{4}},{\displaystyle \frac{7}{4}}\right\},$$

so there are ten different expressions involving $a_5$ for $p\ge 11$. Hence, if we consider the entries that give the smallest number of minors for matrices of order 3 and 4 we will not be able to find the smallest number of minors for matrices of order 5. This property also occurs for larger matrices, which leads to the following remark.

Remark 4.

As we have just shown, in order to find the smallest finite field for a given order we cannot just use the entries that give the smallest number of minors for matrices of smaller order, we have to try all possible combinations of entries. This fact makes the problem of building LT-superregular matrices over small fields a nontrivial problem.

Nevertheless, since we always have $-{\displaystyle \frac{1}{2}}\notin S_5$, we can still use those entries to create superregular matrices, as long as $p\ge 11$. Therefore, we have the following result.

Theorem 4.

Let $p\ge 7$ and $(a_3,a_4)\in \left\{\left({\displaystyle \frac{1}{4}},-{\displaystyle \frac{1}{8}}\right),({\displaystyle \frac{3}{4}},{\displaystyle \frac{3}{8}})\right\}modp$, then the following hold:

1. 

If $p=7$, then $A_5(1,1,a_3,a_4,a_5)$ has 6 different minors involving $a_5$;

2. 

If $p>7$, then $A_5(1,1,a_3,a_4,a_5)$ has 7 different minors involving $a_5$;

3. 

If $p\ge 7$, then $A_5(1,1,a_3,a_4,{\displaystyle \frac{1}{4}})$ is LT-superregular;

4. 

If $p\ge 11$, then $A_5(1,1,{\displaystyle \frac{1}{2}},1,-{\displaystyle \frac{1}{2}})$ is LT-superregular.

Proof. 

The statements $\left(1\right)$ and $\left(2\right)$ were obtained above. Since the conditions (5), (6), and (7) are satisfied, the statements $\left(3\right)$ and $\left(4\right)$ are also true. □

2.3. Case $\gamma =6$

We have $N_6=26$ and if p is sufficiently large, $a_6\in \mathbb{F}$ and

$$\begin{array}{cc}\hfill a_6& \notin S_6=\left\{0,a_5,a_3{a}_4,a_4^2,{\displaystyle \frac{a_5^2}{a_4}},{\displaystyle \frac{a_5^2-2a_3{a}_4{a}_5+a_4^3}{a_4-a_3^2}},{\displaystyle \frac{a_3{a}_5-a_4^2+a_4{a}_5}{a_3}},a_5-a_3{a}_4+a_4^2,\right.\hfill \\ \hfill & a_5-a_3{a}_4+a_3{a}_5,{\displaystyle \frac{a_4{a}_5+a_3^2{a}_5-a_3{a}_4^2-a_5^2}{a_3-a_4}},{\displaystyle \frac{a_3{a}_5+a_4^2-a_3^2{a}_4-a_4{a}_5}{1-a_3}},\hfill \\ \hfill & {\displaystyle \frac{a_5-2a_3{a}_4+a_3^3+2a_4^2-a_3^2{a}_4-a_4{a}_5}{1-a_3}},-1+4a_3-3a_4-3a_3^2+2a_5+2a_3{a}_4,a_3{a}_5,\hfill \\ \hfill & a_3(2a_5-a_3{a}_4),a_3(a_4-a_3^2+a_5),{\displaystyle \frac{a_4{a}_5}{a_3}},-(a_4-a_3^2-a_5+a_3^3-a_3{a}_5),\hfill \\ \hfill & -(a_4-a_3^2-2a_5+2a_3{a}_4-a_4^2),-a_3^2+2a_3{a}_4,-a_4+2a_5,-a_3^2+a_5+a_3{a}_4,\hfill \\ \hfill & -a_4+a_5+a_3{a}_4,a_3-2a_4-a_3^2+2a_5+a_3{a}_4,a_3-a_4-2a_3^2+a_5+2a_3{a}_4,\hfill \\ \hfill & \left.{\displaystyle \frac{2a_3{a}_5+a_4^2-3a_3^2{a}_4+a_3^4-2a_4{a}_5-2a_3^2{a}_5+2a_3{a}_4^2+a_5^2}{1-2a_3+a_4}}\right\},\hfill \end{array}$$

(9)

then all of the minors of $A_6$ involving $a_6$ are nonzero (there are at most 26 minors). We will show that for any prime $p\ge 11$ we can choose $a_3,a_4$, and $a_5$ such that the number of elements of $S_6$ is at most 14 (being smaller than 14 for most of the primes, because it will only be 14 if and only if $p\equiv 87,107mod120$, as we will see below).

We are able to state and formally prove a result about the minimum number of elements of $S_6$ for any $p\ge 11$ (an exhaustive computer search using Maple helped us identify the necessary conditions).

Theorem 5.

Let $p\ge 11$, then the following hold:

1. 

If $p\equiv 1mod4$ and $(a_3,a_4,a_5)=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1\pm \sqrt{-1}}{4}},{\displaystyle \frac{1\pm 2\sqrt{-1}}{8}}\right)$, then $\left|S_6\right|=13$, except when $p=13$ or $p=17$, in which case we have $\left|S_6\right|=12$;

2. 

If $p\equiv \pm 1mod8$ and $(a_3,a_4,a_5)=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{2\pm \sqrt{2}}{8}},{\displaystyle \frac{1\pm \sqrt{2}}{8}}\right)$, then $\left|S_6\right|=13$, except when $p=23$, in which case we have $\left|S_6\right|=12$;

3. 

If $p\equiv 1mod3$ and $(a_3,a_4,a_5)=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{3\pm \sqrt{-3}}{8}},{\displaystyle \frac{2\pm \sqrt{-3}}{8}}\right)$, then $\left|S_6\right|=13$, except when $p=13$, in which case we have $\left|S_6\right|=11$;

4. 

If $p\equiv \pm 1mod5$ and

(a) 

$(a_3,a_4,a_5)=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1\pm \sqrt{5}}{8}},{\displaystyle \frac{\pm \sqrt{5}}{8}}\right)$, then $\left|S_6\right|=13$, except when $p=11$, in which case we have $\left|S_6\right|=11$;

(b) 

$(a_3,a_4,a_5)=\left({\displaystyle \frac{1}{2}},{\left({\displaystyle \frac{1\pm \sqrt{5}}{4}}\right)}^2,{\left({\displaystyle \frac{1\pm \sqrt{5}}{4}}\right)}^3\right)$, then $\left|S_6\right|=13$, except when $p=11$, in which case we have $\left|S_6\right|=10$ when we consider the plus sign for $a_4$ and $a_5$;

5. 

If $p\equiv \pm 1mod12$ and $(a_3,a_4,a_5)=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{-1\pm \sqrt{3}}{4}},{\displaystyle \frac{-3\pm 2\sqrt{3}}{8}}\right)$, then $\left|S_6\right|=14$, except when $p=11$, in which case we have $\left|S_6\right|=10$, when $p=13$, in which case we have $\left|S_6\right|=12$ and $p=23$ or $p=61$ in which case we have $\left|S_6\right|=13$.

Moreover, if $(a_3,a_4,a_5)$ is any of the vectors above for an appropriate prime p, except when $p=11$ and $(a_3,a_4,a_5)=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1+\sqrt{5}}{8}},{\displaystyle \frac{\sqrt{5}}{8}}\right)$, then $A_6(1,1,a_3,a_4,a_5,a_6)$ is LT-superregular, for any $a_6\notin S_{6}modp$.

Proof. 

For each prime $11\le p\le 257$, and using Maple, we found for which values of $(a_3,a_4,a_5)$ we would achieve the minimum of $|S_{6}modp|$ and after identifying which elements of $S_6$ become equal, we deduce the expressions stated in the theorem for $(a_3,a_4,a_5)$. For each prime $11\le p\le 257$, there is at least one vector $(a_3,a_4,a_5)$ with $a_3\equiv {\displaystyle \frac{1}{2}}modp$ (in fact, for many values of p all of the solutions have this value for $a_3$), for which $|S_{6}modp|$ is minimal. Therefore, we will only consider this case. Although we use Maple to help us deduce expressions for $a_4$ and $a_5$ that give the minimum number of minors, it can be easily seen that the following arguments are valid for every $p\ge 11$.

Suppose $p\equiv 1mod4$, with the calculations in Maple we found that the number of elements of $S_{6}modp$ is minimal when the thirteenth element of $S_6$ is null and the fourth and twenty-third elements are equal, i.e.,

$$\begin{array}{cc}\hfill -1+4a_3-3a_4-3a_3^2+2a_5+2a_3{a}_4& =0\hfill \\ \hfill -a_4+a_5+a_3{a}_4& =a_4^2.\hfill \end{array}$$

Considering $a_3\equiv {\displaystyle \frac{1}{2}}modp$ and solving this system of equations, we obtain

$$(a_4,a_5)\in \left\{\left({\displaystyle \frac{1\pm \sqrt{-1}}{4}},{\displaystyle \frac{1\pm 2\sqrt{-1}}{8}}\right)\right\}.$$

Substituting the first solution in $S_6$, we obtain

$$\begin{array}{cc}\hfill S_6=\left\{0,{\displaystyle \frac{3\sqrt{-1}-1}{16}},\right.& {\displaystyle \frac{2\sqrt{-1}+1}{8}},{\displaystyle \frac{\sqrt{-1}+1}{8}},{\displaystyle \frac{4\sqrt{-1}+1}{16}},{\displaystyle \frac{2\sqrt{-1}+1}{16}},{\displaystyle \frac{3\sqrt{-1}+1}{16}},\hfill \\ & \left.{\displaystyle \frac{5\sqrt{-1}+1}{32}},{\displaystyle \frac{7\sqrt{-1}+1}{32}},{\displaystyle \frac{\sqrt{-1}}{4}},{\displaystyle \frac{\sqrt{-1}}{8}},{\displaystyle \frac{3\sqrt{-1}}{8}},{\displaystyle \frac{3\sqrt{-1}}{16}}\right\}.\hfill \end{array}$$

Notice that $S_6$ has at most 13 elements, for every prime for which $\sqrt{-1}$ exists. If $p=13$ and we take $\sqrt{-1}=5$, then ${\displaystyle \frac{5\sqrt{-1}+1}{32}}=0$ and if we take $\sqrt{-1}=8$, then ${\displaystyle \frac{7\sqrt{-1}+1}{32}}={\displaystyle \frac{3\sqrt{-1}}{8}}$. It can be seen that for $p=17$ we also have $\left|S_6\right|=12$, so the statement $\left(1\right)$ is obtained. Notice that the second solution is also in statement $\left(1\right)$, since the solutions of $x^2\equiv -1modp$ are symmetric.

Suppose $p\equiv \pm 1mod8$, with the calculations in Maple we found that the number of elements of $S_{6}modp$ is minimal when the thirteenth element of $S_6$ (in the expression (9)) is null and the third and fifth elements are equal, i.e.,

$$\begin{array}{cc}\hfill -1+4a_3-3a_4-3a_3^2+2a_5+2a_3{a}_4& =0\hfill \\ \hfill {\displaystyle \frac{a_5^2}{a_4}}& =a_3{a}_4.\hfill \end{array}$$

Considering $a_3\equiv {\displaystyle \frac{1}{2}}modp$ and solving this system of equations, we obtain

$$(a_4,a_5)\in \left\{\left(2\pm {\displaystyle \frac{\sqrt{2}}{8}},{\displaystyle \frac{1\pm \sqrt{2}}{8}}\right)\right\}.$$

Substituting the first solution in $S_6$, we obtain

$$\begin{array}{cc}\hfill S_6=\left\{0,{\displaystyle \frac{1}{8}}\sqrt{2},{\displaystyle \frac{1}{16}}\sqrt{2},{\displaystyle \frac{3}{16}}\sqrt{2},{\displaystyle \frac{1}{8}}+{\displaystyle \frac{3}{32}}\sqrt{2},\right.& {\displaystyle \frac{3}{32}}+{\displaystyle \frac{1}{16}}\sqrt{2},{\displaystyle \frac{3}{32}}+{\displaystyle \frac{3}{32}}\sqrt{2},{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{2},\hfill \\ \hfill {\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{8}}\sqrt{2},{\displaystyle \frac{3}{32}}+{\displaystyle \frac{1}{8}}\sqrt{2},& \left.{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{16}}\sqrt{2},{\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{16}}\sqrt{2},{\displaystyle \frac{1}{16}}+{\displaystyle \frac{3}{32}}\sqrt{2}\right\}.\hfill \end{array}$$

Here, $S_6$ has also at most 13 elements, for every prime for which $\sqrt{2}$ exists. If $p=23$ and we consider $\sqrt{2}=5$, then ${\displaystyle \frac{3}{32}}+{\displaystyle \frac{1}{8}}\sqrt{2}=0$ and if we consider $\sqrt{2}=18$, we also obtain $\left|S_6\right|=12$, hence the statement $\left(2\right)$ is proved. Notice that the second solution is also in statement $\left(2\right)$, since the solutions of $x^2\equiv 2modp$ are symmetric.

Suppose $p\equiv 1mod3$, with the calculations in Maple we found that $|S_{6}modp|$ is minimal when the thirteenth element of $S_6$ (in the expression (9)) is null and the fourth and tenth elements are equal, i.e.,

$$\begin{array}{cc}\hfill -1+4a_3-3a_4-3a_3^2+2a_5+2a_3{a}_4& =0\hfill \\ \hfill {\displaystyle \frac{a_4{a}_5+a_3^2{a}_5-a_3{a}_4^2-a_5^2}{a_3-a_4}}& =a_4^2.\hfill \end{array}$$

Solving this system of equations, we obtain

$$(a_4,a_5)\in \left\{\left({\displaystyle \frac{3\pm \sqrt{-3}}{8}},{\displaystyle \frac{2\pm \sqrt{-3}}{8}}\right),\left({\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{8}}\right)\right\},$$

but by Theorem 2 we cannot have $a_4={\displaystyle \frac{1}{4}}$. Substituting the first solution in $S_6$, we obtain

$$\begin{array}{cc}\hfill S_6& =\left\{0,{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{-3},{\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{8}}\sqrt{-3},{\displaystyle \frac{3}{16}}+{\displaystyle \frac{3}{16}}\sqrt{-3},{\displaystyle \frac{3}{32}}+{\displaystyle \frac{5}{32}}\sqrt{-3},{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}\sqrt{-3},\right.\hfill \\ \hfill & {\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{16}}\sqrt{-3},{\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{16}}\sqrt{-3},{\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{16}}\sqrt{-3},{\displaystyle \frac{3}{32}}+{\displaystyle \frac{3}{32}}\sqrt{-3},{\displaystyle \frac{5}{32}}+{\displaystyle \frac{3}{32}}\sqrt{-3},\hfill \\ \hfill & \left.{\displaystyle \frac{5}{32}}+{\displaystyle \frac{5}{32}}\sqrt{-3},{\displaystyle \frac{5}{32}}+{\displaystyle \frac{11}{96}}\sqrt{-3}\right\}.\hfill \end{array}$$

Again, $S_6$ has at most 13 elements, for every prime for which $\sqrt{-3}$ exists. As before, it can be seen that if $p=13$, then $\left|S_6\right|=11$ and so, we obtain the statement $\left(3\right)$. Notice that the second solution is also in statement $\left(3\right)$, since the solutions of $x^2\equiv -3modp$ are symmetric.

Suppose $p\equiv \pm 1mod5$, with the calculations in Maple we found that the number of elements of $S_{6}modp$ is minimal when the thirteenth element of $S_6$ (in the expression (9)) is null and the following hold:

(a) 

The third and sixth elements of $S_6$ are equal, i.e.,

$$\begin{array}{cc}\hfill -1+4a_3-3a_4-3a_3^2+2a_5+2a_3{a}_4& =0\hfill \\ \hfill {\displaystyle \frac{a_4^3-2a_3{a}_4{a}_5+a_5^2}{a_4-a_3^2}}& =a_3{a}_4.\hfill \end{array}$$

Considering $a_3\equiv {\displaystyle \frac{1}{2}}modp$ and solving this system of equations, we obtain

$$(a_4,a_5)\in \left\{\left({\displaystyle \frac{1\pm \sqrt{5}}{8}},{\displaystyle \frac{\pm \sqrt{5}}{8}}\right),\right\}.$$

Substituting the first solution in $S_6$, we obtain

$$\begin{array}{cc}\hfill \left\{0,{\displaystyle \frac{1}{8}}\sqrt{5},\right.& {\displaystyle \frac{1}{16}}\sqrt{5},-{\displaystyle \frac{5}{32}}+{\displaystyle \frac{5}{32}}\sqrt{5},-{\displaystyle \frac{3}{16}}+{\displaystyle \frac{3}{16}}\sqrt{5},-{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{5},{\displaystyle \frac{3}{32}}+{\displaystyle \frac{1}{32}}\sqrt{5},\hfill \\ \hfill -{\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{8}}\sqrt{5},& \left.-{\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{16}}\sqrt{5},{\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{16}}\sqrt{5},{\displaystyle \frac{5}{32}}+{\displaystyle \frac{1}{32}}\sqrt{5},-{\displaystyle \frac{1}{32}}+{\displaystyle \frac{3}{32}}\sqrt{5},{\displaystyle \frac{1}{32}}+{\displaystyle \frac{3}{32}}\sqrt{5}\right\}.\hfill \end{array}$$

So, $S_6$ has at most 13 elements, for every prime for which $\sqrt{5}$ exists. Clearly, if $p=11$, then $\left|S_6\right|\le 11$, but it can be seen that all the elements of ${\mathbb{F}}_{11}$ are in $S_{6}mod11$. So, statement $\left(4\right)a)$ is obtained. Notice that the second solution is also in statement $\left(4\right)$, since the solutions of $x^2\equiv 5modp$ are symmetric.

(b) 

The fourth and fifth elements of $S_6$ are equal, i.e.,

$$\begin{array}{cc}\hfill -1+4a_3-3a_4-3a_3^2+2a_5+2a_3{a}_4& =0\hfill \\ \hfill a_4^3& =a_5^2.\hfill \end{array}$$

This case is similar to the first case studied in the previous section. Let $b_4$ be a square root of $a_4$, which exists since $a_4^3$ is a square. Solving this system of equations (with $a_3={\displaystyle \frac{1}{2}}$), we obtain $b_4={\displaystyle \frac{1\pm \sqrt{5}}{4}}$. Here, we also have that $S_6$ has at most 13 elements, for every prime for which $\sqrt{5}$ exists, and it has 10 elements when $p=11$ and $b_4\equiv {\displaystyle \frac{1+\sqrt{5}}{4}}\equiv 4mod11$.

Finally, suppose $p\equiv \pm 1mod12$, with the calculations in Maple we found that $|S_{6}modp|$ is minimal when the eighth and the thirteenth elements of $S_6$ (in the expression (9)) are null, i.e.,

$$\begin{array}{cc}\hfill a_5-{\displaystyle \frac{a_4}{2}}+a_4^2& =0\hfill \\ \hfill -1+4a_3-3a_4-3a_3^2+2a_5+2a_3{a}_4& =0.\hfill \end{array}$$

Solving this system of equations, we obtain

$$(a_4,a_5)\in \left\{\left(-1\pm {\displaystyle \frac{\sqrt{3}}{4}},{\displaystyle \frac{-3\pm 2\sqrt{3}}{8}}\right),\right\}.$$

Substituting the first solution in $S_6$, we obtain

$$\begin{array}{cc}& \left\{0,-{\displaystyle \frac{15}{32}}+{\displaystyle \frac{9}{32}}\sqrt{3},-{\displaystyle \frac{7}{32}}+{\displaystyle \frac{13}{96}}\sqrt{3},\right.-{\displaystyle \frac{3}{4}}+{\displaystyle \frac{3}{8}}\sqrt{3},-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}\sqrt{3},{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{8}}\sqrt{3},{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{16}}\sqrt{3},\hfill \\ & {\displaystyle \frac{9}{16}}-{\displaystyle \frac{5}{16}}\sqrt{3},-{\displaystyle \frac{7}{16}}+{\displaystyle \frac{1}{4}}\sqrt{3},-{\displaystyle \frac{3}{8}}+{\displaystyle \frac{1}{4}}\sqrt{3},-{\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{8}}\sqrt{3},-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}\sqrt{3},-{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{3},\hfill \\ & \left.-{\displaystyle \frac{5}{16}}+{\displaystyle \frac{3}{16}}\sqrt{3}\right\}.\hfill \end{array}$$

This time $S_6$ has at most 14 elements, for every prime for which $\sqrt{3}$ exists. If $p=11$ and we consider $\sqrt{3}=5$, then

$$\begin{array}{cc}\hfill -{\displaystyle \frac{7}{32}}+{\displaystyle \frac{13}{96}}\sqrt{3}& =0,\hfill \\ \hfill -{\displaystyle \frac{3}{4}}+{\displaystyle \frac{3}{8}}\sqrt{3}& =-{\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{8}}\sqrt{3},\hfill \\ \hfill -{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}\sqrt{3}& ={\displaystyle \frac{9}{16}}-{\displaystyle \frac{5}{16}}\sqrt{3},\hfill \\ \hfill -{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{3}& ={\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{16}}\sqrt{3}.\hfill \end{array}$$

The other exceptions can also be obtained and so statement $\left(5\right)$ is satisfied. Notice that the second solution is also in statement $\left(5\right)$, since the solutions of $x^2\equiv 3modp$ are symmetric.

To complete the proof, we need to show that $a_3={\displaystyle \frac{1}{2}}\notin S_3$, $a_4\notin S_4$, and $a_5\notin S_5$. Clearly ${\displaystyle \frac{1}{2}}\notin S_3$. Since $a_3={\displaystyle \frac{1}{2}}$ in all cases, then $S_4=\{0,{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{4}}\}$.

If ${\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{-1}\in S_4$, then we would obtain $\sqrt{-1}\in \{-1,0,1\}$, which is impossible for $p>2$.

If ${\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}\sqrt{2}\in S_4$, then we would obtain $\sqrt{2}\in \{-2,0,2\}$, which is impossible for $p>2$.

If ${\displaystyle \frac{3}{8}}+{\displaystyle \frac{1}{8}}\sqrt{-3}\in S_4$, then we would obtain $\sqrt{-3}\in \{-3,-1,1\}$, which is impossible for $p>3$.

If ${\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{5}\in S_4$, then we would obtain $\sqrt{5}\in \{-1,1,3\}$, which is impossible for $p>2$.

If $-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{3}\in S_4$, then we would obtain $\sqrt{3}\in \{1,2,3\},$ which is impossible for $p>3$.

Therefore, $a_4\notin S_4$.

In the case $p\equiv 1mod4$, if $a_4={\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{-1}$, then

$$S_5=\left\{0,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}}\sqrt{-1},{\displaystyle \frac{1}{4}}\sqrt{-1},{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{2}}\sqrt{-1},{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{-1},{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{-1},{\displaystyle \frac{1}{8}}+{\displaystyle \frac{3}{8}}\sqrt{-1}\right\}.$$

It is not difficult to see that if $a_5={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{4}}\sqrt{-1}\in S_5$, then $\sqrt{-1}\in \left\{-{\displaystyle \frac{1}{2}},0,{\displaystyle \frac{1}{2}}\right\}$, which is only true if $p=5$.

In the case $p\equiv \pm 1mod8$, if $a_4={\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}\sqrt{2}$, then

$$S_5=\left\{0,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}}\sqrt{2},{\displaystyle \frac{1}{8}}\sqrt{2},{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{2},{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}\sqrt{2},{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{16}}\sqrt{2},{\displaystyle \frac{1}{8}}+{\displaystyle \frac{3}{16}}\sqrt{2},{\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{8}}\sqrt{2},{\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{8}}\sqrt{2}\right\}.$$

If $a_5={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{2}\in S_5$, then $\sqrt{2}\in \left\{-1,0,1\right\}$, which is never true.

If $p\equiv 1mod3$ and $a_4={\displaystyle \frac{3}{8}}+{\displaystyle \frac{1}{8}}\sqrt{-3}$, then

$$\begin{array}{cc}\hfill S_5=\left\{0,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}\sqrt{-3},{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{-3},{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{-3},{\displaystyle \frac{3}{8}}+{\displaystyle \frac{1}{8}}\sqrt{-3},\right.& {\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{16}}\sqrt{-3},\hfill \\ \hfill {\displaystyle \frac{3}{16}}+{\displaystyle \frac{3}{16}}\sqrt{-3},{\displaystyle \frac{5}{16}}+{\displaystyle \frac{1}{16}}\sqrt{-3},& \left.{\displaystyle \frac{5}{16}}+{\displaystyle \frac{3}{16}}\sqrt{-3}\right\}.\hfill \end{array}$$

If $a_5={\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}\sqrt{-3}\in S_5$, then $\sqrt{-3}\in \left\{-2,-1,0,1\right\}$, which is only true if $p=7$.

If $p\equiv \pm 1mod5$ and $a_4={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{5}$, then

$$\begin{array}{cc}\hfill S_5=\left\{0,{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{4}}\sqrt{5},-{\displaystyle \frac{3}{16}}+{\displaystyle \frac{3}{16}}\sqrt{5},-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{5},-{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{5},-{\displaystyle \frac{1}{16}}+{\displaystyle \frac{3}{16}}\sqrt{5},\right.& {\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{5},\hfill \\ \hfill {\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{16}}\sqrt{5},& \left.{\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{16}}\sqrt{5}\right\}.\hfill \end{array}$$

If $a_5={\displaystyle \frac{1}{8}}\sqrt{5}\in S_5$, then $\sqrt{5}\in \left\{0,1,2,3\right\}$, which is never true. The other case is similar.

If $p\equiv \pm 1mod12$ and $a_4=-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{3}$, then

$$\begin{array}{cc}\hfill S_5=\left\{0,{\displaystyle \frac{1}{4}},-1+{\displaystyle \frac{1}{2}}\sqrt{3},-{\displaystyle \frac{5}{4}}+{\displaystyle \frac{3}{4}}\sqrt{3},-{\displaystyle \frac{5}{8}}+{\displaystyle \frac{3}{8}}\sqrt{3},\right.& -{\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{2}}\sqrt{3},-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}\sqrt{3},\hfill \\ \hfill -{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4}}\sqrt{3},& \left.-{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\sqrt{3},{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4}}\sqrt{3}\right\}.\hfill \end{array}$$

If $a_5=-{\displaystyle \frac{3}{8}}+{\displaystyle \frac{1}{4}}\sqrt{3}\in S_5$, then $\sqrt{3}\in \left\{{\displaystyle \frac{3}{2}},{\displaystyle \frac{7}{4}},2,{\displaystyle \frac{5}{2}}\right\}$, which is never true.

Hence, we obtain the last statement. □

Remark 5.

It is not difficult to obtain from Theorem 5 the minimum number of minors of $A_6$ involving $a_6$, for each appropriate prime p. These numbers are detailed in

Table 2. Minimal number of minors of $A_6$ involving $a_6$ for each prime field.

$\mathsf{\gamma }$	${\mathit{N}}_{\mathsf{\gamma }}+1$	Number of Minors	Prime Field Sizes
6	27	10	11
		11	13
		12	17 or 23
		13	$p\ge 19$, $p\ne 23$ and $p\not\equiv$ 83,107$mod120$
		14	$p\equiv$ 83,107$mod120$

.

Remark 6.

For most primes there are other possibilities of obtaining the minimum number of minors, but based on our computations, we conjecture that if p is sufficiently large, the following definitions are true:

(i) 

If $p\equiv 47,53,77mod120$, there are exactly two possibilities;

(ii) 

If $p\equiv 11,17,59,113mod120$, there are exactly four possibilities;

(iii) 

If $p\equiv 29,71,101mod120$, there are exactly six possibilities;

(iv) 

If $p\equiv 19,89,91mod120$, there are exactly eight possibilities.

These are given by Theorem 5.

Theorem 5 shows that, for each prime $p\ge 11$, whenever we choose $a_6$ appropriately, we obtain an LT-superregular matrix. The next question is if it is possible to choose $a_6$ so that $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular for all, or at least many, of the primes in each of the arithmetic progressions above. The next result answers this question.

Corollary 1.

Suppose we have $a_3,a_4$, and $a_5$ as in Theorem 5, for each of the arithmetic progressions considered. Then, the following hold:

1. 

If $p=11$, then the matrix $A_6(1,1,6,1,5,4)$ is LT-superregular.

2. 

If $p=13$, then all the matrices $A_6(1,1,7,8,3,2)$, $A_6(1,1,7,4,12,9)$,

$A_6(1,1,7,6,1,2)$, and $A_6(1,1,7,6,1,4)$ are LT-superregular.

3. 

If $p\equiv 1mod4$ and $p\ge 17$, take $a_6={\displaystyle \frac{1}{4}}$ (if $p=37$, consider $\sqrt{-1}=6$ in the expressions of $a_4$ and $a_5$). Then $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular.

4. 

if $p\equiv \pm 1mod8$ and $p\ge 17$, take $a_6={\displaystyle \frac{1}{4}}$ (if $p=17$, consider $\sqrt{2}=6$ in the expressions of $a_4$ and $a_5$). Then, $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular.

5. 

If $p\equiv 1mod3$ and $p\ge 19$, take $a_6={\displaystyle \frac{1}{4}}$ (if $p=37$, consider $\sqrt{-3}=16$ in the expressions of $a_4$ and $a_5$). Then, $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular.

6. 

If $p\equiv \pm 1mod5$ and $p\ge 19$, take $a_6={\displaystyle \frac{1}{4}}$. Then, $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular.

7. 

If $p\equiv \pm 1mod12$, $p\ge 23$, the following hold:

(a) 

If $p\ne 37$, take $a_6={\displaystyle \frac{1}{4}}$ (if $p=23$ consider $\sqrt{3}=16$ and if $p=73$ consider $\sqrt{3}=52$, in the expressions of $a_4$ and $a_5$). Then, $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular.

(b) 

If $p=37$, take $a_6=10$. Then, $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular.

Proof. 

In the cases $p=11$ or $p=13$, we just wrote all the possibilities for which $a_6\notin S_6$.

If $p\equiv 1mod4$, with $p\ge 17$, the only instance that ${\displaystyle \frac{1}{4}}\in S_6$ is when $p=37$ and $\sqrt{-1}=31$, because ${\displaystyle \frac{1}{4}}={\displaystyle \frac{1}{32}}+{\displaystyle \frac{5}{32}}\sqrt{-1}$.

In the case $p\equiv \pm 1mod8$, with $p\ge 17$, the only instance that ${\displaystyle \frac{1}{4}}\in S_6$ is when $p=17$ and $\sqrt{2}=11$, because ${\displaystyle \frac{1}{4}}={\displaystyle \frac{3}{32}}+{\displaystyle \frac{1}{16}}\sqrt{2}$.

In the case $p\equiv 1mod3$, with $p\ge 17$, the only instance that ${\displaystyle \frac{1}{4}}\in S_6$ is when $p=37$ and $\sqrt{-3}=21$, because ${\displaystyle \frac{1}{4}}={\displaystyle \frac{5}{32}}+{\displaystyle \frac{11}{96}}\sqrt{-3}$.

If $p\equiv \pm 1mod5$ and $p\ge 19$ then ${\displaystyle \frac{1}{4}}\notin S_6$.

When $p\equiv \pm 1mod12$, there are a few instances when ${\displaystyle \frac{1}{4}}\in S_6$. If $p=23$, then we must choose $\sqrt{3}=16$, since when $\sqrt{3}=7$, ${\displaystyle \frac{1}{4}}=-{\displaystyle \frac{7}{32}}+{\displaystyle \frac{13}{96}}\sqrt{3}$. If $p=73$, then we must choose $\sqrt{3}=52$, since when $\sqrt{3}=21$, ${\displaystyle \frac{1}{4}}=-{\displaystyle \frac{7}{16}}+{\displaystyle \frac{1}{4}}\sqrt{3}$. If $p=37$, we always have ${\displaystyle \frac{1}{4}}\in S_6$, because if we choose $\sqrt{3}=15$, then ${\displaystyle \frac{1}{4}}=-{\displaystyle \frac{3}{4}}+{\displaystyle \frac{3}{8}}\sqrt{3}$ and if we choose $\sqrt{3}=22$, then ${\displaystyle \frac{1}{4}}=-{\displaystyle \frac{3}{16}}+{\displaystyle \frac{1}{8}}\sqrt{3}$. Nevertheless, in this case we may take $a_6=10$ (and $a_3=19$, $a_4=33$ and $a_5=19$). □

Remark 7.

There are other possibilities for $a_6$ that make $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ LT-superregular for many primes. For example, if $p\ge 17$, with $p\equiv 1mod4$ and we choose $a_6={\displaystyle \frac{\sqrt{-1}}{2}}$ with $\sqrt{-1}<{\displaystyle \frac{p}{2}}$, then $a_6\notin S_6$. Therefore, $A_6(1,1,{\displaystyle \frac{1}{2}},a_4,a_5,a_6)$ is LT-superregular. Notice that if $p=13$, then $a_6\in S_6$. If we chose $\sqrt{-1}>{\displaystyle \frac{p}{2}}$, then $a_6\in S_6$ when $p\in \{13,17,37,41,61\}$. More explicitly, the following hold:

If $p=13$ and $\sqrt{-1}=8$, then ${\displaystyle \frac{1}{2}}\sqrt{-1}={\displaystyle \frac{1}{16}}+{\displaystyle \frac{3}{16}}\sqrt{-1}$;

If $p=17$ and $\sqrt{-1}=13$, then ${\displaystyle \frac{1}{2}}\sqrt{-1}={\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{4}}\sqrt{-1}$;

If $p=37$ and $\sqrt{-1}=31$, then ${\displaystyle \frac{1}{2}}\sqrt{-1}={\displaystyle \frac{1}{16}}+{\displaystyle \frac{1}{8}}\sqrt{-1}$;

If $p=41$ and $\sqrt{-1}=32$, then ${\displaystyle \frac{1}{2}}\sqrt{-1}={\displaystyle \frac{1}{32}}+{\displaystyle \frac{7}{32}}\sqrt{-1}$;

If $p=61$ and $\sqrt{-1}=50$, then ${\displaystyle \frac{1}{2}}\sqrt{-1}={\displaystyle \frac{1}{32}}+{\displaystyle \frac{5}{32}}\sqrt{-1}$.

We can create more examples of LT-superregular matrices using the values for $a_3,a_4$, and $a_5$ from the previous subsections.

Example 1.

If we take $a_3={\displaystyle \frac{1}{4}}$, $a_4=-{\displaystyle \frac{1}{8}}$ and $a_5={\displaystyle \frac{1}{4}}$, then

$$\begin{array}{cc}\hfill S_6=\left\{0,-{\displaystyle \frac{13}{32}},-{\displaystyle \frac{7}{32}},-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{8}},-{\displaystyle \frac{1}{32}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{16}},{\displaystyle \frac{1}{64}},{\displaystyle \frac{5}{8}},{\displaystyle \frac{5}{32}},{\displaystyle \frac{7}{16}},\right.& {\displaystyle \frac{11}{32}},{\displaystyle \frac{17}{32}},{\displaystyle \frac{17}{128}},{\displaystyle \frac{19}{64}},\hfill \\ & \left.{\displaystyle \frac{23}{32}},{\displaystyle \frac{29}{32}},{\displaystyle \frac{31}{64}},{\displaystyle \frac{49}{64}}\right\}\hfill \end{array}$$

has at most 20 elements. So, if $p\ge 23$ and $a_6=-{\displaystyle \frac{1}{4}}$, which is not in $S_6$, for any $p\ge 23$, then $A_6(1,1,a_3,a_4,a_5,a_6)$ is LT-superregular.

If we take $a_3={\displaystyle \frac{3}{4}}$, $a_4={\displaystyle \frac{3}{8}}$ and $a_5={\displaystyle \frac{1}{4}}$, then

$$\begin{array}{cc}\hfill S_6=\left\{0,-{\displaystyle \frac{1}{32}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{6}},{\displaystyle \frac{1}{8}},{\displaystyle \frac{1}{16}},{\displaystyle \frac{3}{16}},{\displaystyle \frac{3}{32}},{\displaystyle \frac{3}{64}},{\displaystyle \frac{5}{32}},{\displaystyle \frac{7}{32}},{\displaystyle \frac{7}{64}},{\displaystyle \frac{9}{32}},{\displaystyle \frac{9}{64}},{\displaystyle \frac{11}{32}},{\displaystyle \frac{13}{64}},\right.& {\displaystyle \frac{13}{96}},{\displaystyle \frac{17}{64}},\hfill \\ & \left.{\displaystyle \frac{17}{96}},{\displaystyle \frac{21}{128}}\right\}\hfill \end{array}$$

has at most 20 elements. So, if $p\ge 23$ and $a_6=-{\displaystyle \frac{5}{16}}$, which is not in $S_6$, for any $p\ge 23$, then $A_6(1,1,a_3,a_4,a_5,a_6)$ is LT-superregular.

If we take $a_3={\displaystyle \frac{1}{2}}$, $a_4=1$ and $a_5=-{\displaystyle \frac{1}{2}}$, then

$$\begin{array}{cc}\hfill S_6=\left\{-2,-1,0,1,2,-{\displaystyle \frac{13}{8}},-{\displaystyle \frac{11}{4}},-{\displaystyle \frac{9}{4}},-{\displaystyle \frac{7}{2}},-{\displaystyle \frac{7}{4}},-{\displaystyle \frac{5}{4}},-{\displaystyle \frac{3}{4}},-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{4}},\right.& {\displaystyle \frac{1}{8}},{\displaystyle \frac{3}{4}},{\displaystyle \frac{7}{3}},\hfill \\ & \left.{\displaystyle \frac{7}{4}},{\displaystyle \frac{11}{4}},{\displaystyle \frac{37}{16}}\right\}\hfill \end{array}$$

has at most 22 elements. So, if $p\ge 23$ and $a_6={\displaystyle \frac{3}{2}}$, which is not in $S_6$, for any $p\ge 23$, then $A_6(1,1,a_3,a_4,a_5,a_6)$ is LT-superregular.

2.4. When $\gamma \ge 7$

For $\gamma \ge 7$, the count of the minimum number of different minors involving $a_{\gamma }$ for every prime field $\mathbb{F}$ for which $A_{\gamma }$ is LT-superregular, becomes much more complicated, as there are too many different values. Therefore we chose to construct examples of LT-superregular matrices for some of the finite prime fields for which $A_{\gamma }$ is LT-superregular, for each $7\le \gamma \le 10$. For each prime p and each $\gamma$ we created the sets $S_{\mu }$, for $\mu \le \gamma$ and tried recursively, using Maple, all the vectors $(a_3,a_4,\dots ,a_{\gamma -1})\in {\mathbb{F}}^{\gamma -3}$, that satisfied $a_i\notin S_i$, for $3\le i\le \gamma -1$ in order to find the vectors $(a_3,a_4,\dots ,a_{\gamma -1})$ that made $\left|S_{\gamma }\right|$ smallest.

Suppose $\gamma =7$, then $N_{\gamma }=76$. But as one can see from

Table 3. Minimum number of different minors involving $a_7$ for $17\le p\le 97.$.

$\mathsf{\gamma }$	${\mathit{N}}_{\mathsf{\gamma }}+1$	Different Minors	Field Size
7	77	16	17
		18	19
		21	23
		24	29
		25	31
		28	37
		29	41
		30	47
		31	43
		32	53
		35	59 or 61
		36	67 or 73
		37	71
		38	79
		39	83, 89 or 97

, there are too many minimum numbers of different minors of $A_7$ involving $a_7$. The smallest finite prime fields that have an LT-superregular matrix of order 7 have all different minimum numbers and we were not able to find a pattern from which we could deduce general sequences as we did in the case $\gamma =6$. Nevertheless, we are able to exhibit LT-superregular matrices for every $p\ge 17$.

If $p=17$, then it follows that there are 8 LT-superregular matrices $A_7$, one of which is $A_7(1,1,9,3,5,1,3)$. For this example, $\left|S_4\right|=3$, $\left|S_5\right|=8$, $\left|S_6\right|=13$ and $\left|S_7\right|=16$. If $p=19$ there are 82 LT-superregular matrices $A_7$, one of which is $A_7(1,1,10,13,1,18,7)$. The number of elements of $S_4,S_5$, and $S_6$ are also $3,8$ and 13 respectively, and $\left|S_7\right|=18$. If $p=23$, there are only two examples of LT-superregular matrices, which are $A_7(1,1,4,19,6,4,8)$ and $A_7(1,1,4,19,6,4,15)$. It is interesting to notice that $S_4$ and $S_5$ achieve the maximum number of elements in these two examples while $S_6$ has 17 elements. Hence, sometimes $S_{\gamma }$ has the minimum number of elements when some of the $S_{\mu }$, with $\mu <\gamma$ having the maximum.

If we use the vectors already considered in the previous sections, we obtain very large values for the number of elements of $S_7$ (see

Table 4. Maximum size of $S_7$ for some $(a_3,a_4,a_5,a_6)$.

$\mathsf{\gamma }$	${\mathit{N}}_{\mathsf{\gamma }}+1$	$({\mathit{a}}_3,{\mathit{a}}_4,{\mathit{a}}_5,{\mathit{a}}_6)$	$|{\mathit{S}}_7|$
7	77	${\displaystyle \left(\frac{1}{2},1,-\frac{1}{2},\frac{3}{2}\right)}$	56
		${\displaystyle \left(\frac{1}{4},-\frac{1}{8},\frac{1}{4},-\frac{1}{4}\right)}$	67
		${\displaystyle \left(\frac{3}{4},\frac{3}{8},\frac{1}{4},-\frac{5}{16}\right)}$	68
		${\displaystyle \left(\frac{1}{2},\frac{1}{4}+\frac{1}{4}\sqrt{-1},\frac{1}{8}+\frac{1}{4}\sqrt{-1},\frac{1}{4}\right)}$	65
		${\displaystyle \left(\frac{1}{2},\frac{1}{4}+\frac{1}{8}\sqrt{2},\frac{1}{8}+\frac{1}{8}\sqrt{2},\frac{1}{4}\right)}$	57
		${\displaystyle \left(\frac{1}{2},\frac{3}{8}+\frac{1}{8}\sqrt{-3},\frac{1}{4}+\frac{1}{8}\sqrt{-3},\frac{1}{4}\right)}$	65
		${\displaystyle \left(\frac{1}{2},\frac{1}{8}+\frac{1}{8}\sqrt{5},\frac{1}{8}\sqrt{5},\frac{1}{4}\right)}$	55
		${\displaystyle \left(\frac{1}{2},-\frac{1}{4}+\frac{1}{4}\sqrt{3},-\frac{3}{8}+\frac{1}{4}\sqrt{3},\frac{1}{4}\right)}$	71

), in comparison to the ones obtained in Table 3. So, considering $\mu <\nu \le \gamma$, having $|S_{\mu }|$ be small for some $(a_3,\dots ,a_{\mu -1})$ does not imply that $|S_{\nu }|$ is also small for $(a_3,\dots ,a_{\mu -1},a_{\mu },\dots ,a_{\nu })$. Nevertheless, the sequences in Table 4 can be used to construct LT-superregular matrices of order 7, for finite prime fields, when $p\ge 59$, since $59,61$, and 71 are congruent with plus or minus one module 5 and $67\equiv 1mod3$. The

Table 5. Examples of LT-superregular matrices of order 7 for small finite prime fields.

Field Size	Example of $({\mathit{a}}_3,{\mathit{a}}_4,{\mathit{a}}_5,{\mathit{a}}_6,{\mathit{a}}_7)$
17	$(9,3,5,1,3)$
19	$(10,13,1,18,7)$
23	$(4,19,6,4,8)$
29	$(15,19,8,22,1)$
31	$(4,30,22,17,2)$
37	$(8,35,6,25,12)$
41	$(7,22,26,7,1)$
43	$(17,12,25,23,2)$
47	$(24,9,3,18,4)$
53	$(27,42,22,20,3)$

shows examples of vectors $(a_3,a_4,a_5,a_6,a_7)$ such that the size of $S_7$ is minimum and from which we can create $7\times 7$ LT-superregular matrices when $p<59$.

Again, using Maple we were able to compute LT-superregular matrices of order $\gamma$, for $\gamma =8$ and $\gamma =9$ over the two smallest finite prime fields. Using the values of $a_3,a_4,a_5$, and $a_6$ obtained in Theorem 5 and Corollary 1, we also obtained LT-superregular matrices over a few larger fields. These examples are shown in

Table 6. Examples of LT-superregular matrices of order 8 for small finite prime fields.

Field Size	Example of $({\mathit{a}}_3,{\mathit{a}}_4,\dots ,{\mathit{a}}_8)$	Diff. Minors	LT-SR Mat.
31	$(7,22,20,2,13,5)$	30	0
37	$(2,8,28,32,18,16)$	36	2
41	$(21,12,17,31,33,26)$	39	3
43			0
47	$(24,38,32,12,13,17)$	44	11
53	$(27,6,39,40,12,29)$	51	4
59	$(30,38,1,15,58,15)$	56	25
61	$(31,19,57,46,15,7)$	55	65
67	$(34,60,18,17,66,1)$	58	17
71	$(36,55,46,18,52,5)$	62	105
73	$(37,7,16,55,35,1)$	60	172
79	$(40,34,24,20,33,17)$	65	235
83	$42,3,34,21,50,4)$	67	128
89	$(45,59,70,67,37,1)$	70	521
97	$(49,30,42,73,17,1)$	66	1321
101	$(51,28,91,76,69,2)$	72	729
103	$(52,5,95,26,94,4)$	75	923

and

Table 7. Examples of LT-superregular matrices of order 9 for small finite prime fields.

Field Size	Example of $({\mathit{a}}_3,{\mathit{a}}_4,\dots ,{\mathit{a}}_{\mathsf{\gamma }})$	Diff. Minors	LT-SR Mat.
59	$(5,28,58,56,26,18,19)$	58	0
61	$(7,60,55,39,10,12,16)$	60	0
67–89			0
97	$(49,2,14,73,19,32,88)$	96	5
101	$(51,28,91,76,69,35,8)$	100	5
103	$(52,5,95,26,6,92,43)$	102	23
107	$(54,31,71,27,23,58,38)$	106	2
109	$(55,63,22,82,10,4,87)$	108	54
113	$(57,4,18,85,75,97,33)$	110	70
127	$(64,34,18,32,14,85,32)$	125	286
131	$(66,3,52,33,47,32,8)$	128	453
137	$(69,124,4,103,116,18,21)$	134	789
139	$(70,19,71,35,41,125,39)$	135	1005
149	$(75,123,67,112,44,105,16)$	145	1989

. For each prime p, only the ones that have the smaller number of minors are exhibited. In the last column of these tables are the number of LT-superregular matrices obtained in this way. For $p=37$, there are two such matrices, but the example was obtained by trying all possible values for the entries of the matrix. If $\gamma =8$ and $p=43$, and if $\gamma =9$ and $59\le p\le 89$, there are no LT-superregular matrices when we take those values for $a_3,a_4,a_5$, and $a_6$, but they can be found considering other values. If $\gamma =10$ and the values of $a_3,a_4,a_5$, and $a_6$ obtained in Theorem 5 and Corollary 1, the smallest prime field for which there are LT-superregular matrices is ${\mathbb{F}}_{181}$. In

Table 8. Examples of LT-superregular matrices of order 10 for small finite prime fields.

Field Size	Example of $({\mathit{a}}_3,{\mathit{a}}_4,\dots ,{\mathit{a}}_{\mathsf{\gamma }})$	Diff. Minors	LT-SR Mat.
181	$(91,124,56,136,13,112,107,118)$	180	5
191	$(96,169,145,48,86,65,96,171)$	190	8
193	$(97,117,141,145,31,85,154,84)$	192	14
197	$(99,53,176,148,16,18,104,66)$	196	13

we show some examples.

In [21], the authors presented a greedy algorithm able to compute superregular matrices $9\times 9$ and $10\times 10$ over the field ${\mathbb{F}}_{2^8}$. The results presented in the tables lead to the following two conjectures.

Conjecture 1.

For a given $\gamma \ge 2$ and for any odd prime p, there exists a vector $(a_1,a_2,\cdots ,a_{\gamma -1})\in {\mathbb{F}}^{\gamma -1}$ such that $S_{\gamma }$ has at most ${\displaystyle \frac{N_{\gamma }}{2}}+2$ elements.

Conjecture 2.

We also conjecture that for $\gamma \ge 2$, there exists a lower triangular Toeplitz superregular matrix of order $\gamma \times \gamma$ over $\mathbb{F}$ with $\left|\mathbb{F}\right|\ge 2^{{\displaystyle \frac{2\gamma }{3}}}$.

3. Computer Calculations

In this section, we give a brief description of the computer algorithms we used to obtain the superregular matrices described throughout this paper. All the calculations were performed in Maple.

For $4\le \gamma \le 7$, our main goal was to find the minimum number of different minors $A_{\gamma }\in \mathbb{F}$ has, depending on the finite prime field $\mathbb{F}$. So for each $\gamma$, we started with the smallest possible prime number p and tried all the possible combinations of $(a_3,a_4,\dots ,a_{\gamma -1})\in {\mathbb{F}}^{\gamma -3}$ satisfying $a_i\notin S_i$, for $3\le i\le \gamma -1$, following the idea explained in Remark 1. We were unable to fully achieve the main goal for $\gamma =7$, since when $p\ge 101$ the amount of computations is already too large.

For $\gamma =8$ and $\gamma =9$, we just tried to find the two smallest primes p for which exists an LT-superregular Toeplitz matrix over $\mathbb{F}$ and gave those examples. We also gave examples for larger prime fields using Theorem 5 and Corollary 1. For $\gamma =10$, we were unable to find any superregular matrix, using the method of trying all possible values of $(a_3,a_4,\dots ,a_{\gamma -1})\in {\mathbb{F}}^{\gamma -3}$, with p small.

Using a randomized computer search, other LT-superregular matrices of order 10 were obtained, in all cases the number of different minors was $p-1$. In

Table 9. Examples of LT-superregular matrices of order 10 for a few small finite prime fields.

Field Size	Example of $({\mathit{a}}_3,{\mathit{a}}_4,\dots ,{\mathit{a}}_{\mathsf{\gamma }})$
163	$(121,115,98,101,87,129,133,120)$
173	$(156,131,142,64,96,4,107,34)$
179	$(151,64,33,163,178,72,173,143)$
181	$(157,122,177,19,5,90,45,35)$
191	$(106,12,164,120,174,33,115,160)$
193	$(128,144,81,124,95,164,175,171)$
197	$(173,189,74,115,130,75,6,168)$
199	$(179,172,149,3,168,93,129,187)$
211	$(88,64,110,203,208,50,43,139)$
223	$(205,127,9,168,67,92,41,57)$
227	$(6,150,62,124,14,62,161,108)$
229	$(195,120,223,88,46,15,111,210)$
233	$(35,128,23,80,198,101,214,141)$
239	$(179,39,21,23,179,7,162,68)$
241	$(185,32,69,175,96,88,36,63)$
251	$(131,135,195,56,39,64,185,43)$
257	$(182,147,249,62,174,18,50,149)$

we show some examples.

4. Conclusions and Future Work

In this paper, we have continued the study of LT-superregular Toeplitz matrices $A_{\gamma }$. We present new results regarding the minimum number of different minors appearing in $A_{\gamma }$ and the field sizes that allow the construction of these matrices. Based on the work presented, we made the two conjectures. An interesting avenue for further research is to investigate these results using finite field extensions of finite fields of smaller characteristic, e.g., of characteristic 2, which is of particular interest in coding theory. Another interesting open problem left for future research is to know whether, or in what conditions, there exist LT-superregular matrices over finite fields larger than the minimum $\mathbb{F}$ found and smaller than $N_{\gamma }$. For instance, in [20] it was found over ${\mathbb{F}}_{127}$ and in [21] over ${\mathbb{F}}_{256}$, but to the best of our knowledge the existence of LT-superregular matrices was not known over $\mathbb{F}$ when $p\notin \{127,256\}$ and $p<N_{10}=2494$. Also, nothing is known about the case $\gamma \ge 11$.