## 1. Introduction

As it is known, Pascal’s triangle is constructed in the following way: Write the first row “$1\phantom{\rule{4pt}{0ex}}\phantom{\rule{4pt}{0ex}}1$”. Then each member of each subsequent row is given by taking the sum of the just above two members, regarding any blank as 0.

**Example** **1.** Here is the Pascal’s triangle from the first row to the 7-th row: **Remark** **1.** For any integers $n\ge 1$ and $r\ge 0$, we putwhere we put $0!=1$. Then it is well-known that the n-th row in the Pascal’s triangle is equal to the sequenceconsisting of $n+1$ terms. In ([

1], Section 1.4), the construction above is generalized as follows:

**Definition** **1.** Let $k\ge 2$ be any integer. The k-Pascal’s triangle is constructed in the following way: Write the first row “$\stackrel{k}{\overbrace{1\phantom{\rule{4pt}{0ex}}1\phantom{\rule{4pt}{0ex}}\cdots \phantom{\rule{4pt}{0ex}}1}}$”. Then each member of each subsequent row is given by taking the sum of the just above k members regarding the blank as 0.

**Example** **2.** In the case where $k=4$, the 4-Pascal’s triangle from the first row to the 5-th row is the following: **Remark** **2.** $(1)$ In ([1], Section 1.4), for any integers $k\ge 2$ and $n\ge 1$, it is mentioned that the n-th row in the k-Pascal’s triangle consists of $n(k-1)+1$ integerssatisfying the equationof polynomials with indeterminate x and integral coefficients. A detailed proof of this fact is described in ([2], Lemma 1.1). (2) In ([1], Section 9.10), the following formula for ${}_{n}{C}_{i}^{(k)}$ is described:where $\left[\frac{i}{\phantom{\rule{3.33333pt}{0ex}}k\phantom{\rule{3.33333pt}{0ex}}}\right]$ is the greatest integer that is less than or equal to $\frac{i}{\phantom{\rule{3.33333pt}{0ex}}k\phantom{\rule{3.33333pt}{0ex}}}$. In Example 1, we can see that the n-th row consists of odd integers when n is equal to the Mersenne number 1, 3 or 7. Actually, Lucas showed the following

**Theorem** **1** ([

3], Exemple I in Section 228).

Let $n\ge 1$ be any integer. Then ${}_{n}{C}_{r}$ is odd for any $0\le r\le n$ if and only if n is a Mersenne number, i.e., n is of the form ${2}^{m}-1$ with some integer $m\ge 1$.In

Section 2 in this article, we generalize the Lucas’ result above as the following

**Theorem** **2.** Let p be any prime number and e any positive integer. For any integer $n\ge 1$, the n-th row in the ${p}^{e}$-Pascal’s triangle consists of integers which are congruent to 1 modulo p if and only if n is of the form $\frac{{p}^{em}-1}{{p}^{e}-1}$ with some integer $m\ge 1$.

**Remark** **3.** (1) Theorem 2 is a generalization of ([2], Theorem 0.2) which is in the case where $e=1$. (2) We can see that Example 2 gives a partial example of Theorem 2 in the case where $p=2$, $e=2$ and $m=1,2$.

As an application of Theorem 2, we can prove that ([

2], Conjecture 0.3) holds for

$k=4$, i.e., there exists no row in the 4-Pascal’s triangle consisting of integers which are congruent to 1 modulo 4 except the first row as follows:

By Theorem 2, in the case where $k=4$, we see that for any integer $n\ge 1$, the n-th row in the 4-Pascal’s triangle consists of odd integers if and only if n is of the form $\frac{{4}^{m}-1}{3}$ with some integer $m\ge 1$.

Moreover, we can see an essential property of the

$\frac{{4}^{m}-1}{3}$-th row in the 4-Pascal’s triangle for any integer

$m\ge 2$ as in the following theorem proved in

Section 3.2:

**Theorem** **3.** For any integer $m\ge 2$, the $\frac{{4}^{m}-1}{3}$-th row in the 4-Pascal’s triangle is congruent to the sequencemodulo 4, which consists of the repeated 1133’s and 3311’s whose numbers are the same ${2}^{2m-3}$. Therefore we can obtain the following

**Corollary** **1.** ([2], Conjecture 0.3) holds for $k=4$, i.e., there exists no row in the 4-Pascal’s triangle consisting of integers which are congruent to 1 modulo 4 except the first row. **Remark** **4.** (1) By Example 2, in the case where $m=2$, we can see that the 5-th row in the 4-Pascal’s triangle is congruent to the sequencemodulo 4, which matches the assertion of Theorem 3. (2) It seems that one could obtain the forms of the sequenece to which the $\left(\frac{{4}^{m}-1}{3}\pm \ell \right)$-th row in the 4-Pascal’s triangle is congruent modulo 4 for some positive integers ℓ by means of Theorem 3. We would like to do these calculations in the future.

**Lemma** **1.** For any prime number p and any positive integer e, we have the following coefficient-wise congruenceof binomial expansions with indetermiate x. ## 2. A Proof of Theorem 2

Although Theorem 2 can be proved by the same argument as the proof of ([

2], Theorem 0.2), we shall describe its detailed proof here to make this article self-contained.

Let n and e be any positive integers and p be any prime number.

Firstly, we assume that

n is of the form

$n=\frac{{p}^{em}-1}{{p}^{e}-1}$ with some integer

$m\ge 1$. In the algebra

${\mathbb{F}}_{p}\left[x\right]$ of polynomials of one varible

x with coefficients in the finite field

${\mathbb{F}}_{p}=\mathbb{Z}/p\mathbb{Z}$ of

p elements, we see that for any positive integer

ℓ,

Therefore we see that

in

${\mathbb{F}}_{p}\left[x\right]$. By Remark 2 (1), this implies that the

n-th row in the

${p}^{e}$-Pascal’s triangle consists of integers which are congruent to 1 modulo

p as desired.

Conversely, we now assume that

n is of the form

with some integers

$m\ge 1$ and

$1\le k\le {p}^{em}-1$. Moreover, we assume that we have

in

${\mathbb{F}}_{p}\left[x\right]$ to obtain some contradiction. Since the left hand side is equal to

${(x-1)}^{n({p}^{e}-1)}$ and the right hand side is equal to

$\frac{{x}^{n({p}^{e}-1)+1}-1}{x-1}$, we then have the equality

in

${\mathbb{F}}_{p}\left[x\right]$. Since

$n=\frac{{p}^{em}-1}{{p}^{e}-1}+k$, this implies that

Let

${v}_{p}(a)$ be the

p-adic valuation of any non-zero integer

a, i.e.,

${p}^{{v}_{p}(a)}\mid a$ and

${p}^{{v}_{p}(a)+1}\nmid a$. Since

$1\le k\le {p}^{em}-1$, we see that

${v}_{p}(k)<em$ and then

Therefore we can put

with some positive integer

t which is prime to

p. Then we have

which implies that

since

${\mathbb{F}}_{p}\left[x\right]$ is an integral domain. Since

${p}^{{v}_{p}(k)}<{p}^{em}$, we see that

$t\ge 2$. Therefore substituting

$x=1$ leads a contradiction

$t=0$ in

${\mathbb{F}}_{p}$ as desired, and Theorem 2 is proved.

## 3. An Application to the 4-Pascal’s Triangle

By Theorem 2, in the case where $p=2$ and $e=2$, we see that for any integer $n\ge 1$, the n-th row in the 4-Pascal triangle consists of odd integers if and only if n is of the form $\frac{{4}^{m}-1}{3}$ with some integer $m\ge 1$.

In this section, we shall prove Theorem 3 asserting that for any integer

$m\ge 2$, the

$\frac{{4}^{m}-1}{3}$-th row in the 4-Pascal’s triangle is congruent to the sequence

modulo 4. Here we should note that

${2}^{2m-3}$ is the number of 1133’s and 3311’s, respectively.

Then Theorems 2 and 3 imply that ([

2], Conjecture 0.3) holds in the case where

$k=4$, i.e., there exists no row in the 4-Pascal’s triangle consisting of integers which are congruent to 1 modulo 4 except the first row as we have seen in Corollary 1.

#### 3.1. On a Congruence of Binomial Expansions

Before proving Theorem 3, we shall prove Lemma 1 on a congruence of binomial expansions in this subsection.

Let

p be any prime number and

e any positive integer. In order to prove the congruence

of binomial expansions with indeterminate

x, it suffices to see the following two congruences hold:

(1) For any integer

$1\le \ell \le {p}^{e}-1$ which is prime to

p,

(2) In the case where

$e\ge 2$, for any integers

$0\le f\le e-2$ and

i such that

$1\le i{p}^{f}\le {p}^{e-1}-1$ and

$(i,p)=1$,

Firstly, we shall prove the part (1). In the case where

$\ell =1$, we see that

Moreover, in the case where

$2\le \ell \le {p}^{e}-1$, we see that

Since

${v}_{p}({p}^{e}-j)={v}_{p}(j)$ for any

$1\le j\le \ell -1<{p}^{e}$ and

ℓ is prime to

p, we then see that

Therefore ${}_{{p}^{e}}{C}_{\ell}\equiv 0\phantom{\rule{4pt}{0ex}}(\mathrm{mod}\phantom{\rule{4pt}{0ex}}{p}^{e})$, and part (1) is proved.

Secondly, we shall prove part (2). We see that

and that

Since $(i,p)=1$, we then see that ${}_{{p}^{e}}{C}_{i{p}^{f+1}}-{}_{{p}^{e-1}}{C}_{i{p}^{f}}$ is divisible by ${p}^{e-f-1}\xb7{p}^{f+1}={p}^{e}$ as desired.

#### 3.2. A Proof of Theorem 3

Now we shall prove Theorem 3 by means of Lemma 1 with

$p=2$ and

$e=2$, i.e., the congruence of binomial expansions

By Remark 2 (1), proving Theorem 3 is equivalent to proving that for any integer

$m\ge 2$, the coefficient-wise congruence

holds with indeterminate

x by the induction on

m.

Before doing this, we see the following

**Lemma** **2.** The polynomial in the right hand side of the congruence relation $(\ast \ast )$ can be decomposed as **Proof.** By a direct calculation, we can see that there exists some positive integer

ℓ such that the polynomial in the right hand side of the congruence relation

$(\ast \ast )$ can be decomposed as

Since the degree of the polynomial in the right hand side of the congruence relation

$(\ast \ast )$ is equal to

${4}^{m}-1$, we then see that

which implies that

$\ell =2m-3$ as desired. □

Let us start to prove Theorem 3 by the induction on

$m\ge 2$. Firstly, in the case where

$m=2$, since

and

by the congruence relation

$(\ast )$, we see that

Therefore the congruence relation $(\ast \ast )$ holds for $m=2$ by Lemma 2.

Secondly, we assume that the congruence relation

$(\ast \ast )$ holds for some

$m\ge 2$. By the congruence relation

$(\ast )$, we see that

By Lemma 2, we then see that

i.e., the congruence relation

$(\ast \ast )$ also holds for

$m+1$ as desired. This proves Theorem 3.