Tian’s Conjecture on the Prime Factorization of the Binomial Coefficient $\mathbf{\left(}{\displaystyle \genfrac{}{}{0pt}{}{\mathit{n}\mathbf{+}\mathbf{1}}{\mathbf{2}}}\mathbf{\right)}$

Abstract

Tian’s conjecture states that for any fixed distinct prime numbers $p_1,\dots ,p_m$, the Diophantine equation $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=p_1^{{\alpha }_1}·p_2^{{\alpha }_2}···p_m^{{\alpha }_m}$ in positive integers $n,{\alpha }_1,\dots ,{\alpha }_m$ has at most m solutions. In this paper, we develop a computational method to verify some special cases of this conjecture. We also give an alternative proof using the classical Zsigmondy theorem. For $m=2$ and 3, a sharp absolute upper bound for the number of solutions is given.

1. Introduction

In January 2025, Jianjun Paul Tian posted a conjecture on social media about the cubic sum of consecutive natural numbers. The conjecture was originally inspired by an observation that $2025={(1+2+\cdots +9)}^2=1^3+2^3+\cdots +9^3$ on a New Year card sent from a mathematician friend. He observed that $1^3+2^3+3^3+4^3+5^3={(1+2+3+4+5)}^2=225=3^2·5^2$ and that $2025=3^4·5^2$. Then, he made a conjecture that 225 and 2025 are the only two positive integers that can be written as a cubic sum of consecutive natural numbers and have 3 and 5 as their two prime factors. Later, this conjecture was extended to two general cases. We may consider it to be three separate sub-conjectures. The first one is the original conjecture:

(1)

For the exponential Diophantine equation

$$1^3+2^3+3^3+\dots +n^3=3^{{\beta }_1}·5^{{\beta }_2},$$

where $n,{\beta }_1$ and ${\beta }_2$ are unknown positive integers, there are only two solutions, namely, $(n,{\beta }_1,{\beta }_2)=(5,2,2)$ and $(9,4,2)$.

One can generalize this problem to arbitrary primes $p_1$ and $p_2$.

(2)

Let $p_1$ and $p_2$ be distinct fixed prime numbers. The exponential Diophantine equation

$$1^3+2^3+3^3+\dots +n^3=p_1^{{\beta }_1}·p_2^{{\beta }_2}$$

has at most two solutions in positive integers $n,{\beta }_1$ and ${\beta }_2$.

Finally, the hardest conjecture concerns m fixed primes.

(3)

Let $p_1,\dots ,p_m$ be distinct fixed prime numbers. The exponential Diophantine equation

$$1^3+2^3+3^3+\dots +n^3=p_1^{{\beta }_1}·p_2^{{\beta }_2}\dots p_m^{{\beta }_m}$$

has at most m positive integer solutions in positive integers $n,{\beta }_1,\dots ,{\beta }_m$. Here, a solution is the $m+1$-tuple $(n,{\beta }_1,{\beta }_2,\dots ,{\beta }_m)$.

Since

$$1^3+2^3+\dots +n^3={\left({\displaystyle \frac{1}{2}}n(n+1)\right)}^2,$$

we will consider the factorization of the binomial coefficient $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)$ instead of the sum of cubes.

The study of the greatest prime factor and the number of distinct prime divisors of the binomial coefficients $\left({\displaystyle \genfrac{}{}{0pt}{}{n+k}{k}}\right)$ has an extensive literature, for example, results mentioned in the notes of Chapter 7 in [1]. The proofs of these results are mainly based on the theory of linear forms in the logarithms of algebraic numbers [2]; accordingly, they contain enormous constants. For $k=2$, we refer to the computational result Theorem 1.2 in [3]. For elliptic curves, we refer to the book [4]. These results are related to Tian’s conjecture. However, there is a difference between Tian’s conjecture and the existing studies. The difference lies in the fact that Tian’s conjecture specifies the prime factors, while these studies consider the prime factors unknown and search for them.

The study of Tian’s conjecture is of some significance. It provides information about the distribution of prime factors in the sequence of cubic sums of consecutive natural numbers. It is related to Oppermann’s conjecture in number theory.

The rest of this paper is organized as follows. In Section 2, we develop a general computational method based on estimating linear forms in elliptic logarithms; see [5]. Using the computational algorithm, we verify several special cases where the number of prime factors is two or three, including the first sub-conjecture. In Section 3, we give a proof of the first sub-conjecture and some upper bounds for the number of solutions to the second sub-conjecture using the classical Zsigmondy theorem. The paper ends with a conclusion section.

2. A General Computational Approach

We consider the Diophantine equation

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=p_1^{{\alpha }_1}\cdots p_m^{{\alpha }_m},$$

(1)

where $p_1,\dots ,p_m$ are fixed prime numbers and $n,{\alpha }_1,\dots ,{\alpha }_m$ are unknown positive integers. One can take ${\alpha }_i=3{\beta }_i+{\gamma }_i$, where ${\gamma }_i\in \{0,1,2\},i=1,\dots ,m$. From Equation (1), we have

$$4n(n+1)+1=8p_1^{{\gamma }_1}\cdots p_m^{{\gamma }_m}{(p_1^{{\beta }_1}\cdots p_m^{{\beta }_m})}^3+1$$

and

$$p_1^{2{\gamma }_1}\cdots p_m^{2{\gamma }_m}{(2n+1)}^2={(2p_1^{{\beta }_1+{\gamma }_1}\cdots p_m^{{\beta }_m+{\gamma }_m})}^3+p_1^{2{\gamma }_1}\cdots p_m^{2{\gamma }_m},$$

and after substituting $X=2p_1^{{\beta }_1+{\gamma }_1}\cdots p_m^{{\beta }_m+{\gamma }_m}$ and $Y=p_1^{{\gamma }_1}\cdots p_m^{{\gamma }_m}(2n+1)$, we obtain $3^m$ so-called Mordell equations,

$$Y^2=X^3+p_1^{2{\gamma }_1}\cdots p_m^{2{\gamma }_m},$$

in positive integers $X,Y$. There is an extensive literature on Mordell equations; for classical and computational results, we refer to [6]. We remark that ${\beta }_i+{\gamma }_i>0$, and so X is divisible by $2p_1\cdots p_m$; there are no prime factors other than $2,p_1,p_2,\dots ,p_m$. These facts provide a strong sieve for the possible solutions of Equation (1). Finally, one can solve these Mordell equations by using the MAGMA V2.29-4 online calculator. In our future work, we plan to extend the computational approach to $m\ge 4$ by solving $3^m$ Mordell equations; while the cases $m=2$ and $m=3$ can be handled by hand, larger values of m require a fully computerized implementation of the sieve.

2.1. Two Special Cases: The Equations $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=3^{{\alpha }_1}{5}^{{\alpha }_2}$ and $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=2^{{\alpha }_1}{3}^{{\alpha }_2}{5}^{{\alpha }_3}$

To generate the corresponding Mordell equations and their solutions, we use the following commands:

for i in [0..2] do; for j in [0..2] do; for k in [0..2] do;

E:=EllipticCurve([0,$2^{2\ast i}\ast 3^{2\ast j}\ast 5^{2\ast k}$]); print($2^i\ast 3^j\ast 5^k$); E;

IntegralPoints(E);

end for; end for; end for;

The CPU time is 2.08 s, and the total memory usage is 85.16 MB.

For the equation

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=3^{{\alpha }_1}{5}^{{\alpha }_2},$$

we have ${\gamma }_1=0$; the solutions x are divisible by 30, and their 2-order is 1. From

Table 1. $p_1=2,p_2=3,p_3=5$: corresponding Mordell equations and their solutions.

$2^{{\mathit{\gamma }}_1}{3}^{{\mathit{\gamma }}_2}{5}^{{\mathit{\gamma }}_3}$	E	Solutions $(\mathit{x},\mathit{y})$ of E
1	$y^2=x^3+1$	(−1, 0); (0, 1); (2, 3)
5	$y^2=x^3+25$	$(0,5)$
25	$y^2=x^3+625$	$(0,25);(6,29);(75,650)$
3	$y^2=x^3+9$	(−2, 1); (0, 3); (3,6); (6, 15); (40, 253)
15	$y^2=x^3+225$	(−6, 3); (−5, 10); (0, 15); (4, 17); (6, 21); (10, 35); (15, 60); (30, 165); (60, 465); (180, 2415); (336, 6159); (351, 6576); (720,114, 611,085,363)
75	$y^2=x^3+5625$	$(0,75)$
9	$y^2=x^3+81$	(0, 9)
45	$y^2=x^3+2025$	(−9,36); (0, 45); (10, 55); (90, 855)
225	$y^2=x^3+\mathrm{50,625}$	(−36, 63); (0, 225); (100, 1025)
2	$y^2=x^3+4$	$(0,2)$
10	$y^2=x^3+100$	(−4, 6); (0, 10); (5, 15); (20,90); (24, 118); (2660, 137,190)
50	$y^2=x^3+2500$	(0,50)
6	$y^2=x^3+36$	(−3, 3); (0, 6); (4, 10); (12, 42)
30	$y^2=x^3+900$	(0, 30)
150	$y^2=x^3+\mathrm{22,500}$	(0, 150)
18	$y^2=x^3+324$	(0, 18)
90	$y^2=x^3+8100$	(−20, 10); (0,90); (36, 234); (45, 315); (3640, 219,610)
450	$y^2=x^3+\mathrm{202,500}$	(0, 450); (189, 2637)
4	$y^2=x^3+16$	(0, 4)
20	$y^2=x^3+400$	(0, 20)
100	$y^2=x^3+\mathrm{10,000}$	(0, 100)
12	$y^2=x^3+144$	(0, 12)
60	$y^2=x^3+3600$	(−15, 15); (0, 60); (24, 132); (40, 260)
300	$y^2=x^3+\mathrm{90,000}$	(−24, 276); (0, 300); (25, 325); (600, 14,700)
36	$y^2=x^3+1296$	(−8, 28); (0, 36); (9, 45); (72, 612)
180	$y^2=x^3+\mathrm{32,400}$	(0, 180)
900	$y^2=x^3+\mathrm{810,000}$	(0, 900)

, the solutions are

$$(3^{{\gamma }_2}{5}^{{\gamma }_3},x,y)=(15,30,165),(45,90,855),$$

and the corresponding values are $n=5$ and 9, respectively.

Next, we consider the equation

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=2^{{\alpha }_1}{3}^{{\alpha }_2}{5}^{{\alpha }_3}.$$

Here, the solutions x are divisible by 60, and Table 1 gives the solutions

$$(2^{{\gamma }_1}{3}^{{\gamma }_2}{5}^{{\gamma }_3},x,y)=(15,60,465),(15,180,2415),(300,600,14700)$$

and the corresponding values of $n=15,80$ and 24, respectively.

2.2. Another Examples: The Equations $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=3^{{\alpha }_2}{7}^{{\alpha }_3}$ and $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=2^{{\alpha }_1}{3}^{{\alpha }_2}{7}^{{\alpha }_3}$

For the equations $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=3^{{\alpha }_1}{7}^{{\alpha }_2}$ and $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=2^{{\alpha }_1}{3}^{{\alpha }_2}{7}^{{\alpha }_3}$, the possible solutions are

$$(3^{{\gamma }_2}{7}^{{\gamma }_3},x,y)=(21,42,273),(63,630,15813)$$

and

$$(2^{{\gamma }_1}{3}^{{\gamma }_2}{7}^{{\gamma }_3},x,y)=(147,588,14259),(14,84,770),(252,1008,32004),$$

respectively; see

Table 2. $p_1=2,p_2=3,p_3=7$: corresponding Mordell equations and their solutions.

$2^{{\mathit{\gamma }}_1}{3}^{{\mathit{\gamma }}_2}{7}^{{\mathit{\gamma }}_3}$	E	Solutions $(\mathit{x},\mathit{y})$ of E
1	$y^2=x^3+1$	(−1, 0); (0, 1); (2, 3)
7	$y^2=x^3+49$	(0, 7)
49	$y^2=x^3+2401$	(0, 49); (15, 76)
3	$y^2=x^3+9$	(−2, 1); (0, 3); (3, 6); (6, 15); (40, 253)
21	$y^2=x^3+441$	(−6, 15); (0, 21); (7, 28); (42, 273)
147	$y^2=x^3+\mathrm{21,609}$	(−12, 141); (0, 147); (588, 14,259)
9	$y^2=x^3+81$	(0, 9)
63	$y^2=x^3+3969$	(−14, 35); (−5, 62); (0, 63); (18, 99); (28, 161); (36, 225); (63, 504); (270, 4437); (630, 15,813)
441	$y^2=x^3+\mathrm{194,481}$	$(0,441)$
2	$y^2=x^3+4$	$(0,2)$
14	$y^2=x^3+196$	(−3, 13); (0, 14); (84, 770)
98	$y^2=x^3+9604$	(0, 98)
6	$y^2=x^3+36$	(−3, 3); (0, 6); (4, 10); (12, 42)
42	$y^2=x^3+1764$	(−12, 6); (0, 42); (21, 105); (28, 154); (1320, 47,958)
294	$y^2=x^3+\mathrm{86,436}$	(0, 294); (72, 678)
18	$y^2=x^3+324$	(0, 18)
126	$y^2=x^3+\mathrm{15,876}$	(0, 126)
882	$y^2=x^3+\mathrm{777,924}$	(0, 882)
4	$y^2=x^3+16$	(0, 4)
28	$y^2=x^3+784$	(−7, 21); (0, 28); (8, 36); (56, 420)
196	$y^2=x^3+\mathrm{38,416}$	(0, 196); (1617, 65,023)
12	$y^2=x^3+144$	(0, 12)
84	$y^2=x^3+7056$	(0, 84)
588	$y^2=x^3+\mathrm{345,744}$	(0, 588)
36	$y^2=x^3+1296$	(−8, 28); (0, 36); (9, 45); (72, 612)
252	$y^2=x^3+\mathrm{63,504}$	(0, 252); (16, 260); (1008, 32,004)
1764	$y^2=x^3+\mathrm{3,111,696}$	(−143, 433); (0, 1764)

. In the case of $(63,630,15813)$, the prime factors of 630 are $2,3,5$ and 7; this does not give a solution of the first equation owing to the presence of 5. Thus, the unique solution of the first equation is $\left({\displaystyle \genfrac{}{}{0pt}{}{7}{2}}\right)=3·7$. For the second equation, we have three real solutions, namely, $n=48,27$ and 63, respectively.

The previous numerical results can be summarized as follows.

Theorem 1.

The only solutions of the equations

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=3^{{\alpha }_1}·5^{{\alpha }_2}\mathrm{and}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=2^{{\alpha }_1}·3^{{\alpha }_2}·5^{{\alpha }_3}$$

in positive integers $n,{\alpha }_1,{\alpha }_2$ and $n,{\alpha }_1,{\alpha }_2,{\alpha }_3$, respectively, are

$$(n,{\alpha }_1,{\alpha }_2)=(5,1,1),(9,2,1)$$

and

$$(n,{\alpha }_1,{\alpha }_2,{\alpha }_3)=(15,3,1,1),(24,2,1,2),(80,3,4,1),$$

respectively. Further, the only solutions of the equations

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=3^{{\alpha }_1}·7^{{\alpha }_2}\mathrm{and}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=2^{{\alpha }_1}·3^{{\alpha }_2}·7^{{\alpha }_3}$$

in positive integers $n,{\alpha }_1,{\alpha }_2$ and $n,{\alpha }_1,{\alpha }_2,{\alpha }_3$, respectively, are

$$(n,{\alpha }_1,{\alpha }_2)=(6,1,1)$$

and

$$(n,{\alpha }_1,{\alpha }_2,{\alpha }_3)=(27,1,3,1),(48,3,1,2),(63,5,2,1),$$

respectively.

Now, Tian’s sub-conjecture 1 is proved, and the theorem above shows that sub-conjecture 3 is sharp for $m=2$ and 3; in these cases, we can give two and three primes, respectively, such that the corresponding equations have two and three solutions, respectively. Our next result is based on

Table 3. The solutions of the equation $\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}$ in positive integers ${\alpha }_1$ and ${\alpha }_2$.

${\mathit{p}}_1,{\mathit{p}}_2$	Solutions
2, 3	$\left({\displaystyle \genfrac{}{}{0pt}{}{4}{2}}\right),\left({\displaystyle \genfrac{}{}{0pt}{}{9}{2}}\right)$
2, 5	$\left({\displaystyle \genfrac{}{}{0pt}{}{5}{2}}\right)$
2, 7	$\left({\displaystyle \genfrac{}{}{0pt}{}{8}{2}}\right)$
2, 11	–
2, 13	–
2, 17	$\left({\displaystyle \genfrac{}{}{0pt}{}{17}{2}}\right)$
3, 5	$\left({\displaystyle \genfrac{}{}{0pt}{}{6}{2}}\right),\left({\displaystyle \genfrac{}{}{0pt}{}{10}{2}}\right)$
3, 7	$\left({\displaystyle \genfrac{}{}{0pt}{}{7}{2}}\right)$
3, 11	$\left({\displaystyle \genfrac{}{}{0pt}{}{243}{2}}\right)$
3, 13	$\left({\displaystyle \genfrac{}{}{0pt}{}{27}{2}}\right)$
3, 17	$\left({\displaystyle \genfrac{}{}{0pt}{}{18}{2}}\right)$
5, 7	$\left({\displaystyle \genfrac{}{}{0pt}{}{50}{2}}\right)$
5, 11	$\left({\displaystyle \genfrac{}{}{0pt}{}{11}{2}}\right)$
5, 13	$\left({\displaystyle \genfrac{}{}{0pt}{}{26}{2}}\right)$
5, 17	–
7, 11	–
7, 13	$\left({\displaystyle \genfrac{}{}{0pt}{}{14}{2}}\right)$
7, 17	–
11, 13	–
11, 17	–
13, 17	–

, generated by our algorithm, and gives support for sub-conjecture 2.

Theorem 2.

The equation

$$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{2}}\right)=p_1^{{\alpha }_1}·p_2^{{\alpha }_2}$$

in positive integers $n,{\alpha }_1,{\alpha }_2$ and the same equation in prime numbers $p_1,p_2$ with $2\le p_1<p_2\le 17$ imply

$$n\in \{3,4,5,6,7,8,9,10,13,16,17,25,26,49,242\}.$$

3. Proofs with the Classical Zsigmondy Theorem

In this section, we give an alternative proof of Tian’s sub-conjecture 1 and some results concerning sub-conjecture 2. We will apply the Zsigmondy theorem, which was published in [7]. It is a generalization of an earlier result appearing in [8,9], which is now called Bang’s theorem. Bang’s theorem states that if $n>1$ and n is not equal to 6, then $2^n-1$ has a prime divisor not dividing any $2^k-1$ with $k<n$. Later, many extensions of these results were published; examples include [10,11,12].

3.1. Proof of the First Sub-Conjecture

Tian’s first sub-conjecture is to find all natural numbers $n,{\beta }_1,{\beta }_2$ so that

$$1^3+2^3+3^3+\dots +n^3=3^{{\beta }_1}·5^{{\beta }_2}.$$

(2)

Since

$$1^3+2^3+\dots +n^3={\left({\displaystyle \frac{1}{2}}n(n+1)\right)}^2,$$

we can transform Equation (2) to the following one:

$${\left({\displaystyle \frac{1}{2}}n(n+1)\right)}^2=3^{{\beta }_1}·5^{{\beta }_2}.$$

Therefore, ${\beta }_1,{\beta }_2$ must be even numbers. We can assume that ${\beta }_1=2{\alpha }_1\ge 2,n=2\alpha \ge 2$. Thus, we can transform Equation (2) to

$$n(n+1)=2·3^{{\alpha }_1}·5^{{\alpha }_2}.$$

(3)

This implies that

$$n=2^x·3^y·5^z,n+1=2^{1-x}·3^{{\alpha }_1-x}·5^{{\alpha }_2-z},$$

where $x,y,z$ are integers, and

$$0\le x\le 1,0\le y\le {\alpha }_1,0\le z\le {\alpha }_2.$$

By combining the above with the fact that $gcd(n,n+1)=1$, we obtain

$$x=\{0,1\},y=\{0,{\alpha }_1\},z=\{0,{\alpha }_2\}.$$

It is easy to see that the cases $(x,y,z)=(0,0,0),(1,0,0),(0,{\alpha }_1,{\alpha }_2)$ and $(1,{\alpha }_1,{\alpha }_2)$ are impossible. Thus, Equation (3) can be transformed to the following equations:

$$\begin{array}{cc}\hfill n=2·3^{{\alpha }_1},n+1=5^{{\alpha }_2}& ⟹5^{{\alpha }_2}-1=2·3^{{\alpha }_1},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill n=2·5^{{\alpha }_2},n+1=3^{{\alpha }_1}& ⟹3^{{\alpha }_1}-1=2·5^{{\alpha }_2},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill n=3^{{\alpha }_1},n+1=2·5^{{\alpha }_2}& ⟹3^{{\alpha }_1}+1=2·5^{{\alpha }_2},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill n=5^{{\alpha }_2},n+1=2·3^{{\alpha }_1}& ⟹5^{{\alpha }_2}+1=2·3^{{\alpha }_1}.\hfill \end{array}$$

(7)

3.1.1. The Simple Cases

We have the following lemmas:

Lemma 1.

(1) $({\alpha }_1,{\alpha }_2)=(2,1)$ is a solution of Equation (6).

(2) $({\alpha }_1,{\alpha }_2)=(1,1)$ is a solution of Equation (7).

Proof. 

Obvious. □

Lemma 2.

Equation (4) has no solution.

Proof. 

From (4), we have

$$5^{{\alpha }_2}-1=2·3^{{\alpha }_1}.$$

However, $5^{{\alpha }_2}-1\equiv 0(mod4)$, and $2·3^{{\alpha }_1}\equiv 2(mod4)$. □

Lemma 3.

Equation (5) has no solution.

Proof. 

If $({\alpha }_1,{\alpha }_2)$ is a solution of Equation (5), $3^{{\alpha }_1}-1=2·5^{{\alpha }_2}$, then 2 is a divisor, but 4 is not, i.e., ${\alpha }_1$ satisfies the following relation:

$$3^{{\alpha }_1}-1\not\equiv 0(mod4).$$

This implies that ${\alpha }_1$ is an odd number. Hence, ${\alpha }_1=2r+1$ for some $r\ge 0$. Thus, we have ${\alpha }_1=4s+1$ or ${\alpha }_1=4s+3$ for some $s\ge 0$. It is easy to check that

$$3^{4r+1}-1\equiv 2(mod5),3^{4r+3}-1\equiv 1(mod5).$$

Thus, for odd ${\alpha }_1$, we have $3^{{\alpha }_1}\not\equiv 0(mod5)$. This contradiction means that (5) has no solution. □

3.1.2. The Difficult Cases

We have the following result:

Theorem 3.

Each equation of (6), (7) has at most one solution.

To prove this theorem, we need the following result from Zsigmondy.

Lemma 4

(Zsigmondy Theorem). 1. If $a>b>0$ are coprime integers, then for any integer $n\ge 1$, there is a prime number p (called a primitive prime divisor) that divides $a^n-b^n$ and does not divide $a^k-b^k$ for any positive integer $k<n$, with the following exceptions:

(1) 

$n=1,a-b=1$; then $a^n-b^n=1$, which has no prime divisors.

(2) 

$n=2$, $a+b=2^u$; then any odd prime factors of $a^2-b^2=(a+b)(a^1-b^1)$ must be contained in $a^1-b^1$, which is also even.

(3) 

$n=6,a=2,b=1$; then

$$a^6-b^6=63=3^2\times 7={(a^2-b^2)}^2(a^3-b^3).$$

2. If $a>b>0$ are coprime integers, then for any integer $n\ge 1$, $a^n+b^n$ has at least one primitive prime divisor with the exception of $2^3+1^3=9$.

The above result from Zsigmondy was reported in [7]; see also [10,11,12] for further materials.

Proof of Theorem 3.

Here, we simply show that Equation (6) has only one solution; one can repeat the proof for Equation (7). As we have seen,

$$3^2+1=2·5.$$

Therefore, for any ${\alpha }_1>2$, $3^{{\alpha }_1}+1$ has a prime factor $u\left({\alpha }_1\right)$ not dividing $3^2+1=10$, i.e., $u\left({\alpha }_1\right)\notin \{2,5\}$. It is apparent that $u\left({\alpha }_1\right)\ne 3$. Thus, $u\left({\alpha }_1\right)>5$, and immediately,

$$3^{{\alpha }_1}+1\ne 2·5^{{\alpha }_2}$$

for any ${\alpha }_2\in \mathbb{N}$. Thus, we have proven that $3^{{\alpha }_1}+1=2·5^{{\alpha }_2}$ has no solution for ${\alpha }_1>2$.

Now we have determined all solution of Equations (4)–(7), and one can check that they correspond to the solutions $(n,{\alpha }_1,{\alpha }_2)=(5,1,1)$ and $(9,2,1)$ of (3). □

3.2. Results Concerning the Second Sub-Conjecture

Let $p_1,\dots ,p_m$ be distinct (positive) prime numbers, and set

$$S=\{p_1^{{\alpha }_1}\cdots p_m^{{\alpha }_m}:{\alpha }_i\ge 0\mathrm{integers}\}.$$

We consider the exponential Diophantine equation

$$ax+by=c$$

(8)

where $a,b,c$ are non-zero fixed integers and $x,y\in S$ are unknowns. The Diophantine equations in the form of (8) are called S-unit equations. We note that the number of solutions of (8) is exponential in m; see Chapter 6 of [13]. Below, we transform our original problems to S-unit equations.

Clearly, from Theorem 3 we see that the first sub-conjecture is true. We gave the alternative proof without a computer to illustrate the power of the Zsigmondy theorem. By applying it, we can determine a sharp upper bound for the number of solutions.

Theorem 4.

1. For any distinct prime numbers $p_1,p_2$, the Diophantine equation

$$1^3+2^3+\dots +n^3=p_1^{{\beta }_1}·p_2^{{\beta }_2}$$

(9)

in positive integers $n,{\beta }_1,{\beta }_2$ has at most 4 solutions.

2. For any three distinct prime numbers $p_1=2,p_2,p_3$, the equation

$$n(n+1)=2^{{\alpha }_1}·p_2^{{\alpha }_2}·p_3^{{\alpha }_3}$$

(10)

in positive integers $n,{\alpha }_1,{\alpha }_2,{\alpha }_3$ has at most 6 solutions.

Proof. 

We follow the argument of the alternative proof. For statement 1, Equation (9) can be transformed into

$${\left({\displaystyle \frac{1}{2}}n(n+1)\right)}^2=p_1^{{\beta }_1}·p_2^{{\beta }_2},$$

and ${\beta }_1=2{\alpha }_1,{\beta }_2=2{\alpha }_2$ for ${\alpha }_1,{\alpha }_2\in \mathbb{N}$. Therefore, we have

$$n(n+1)=2·p_1^{{\alpha }_1}·p_2^{{\alpha }_2}.$$

Since $2,p_1,p_2$ are prime numbers, we have

$$n=2^x{p}_1^y{p}_2^z,n+1=2^{1-x}{p}_1^{{\alpha }_1-x}{p}_2^{{\alpha }_2-z}$$

for $x,y,z\in \mathbb{N}$ with

$$0\le x\le 1,0\le y\le {\alpha }_1,0\le z\le {\alpha }_2.$$

By combining the above with the fact $gcd(n,n+1)=1$, we can infer that

$$x\in \{0,1\},y\in \{0,{\alpha }_1\},z\in \{0,{\alpha }_2\}.$$

Similar to what we discussed in Section 3.1 for $p_1=3,p_2=5$, the combination $(x,y,z)=(0,0,0),(1,0,0),(0,{\alpha }_1,{\alpha }_2)$ and $(1,{\alpha }_1,{\alpha }_2)$ can be excluded. Therefore, the solution of Equation (9) can be expressed as

$$\begin{array}{cc}\hfill n=2p_1^{{\alpha }_1},n+1=p_2^{{\alpha }_2}& ⟹p_2^{{\alpha }_2}-1=2p_1^{{\alpha }_1};\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill n=2p_2^{{\alpha }_2},n+1=p_1^{{\alpha }_1}& ⟹p_1^{{\alpha }_1}-1=2p_2^{{\alpha }_2};\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill n=p_1^{{\alpha }_1},n+1=2p_2^{{\alpha }_2}& ⟹p_1^{{\alpha }_1}+1=2p_2^{{\alpha }_2};\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill n=p_2^{{\alpha }_2},n+1=2p_1^{{\alpha }_1}& ⟹p_2^{{\alpha }_2}+1=2p_1^{{\alpha }_1}.\hfill \end{array}$$

(14)

Now, using the Zsigmondy theorem, it is easy to prove that each of the above four equations has at most one solution, and therefore, the original Equation (9) also has at most four solutions. Here, we show that if Equation (11) has a solution, then that solution is the unique solution of the equation. The proof for the remaining three equations is very similar, and we leave it to the readers. It is easy to see that if $({\alpha }_1^{\prime },{\alpha }_2^{\prime }),({\alpha }_1^{″},{\alpha }_2^{″})$ are solutions of the equation and ${\alpha }_2^{\prime }<{\alpha }_2^{″}$, then ${\alpha }_1^{\prime }<{\alpha }_1^{″}$. Thus, we define an order ≺ on the set of all solutions:

$$({\alpha }_1^{\prime },{\alpha }_2^{\prime })\prec ({\alpha }_1^{″},{\alpha }_2^{″})⟺{\alpha }_2^{\prime }<{\alpha }_2^{″}.$$

Assuming that $({\alpha }_1,{\alpha }_2)\in {\mathbb{N}}^2$ is the minimal solution in this order and that $({\alpha }_1^{\prime },{\alpha }_2^{\prime })$ is any other solution of the same equation, it follows that

$$p_2^{{\alpha }_2}-1=2p_1^{{\alpha }_1},p_2^{{\alpha }_2^{\prime }}-1=2p_1^{{\alpha }_1^{\prime }}.$$

Then, accordingly, $p_2^{{\alpha }_2^{\prime }}-1$ has a divisor u not in the set of divisors of $p_2^{{\alpha }_2}-1$, which is $\{2,p_1\}$. However, u is a divisor of $p_2^{{\alpha }_2^{\prime }}-1=2p_1^{{\alpha }_a^{\prime }re1}$, and the divisors of $2p_1^{{\alpha }_1^{\prime }}$ are also $\{2,p_1\}$, i.e., $u\in \{2,p_1\}$; thus, we arrive at a contradiction.

For statement 2, if $2,p_2,p_3$ are prime numbers, we can decompose Equation (10) into six equations:

• $p_3^{{\alpha }_3}-1=2^{{\alpha }_1}·p_2^{{\alpha }_2},p_2^{{\alpha }_2}-1=2^{{\alpha }_1}·p_3^{{\alpha }_3},2^{{\alpha }_1}-1=p_2^{{\alpha }_2}·p_3^{{\alpha }_3},$
• $p^{{\alpha }_3}+1=2^{{\alpha }_1}·p_2^{{\alpha }_2},p_2^{{\alpha }_2}+1=2^{{\alpha }_1}·p_3^{{\alpha }_3},2^{{\alpha }_1}+1=p_2^{{\alpha }_2}·p_3^{{\alpha }_3}.$

Similarly, the Zsigmondy theorem can be applied here to prove that each equation has at most one solution. □

4. Conclusions

Herein, a general computational method for solving special cases of Tian’s sub-conjectures is developed. Using this method, the first sub-conjecture is verified, and the second sub-conjecture is supported. We also give an alternative theoretical proof for the first sub-conjecture, and we establish upper bounds on the number of solutions for the second sub-conjecture using a classical theorem. Although our numerical results provide support for the second and third sub-conjectures, a general proof is beyond our current knowledge. The study of the prime factorization of binomial coefficients $\left({\displaystyle \genfrac{}{}{0pt}{}{n+k}{k}}\right),(k\ge 2)$ is based on the special case $k=2$. However, due to the divisibility relation, the corresponding exponents can also be zero, resulting in many additional possibilities that will require a computational combinatorial approach. This will be addressed in future work.