1. Introduction
In 1894, Franel [Reference Franel2] found that the numbers
satisfy the recurrence relation (cf. [Reference Sloane14, A000172]):
These numbers are now called Franel numbers. Callan [Reference Callan1] found a combinatorial interpretation of the Franel numbers. The Franel numbers play important roles in combinatorics and number theory. The sequence $\{f_n\}_{n \geqslant 0}$ is one of the five sporadic sequences (cf. [Reference Zagier23, § 4]) which are integral solutions of certain Apéry-like recurrence equations and closely related to the theory of modular forms. In 2013, Sun [Reference Sun19] revealed some unexpected connections between the numbers $f_n$ and representations of primes $p \equiv 1 (\text{mod} 3)$ in the form $x^2+3y^2$ with $x,y\in \mathbb {Z}$, for example, Sun [Reference Sun19, (1.2)] showed that
and in the same paper, Sun proposed some conjectures involving Franel numbers, one of which is
Conjecture 1.1 Let $p>2$ be a prime. If $p=x^2+3y^2$ with $x,y\in \mathbb {Z}$ and $x\equiv 1 ({\rm {mod}}\ 3)$, then
For more details on Franel numbers, we refer the readers to [Reference Guo3, Reference Guo4, Reference Liu6, Reference Mao8, Reference Mao9, Reference Sun18, Reference Sun20] and so on.
In this paper, our first goal is to prove the above conjecture.
Theorem 1.1 Conjecture 1.1 is true.
Combining (1.1) and theorem 1.1, we immediately obtain the following result.
Corollary 1.1 For any prime $p\equiv 1\pmod 3$, we have
Sun [Reference Sun19] also gave the following conjecture.
Conjecture 1.2 Let $p>2$ be a prime. If $p\equiv 1\pmod 3$, then
Our last goal is to prove this conjecture.
Theorem 1.2 Conjecture 1.2 is true.
We are going to prove theorem 1.1 in §2. Section 3 is devoted to proving theorem 1.2. Our proofs make use of some combinatorial identities which were found by the package Sigma [Reference Schneider13] via software Mathematica and the $p$-adic gamma function. The proof of theorem 1.2 is somewhat difficult and complex because it is rather convoluted. Throughout this paper, prime $p$ always $\equiv 1\pmod 3$, so in the following lemmas $p>5$ or $p>3$ or $p>2$ is the same, we mention it here first.
2. Proof of theorem 1.1
For a prime $p$, let $\mathbb {Z}_p$ denote the ring of all $p$-adic integers and let $\mathbb {Z}_p^{\times }:=\{a\in \mathbb {Z}_p:\,a\text { is prime to }p\}.$ For each $\alpha \in \mathbb {Z}_p$, define the $p$-adic order $\nu _p(\alpha ):=\max \{n\in \mathbb {N}:\, p^n\mid \alpha \}$ and the $p$-adic norm $|\alpha |_p:=p^{-\nu _p(\alpha )}$. Define the $p$-adic gamma function $\Gamma _p(\cdot )$ by
and
In particular, we set $\Gamma _p(0)=1$. In the following, we need to use the most basic properties of $\Gamma _p$, and all of them can be found in [Reference Murty11, Reference Robert12]. For example, we know that
where $a_0(x)\in \{1,2,\ldots,p\}$ such that $x\equiv a_0(x)\ ({\rm {mod}}\ p)$. And a property we need here is the fact that for any positive integer $n$,
Lemma 2.1 ([Reference Sun19, lemma 2.2]) For any $n\in \mathbb {N}$ we have
and
For $n,m\in \{1,2,3,\ldots \}$, define
these numbers with $m=1$ are often called the classic harmonic numbers. Recall that the Bernoulli polynomials are given by
Lemma 2.2 ([Reference Sun15, Reference Sun16]) Let $p>5$ be a prime. Then
where $q_p(a)=(a^{p-1}-1)/p$ stands for the Fermat quotient.
Lemma 2.3 Let $p>5$ be a prime. If $0\leq j\leq (p-1)/2$, then we have
Proof. If $0\leq j\leq (p-1)/2$ and $j\neq (p-1)/3$, then we have
If $j=(p-1)/3$, then by lemma 2.2, we have
and
Now the proof of lemma 2.3 is complete.
Proof of theorem 1.1 With the help of (2.4), we have
By loading the package Sigma in software Mathematica, we find the following identity:
Thus, replacing $n$ by $p$ in the above identity and then substitute it into (2.6), we have
Hence, we immediately obtain the following result by lemma 2.3,
It is easy to verify that
where
and
Applying the famous partial fraction identity
with $x=1/3, n=(p-1)/2$ and $x=2/3,n=(p-1)/2$, we may simplify (2.9) as
where $(a)_n=a(a+1)\cdots (a+n-1)$ is the rising factorial or the Pochhammer symbol.
In view of (2.2), we have
where $\Gamma (\cdot )$ is the gamma function. In view of [Reference Long and Ramakrishna7, theorem 14] and [Reference Liu5, (2.4)] (or [Reference Mao and Pan10, (3.2)]), for $\alpha,s\in \mathbb {Z}_p$, we have
and
where $\Gamma _p'(x)$ denotes the $p$-adic derivative of $\Gamma _p(x)$, $\langle \alpha \rangle _n$ denotes the least non-negative residue of $\alpha$ modulo $n$, i.e. the integer lying in $\{0,1,\ldots,n-1\}$ such that $\langle \alpha \rangle _n\equiv \alpha \ ({\rm {mod}}\ n)$.
Therefore,
In view of (2.1) and (2.2), we have
In view of [Reference Yeung22, proposition 4.1], we have
Then with the help of [Reference Yeung22, theorem 4.12] and lemma 2.2, we have
and
Hence,
Lemma 2.4 Let $p>3$ be a prime. For any $p$-adic integer $t$, we have
Proof. Set $m=(p-1)/2$. It is easy to check that
So lemma 2.4 is finished.
Now we evaluate $S_2$ modulo $p^2$. It is easy to obtain that
with the help of lemmas 2.2, 2.4 and [Reference Yeung22, theorem 4.12].
Therefore, in view of (2.7), (2.8), (2.15) and (2.17), we immediately get the desired result
On the contrary, we use equation (2.5) to obtain
By using the package Sigma again, we find the following identity:
Thus,
Lemma 2.5 Let $p>5$ be a prime. If $0\leq j\leq (p-1)/2$ and $j\neq (p-1)/3$, then
If $(p+1)/2\leq j\leq p-1$, then
Proof. If $0\leq j\leq (p-1)/2$ and $j\neq (p-1)/3$, then we have
If $(p+1)/2\leq j\leq p-1$, then
Now the proof of lemma 2.5 is complete.
It is known that $\binom {2k}k\equiv 0\ ({\rm {mod}}\ p)$ for each $(p+1)/2\leq k\leq p-1$, and it is easy to check that for each $0\leq j\leq (p-1)/2$:
These, with (2.18) yield
where
As above, with (2.10), (2.13), (2.14), lemma 2.2 and [Reference Yeung22, theorem 4.12], we have the following congruence modulo $p^2$:
Now we evaluate $S_3$. It is easy to see that
In view of (2.17) and [Reference Yeung22, theorem 4.12], we immediately obtain
This, with (2.19) and (2.20) yields
Now the proof of theorem 1.1 is complete.
3. Proof of theorem 1.2
Proof of theorem 1.2 With the help of (2.4), we have
By loading the package Sigma in software Mathematica, we have the following identity:
Thus, replace $n$ by $p$ in the above identity and then substitute it into (3.1), we have
Hence, we immediately obtain the following result by lemma 2.3:
where
It is easy to verify that
where
So,
It is easy to see that
On the other hand, we have
So by lemma 2.5 and the fact that for each $0\leq k\leq (p-1)/2$,
and for each $(p+1)/2\leq j\leq p-1$,
we have the following congruence modulo $p^3$:
where
Hence, we have
where
By Sigma, we can find and prove the following identity:
In view of [Reference Sun17, lemma 3.1] and lemma 2.2, we have
It is easy to check that
And by [Reference Sury, Wang and Zhao21, (6)], we have
By Sigma, we find the following identity which can be proved by induction on $n$:
So by setting $n=(p-1)/2$ in the above identity and with lemma 2.2, we have
And by (3.10), we have
It is easy to check that
By Sigma again, we have
So in view of lemma 2.2 and [Reference Yeung22], we have
Thus, by (3.10), we have
Also it is easy to see that
And by Sigma, we have
So in view of lemma 2.2, we have
Hence,
This, with (3.11) and (3.12) yields
By Sigma, we find the following identity which can be proved by induction on $n$:
So in view of [Reference Yeung22], we have
This, with (3.9) and (3.13) yields
Thus, with (3.8) we have
So by (3.7), we have
Therefore, by (3.6) and (3.4), we deduce
Now, we evaluate the second sum on the right-hand side of (3.5). It is easy to see
By (3.14), we have
Now we consider the first sum of the right-hand side in (3.17):
The following identity is very important to us:
This, with [Reference Yeung22] yields
And by (3.19), we have
In view of (3.19) and [Reference Yeung22], we have
This, with (3.21) yields
Combining this with (3.20), we have
Thus, by (3.17) and (3.18), we have
This, with (3.5) and (3.16) yields
While
It is easy to check that
and
So by lemma 2.2 and the fact that $H_{p-1-k}^{(2)}\equiv -H_k^{(2)}\ ({\rm {mod}}\ p)$ for each $0\leq k\leq p-1$, we have
and
So by $\binom {-\frac 12}{\frac {p-1}3}\equiv \binom {\frac {p-1}2}{\frac {p-1}3}\ ({\rm {mod}}\ p)$ and $\binom {p-1}{\frac {p-1}3}\equiv (-1)^{\frac {p-1}3}=1\ ({\rm {mod}}\ p)$, we can immediately obtain that
This, with (3.22) yields
Now the proof of theorem 1.2 is complete.
Acknowledgements
The authors would like to thank the anonymous referee for careful reading of this manuscript and valuable comments, which make the paper more readable. This research was supported by the National Natural Science Foundation of China (Grant 12001288).