On the divisibility of the truncated hypergeometric function ${}_{3}F_2$

We prove a result on the divisibility of the truncated hypergeometric function ${}_{3}F_2\Big[\genfrac{}{}{0pt}{}{\alpha}{}\genfrac{}{}{0pt}{}{\beta}{1}\genfrac{}{}{0pt}{}{1-\alpha-\beta}{1}\Big|1\Big]$.

The hypergeometric function ${}_{q+1}F_q\Big[\genfrac{}{}{0pt}{}{\alpha_1}{}\genfrac{}{}{0pt}{}{\alpha_2}{\beta_1}\genfrac{}{}{0pt}{}{\ldots}{\ldots}\genfrac{}{}{0pt}{}{\alpha_{q+1}}{\beta_q}\Big|x\Big]$ is given by $${}_pF_q\bigg[\begin{matrix}\alpha_1&\alpha_2&\ldots&\alpha_p\\\beta_1&\beta_2&\ldots&\beta_q\end{matrix}\bigg|x\bigg]=\sum_{k=0}^\infty\frac{(\alpha_1)_k(\alpha_2)_k\cdots (\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\cdots (\beta_q)_k}\cdot\frac{x^k}{k!},$$ where $$(\alpha)_k=\begin{cases} \alpha(\alpha+1)\cdots(\alpha+k-1),\quad&\text{if }k\geq 1,\\
1,&\text{if }k=0.
\end{cases}$$

Define the truncated hypergeometric function $${}_pF_q\bigg[\begin{matrix}\alpha_1&\alpha_2&\ldots&\alpha_p\\\beta_1&\beta_2&\ldots&\beta_q\end{matrix}\bigg|x\bigg]_n=
\sum_{k=0}^{n-1}\frac{(\alpha_1)_k(\alpha_2)_k\cdots (\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\cdots (\beta_q)_k}\cdot\frac{x^k}{k!}.$$ In @M, Mortenson studied the arithmetical properties of ${}_{2}F_1\Big[\genfrac{}{}{0pt}{}{m/d}{}\genfrac{}{}{0pt}{}{(d-m)/d}{1}\Big|x\Big]$, where $1\leq m<d$ and $p$ is a prime with $p\equiv1\pmod{d}$. With help of the Gross-Koblitz formula, he proved that for prime $p\geq 5$, $${}_2F_1\bigg[\begin{matrix}1/2&1/2\\&1\end{matrix}\bigg|1\bigg]_p\equiv\jacob{-1}{p}\pmod{p^2},\qquad{}_2F_1\bigg[\begin{matrix}1/3&2/3\\&1\end{matrix}\bigg|1\bigg]_p\equiv\jacob{-3}{p}\pmod{p^2},$$$${}_2F_1\bigg[\begin{matrix}1/4&3/4\\&1\end{matrix}\bigg|1\bigg]_p\equiv\jacob{-2}{p}\pmod{p^2},\qquad{}_2F_1\bigg[\begin{matrix}1/6&5/6\\&1\end{matrix}\bigg|1\bigg]_p\equiv\jacob{-1}{p}\pmod{p^2},$$ where $\jacob{\cdot}{p}$ denotes the Legendre symbol modulo $p$. Subsequently, Sun @S gave the elementary proofs for the above congruences. However, in fact, in @S2, Sun proved that $${}_2F_1\bigg[\begin{matrix}\alpha&1-\alpha\\&1\end{matrix}\bigg|1\bigg]_p\equiv (-1)^{\{\alpha\}_p-1}\pmod{p^2},$$ where $\{\alpha\}_p$ denotes the least non-negative residue of $\alpha$ modulo $p$.

Motivated by the above results, in this note, we shall study the truncated hypergeometric function ${}_{3}F_2\Big[\genfrac{}{}{0pt}{}{\alpha}{}\genfrac{}{}{0pt}{}{\beta}{1}\genfrac{}{}{0pt}{}{1-\alpha-\beta}{1}\Big|1\Big]$.

\[t2\] Suppose that $p$ is an odd prime and $\alpha,\beta$ are $p$-integral. If $\{\alpha\}_p,\{\beta\}_p\geq 1$ and $\{\alpha\}_p+\{\beta\}_p\leq p$, then $${}_3F_2\bigg[\begin{matrix}\alpha&\beta&1-\alpha-\beta\\&1&1\end{matrix}\bigg|1\bigg]_p\equiv0\pmod{p^2}.$$

By replacing $\alpha$ and $\beta$ by $-\alpha$ and $-\beta$, (\[t1e\]) is equivalent to $$\sum_{k=0}^{p-1}(-1)^k\binom{\alpha}{k}\binom{\beta}{k}\binom{-1-\alpha-\beta}{k}\equiv0\pmod{p^2},$$ where $\{\alpha\}_p+\{\beta\}_p\geq p$. Let us recall the Saalschütz identity: $${}_3F_2\bigg[\begin{matrix}a&b&c\\ &d&e\end{matrix}\bigg|1\bigg]=\frac{(d-a)_{|c|}(d-b)_{|c|}}{(d)_{|c|}(d-a-b)_{|c|}},$$ where $c$ is a non-positive integer and $
d+e=a+b+c+1$.

Below assume that $\alpha+\beta\geq p$. If $1\leq \alpha,\beta\leq p-1$, then by the Saalschütz identity, we have $$\begin{aligned}
\sum_{k=0}^{p-1}(-1)^k\binom{\alpha}{k}\binom{\beta}{k}\binom{-1-\alpha-\beta}{k}\equiv&.\sum_{k=0}^{\infty}\frac{(-\alpha)_k(-\beta)_k(1+\alpha+\beta)_k}{(k!)^3}\\
=&\frac{(1+\beta)_{\alpha}(-\alpha-\beta)_{\alpha}}{(1)_{\alpha}(-\alpha)_{\alpha}}=\binom{\alpha+\beta}{\alpha}^2\equiv0\pmod{p^2}.\end{aligned}$$ Thus it suffices to prove that for any integer $s$ $$\sum_{k=0}^{p-1}(-1)^k\binom{\alpha}{k}\binom{\beta}{k}\binom{-1-\alpha-\beta}{k}\equiv \sum_{k=0}^{p-1}(-1)^k\binom{\alpha}{k}\binom{\beta+sp}{k}\binom{-1-\alpha-\beta-sp}{k}\pmod{p^2},$$ provided that $0\leq\alpha,\beta\leq p-1$ and $\alpha+\beta\geq p$.

Evidently, $$\begin{aligned}
\binom{\alpha+sp}{k}&=\sum_{j=0}^{k}\binom{\alpha}{k-j}\binom{sp}{j}=\binom{\alpha}{k}+\sum_{j=1}^{k}\binom{\alpha}{k-j}\binom{(s-1)p+p-1}{j-1}\frac{sp}{j}\notag\\
&\equiv{\binom{\alpha}{k}+\sum_{j=1}^{k}\binom{\alpha}{k-j}\binom{s-1}{0}\binom{p-1}{j-1}\frac{sp}{j}}\notag\\
&\equiv{\binom{\alpha}{k}+sp\sum_{j=1}^{k}\binom{\alpha}{k-j}\frac{(-1)^{j-1}}{j}}\pmod{p^2}.\end{aligned}$$ Then we have $$\begin{aligned}
&\sum_{k=0}^{\alpha}(-1)^k\binom{\alpha}{k}\binom{\beta+sp}{k}\binom{-1-\alpha-\beta-sp}{k}-\sum_{k=0}^{\alpha}(-1)^k\binom{\alpha}{k}\binom{\beta}{k}\binom{-1-\alpha-\beta}{k}\\
\equiv&
sp\sum_{k=1}^{\alpha}(-1)^k\binom{\alpha}{k}\sum_{j=1}^k\frac{(-1)^j}{j}\bigg(\binom{\beta}{k}\binom{-1-\alpha-\beta}{k-j}-\binom{-1-\alpha-\beta}{k}\binom{\beta}{k-j}\bigg)\pmod{p^2}.\end{aligned}$$ Applying the Saalschütz identity again, we get that $$\begin{aligned}
&\sum_{j=1}^\alpha\frac{1}{j}\sum_{k=j}^{\alpha}(-1)^{k-j}\binom{\alpha}{k}\binom{-1-\alpha-\beta}{k}\binom{\beta}{k-j}\\
=&\sum_{j=1}^\alpha\frac{(-\alpha)_j(1+\alpha+\beta)_j}{j\cdot(j!)^2}\sum_{k=j}^{\alpha}\frac{(j-\alpha)_{k-j}(-\beta)_{k-j}(j+1+\alpha+\beta)_{k-j}}{((j+1)_{k-j})^2\cdot(k-j)!}\\
=&\sum_{j=1}^\alpha\frac{(-\alpha)_j(1+\alpha+\beta)_j}{j\cdot(j!)^2}\cdot\frac{(j+1+\beta)_{\alpha-j}(-\alpha-\beta)_{\alpha-j}}{(j+1)_{\alpha-j}(-\alpha)_{\alpha-j}}\\
=&\sum_{j=1}^\alpha\frac{(-1)^\beta}{j}\cdot\binom{-\alpha-1}{j+\beta}\binom{\alpha+\beta}{j+\beta}.\end{aligned}$$ Since $\alpha+\beta\geq p$, we have $$\begin{aligned}
\sum_{j=\beta+1}^{\alpha+\beta}\frac{(-1)^\beta}{j-\beta}\cdot\binom{-\alpha-1}{j}\binom{\alpha+\beta}{j}
\equiv&
\sum_{j=p}^{\alpha+\beta}\frac{(-1)^\beta}{j-\beta}\cdot\binom{-\alpha-1}{j}\binom{\alpha+\beta}{j}\\
\equiv&
\sum_{j=p}^{\alpha+\beta}\frac{(-1)^\beta}{j-\beta}\cdot\binom{-\alpha-1}{j-p}\binom{\alpha+\beta-p}{j-p}\\
\equiv&
\sum_{j=0}^{\alpha-(p-\beta)}\frac{(-1)^\beta}{j+p-\beta}\cdot\binom{-\alpha-1}{j}\binom{\alpha-(p-\beta)}{j}\pmod{p}.\end{aligned}$$

\[l1\] Suppose that $\alpha,\beta$ are positive integers and $\alpha\geq\beta$. Then $$\begin{aligned}
\label{abj}
\sum_{j=0}^{\alpha-\beta}\frac{1}{j+\beta}\cdot\binom{-\alpha-1}{j}\binom{\alpha-\beta}{j}
=\sum_{j=0}^{\beta-1}\frac{(-1)^{1+\alpha}}{j+1+\alpha-\beta}\cdot\binom{-\alpha-1}{j}\binom{\beta-1}{j}.\end{aligned}$$

We shall use induction on $\alpha$ and $\beta$. When $\beta=1$, we have $$\begin{aligned}
\sum_{j=0}^{\alpha-1}\frac{1}{j+1}\cdot\binom{-\alpha-1}{j}\binom{\alpha-1}{j}=
-\frac1{\alpha}\sum_{j=0}^{\alpha-1}\cdot\binom{-\alpha}{j+1}\binom{\alpha-1}{\alpha-1-j}=
-\frac1{\alpha}\binom{-1}{\alpha}.\end{aligned}$$ So (\[abj\]) is true for $\beta=1$.

Assume that $\beta>1$ and (\[abj\]) holds for any smaller value of $\beta$. It is easy to check that $$\begin{aligned}
&\frac{1}{j+1+\alpha-\beta}\cdot\binom{-\alpha-1}{j}\binom{\beta-1}{j}\\
=&\frac{\alpha+\beta-1}{\alpha^2}\cdot\binom{-\alpha}{j}\binom{\beta-1}{j}+\frac{(\beta-1)^2}{\alpha^2}\cdot\frac1{j+1+\alpha-\beta}\cdot\binom{-\alpha}{j}\binom{\beta-2}{j}.\end{aligned}$$ It follows for the induction hypothesis that $$\begin{aligned}
\label{abb}
&\sum_{j=0}^{\beta-1}\frac{1}{j+1+\alpha-\beta}\cdot\binom{-\alpha-1}{j}\binom{\beta-1}{j}\notag\\
=&\binom{\beta-1-\alpha}{\beta-1}\cdot\frac{\alpha+\beta-1}{\alpha^2}+
\frac{(\beta-1)^2}{\alpha^2}\sum_{j=0}^{\beta-2}\frac{1}{j+1+\alpha-\beta}\cdot\binom{-\alpha}{j}\binom{\beta-2}{j}\\
=&\binom{\beta-1-\alpha}{\beta-1}\cdot\frac{\alpha+\beta-1}{\alpha^2}+
\frac{(\beta-1)^2}{\alpha^2}\sum_{j=0}^{\alpha-\beta}\frac{(-1)^\alpha}{j+\beta-1}\cdot\binom{-\alpha}{j}\binom{\alpha-\beta}{j}\notag\end{aligned}$$ Now letting $n=\alpha-\beta$, we only need to show that $$\begin{aligned}
\label{abn}
&\sum_{j=0}^{n}\binom{n}{j}\bigg(\frac{(\alpha-n-1)^2}{j+\alpha-n-1}\cdot\binom{-\alpha}{j}+\frac{\alpha^2}{j+\alpha-n}\cdot\binom{-\alpha-1}{j}\bigg)\notag\\
=&(-1)^{n}(2\alpha-n-1)\cdot\binom{\alpha-1}{n}.\end{aligned}$$ We shall use induction on $n$. There is nothing to do if $n=0$. And in view of (\[abb\]), we have $$\begin{aligned}
\sum_{j=0}^{n}\frac{1}{j+\alpha-n}\cdot\binom{n}{j}\binom{-\alpha-1}{j}
=\binom{n-\alpha}{n}\cdot\frac{\alpha+n}{\alpha^2}+
\frac{n^2}{\alpha^2}\sum_{j=0}^{n-1}\frac{1}{j+\alpha-n}\cdot\binom{-\alpha}{j}\binom{n-1}{j}.\end{aligned}$$ So $$\begin{aligned}
&\sum_{j=0}^{n}\binom{n}{j}\bigg(\frac{(\alpha-n-1)^2}{j+\alpha-n-1}\cdot\binom{-\alpha}{j}+\frac{\alpha^2}{j+\alpha-n}\cdot\binom{-\alpha-1}{j}\bigg)\\
=&\binom{n-\alpha}{n}\cdot(\alpha+n)+\binom{n-\alpha+1}{n}\cdot\frac{(\alpha+n-1)\cdot(\alpha-n-1)^2}{(\alpha-1)^2}\\
&+n^2\sum_{j=0}^{n-1}\frac{1}{j+\alpha-n}\cdot\binom{-\alpha}{j}\binom{n-1}{j}+
\frac{n^2(\alpha-n-1)^2}{(\alpha-1)^2}\sum_{j=0}^{n-1}\frac{1}{j+\alpha-n-1}\cdot\binom{1-\alpha}{j}\binom{n-1}{j}\\
=&\binom{n-\alpha}{n}\cdot(\alpha+n)+\binom{n-\alpha+1}{n}\cdot\frac{(\alpha+n-1)\cdot(\alpha-n-1)^2}{(\alpha-1)^2}\\
&+
\frac{n^2}{(\alpha-1)^2}\cdot (-1)^{n-1}(2\alpha-n-2)\cdot\binom{\alpha-2}{n-1}=(-1)^{n}(2\alpha-n-1)\cdot\binom{\alpha-1}{n}.\end{aligned}$$

With help of Lemma \[l1\], we get that $$\begin{aligned}
\sum_{j=1}^\alpha\frac{(-1)^\beta}{j}\cdot\binom{-\alpha-1}{j+\beta}\binom{\alpha+\beta}{j+\beta}
\equiv&\sum_{j=0}^{p-\beta-1}\frac{(-1)^{1+\alpha}}{j+1+\alpha+\beta-p}\cdot\binom{-\alpha-1}{j}\binom{p-\beta-1}{j}\\
\equiv&\sum_{j=1}^\alpha\frac{(-\alpha)_j(-\beta)_j}{j\cdot(j!)^2}\cdot\frac{(j-\alpha-\beta)_{\alpha-j}(1+\beta)_{\alpha-j}}{(j+1)_{\alpha-j}(-\alpha)_{\alpha-j}}\\
=&\sum_{j=1}^\alpha\frac{1}{j}\sum_{k=j}^{\alpha}(-1)^{k-j}\binom{\alpha}{k}\binom{-1-\alpha-\beta}{k}\binom{\beta}{k-j}
\pmod{p}.\end{aligned}$$ This is the desired result.

<span>9</span>

E. Mortenson, *Supercongruences between truncated ${}_2F_1$ by geometric functions and their Gaussian analogs*, Trans. Amer. Math. Soc., <span>**355**</span>(2003), 987-1007.

Z. W. Sun, *Supecongruences involving products of two binomial coefficients*, Finite Fields Appl., <span>**22**</span>(2013), 24-44.

Z. H. Sun, *Generalized Legendre polynomials and related congruences modulo $p^2$* , preprint, arXiv:1101.5386.

{'source': 'arXiv_out/1309'}