Hypergeometric polynomials over finite fields (Q1292771)

Let \(p\geq 5\) be a prime. For elements \(a,b,c\) of the finite field \(\mathbb{F}_p\) the hypergeometric polynomial \(F(a,b,c;x)\) is defined by \[ F(a,b,c;x)=\sum_{n=0} \frac{(a)_n(b)_n}{(1)_n(c)_n}x^n, \] where \((a)_n=a(a+1)\cdots (a+n-1)\) and the sum stops as soon as the numerator vanishes under the assumption that the denominator does not vanish before the numerator vanishes. Moreover, for \(\varepsilon, \varepsilon'\in \{-1,1\}\) let \[ F_{\varepsilon,\varepsilon'}(x)=\prod_a(x-a), \] where \(a\) runs over all elements \(a\in \mathbb{F}_p^*\) satisfying \(\left(\frac{a}{p}\right)=\varepsilon\) and \((\frac{1-a}{p})=\varepsilon',\) where \((\frac{\cdot}{p})\) denotes the Legendre symbol. \textit{T. Honda} [Algebraic geometry, Roma 1979, Symp. Math. 24, 169-204 (1981; Zbl 0464.12013)] evaluated \(F_{-1,\pm 1}(x)\) in terms of hypergeometric polynomials. In the paper under review the author evaluates the remaining polynomials \(F_{1,\pm 1}(x)\): \[ F_{1,-1}(x)=a_{1,-1}^{(p)}F\left(\frac{1}{4},\frac{3}{4},\frac{1}{2};x\right)\text{ and } F_{1,1}(x)=a_{1,1}^{(p)}F\left(\frac{3}{4},\frac{5}{4},\frac{3}{2};x\right), \] where \[ a_{1,-1}^{(p)}=\begin{cases} 1 &\text{if }p\equiv 1\bmod 4,\\ -2&\text{if }p\equiv 3\bmod 4,\end{cases}\quad\text{ and }\quad a_{1,1}^{(p)}=\begin{cases} 1 &\text{if }p\equiv 1\bmod 4,\\ -1/2&\text{if }p\equiv 3\bmod 4.\end{cases}. \]