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Can you prove or disprove the following claim:

Let $P_n^{(\alpha,\beta)}(x)$ be Jacobi polynomial . If $p$ is a prime number such that $\alpha , \beta$ are natural numbers and $\alpha + \beta <p$, then $$P_p^{(\alpha,\beta)}(a) \equiv a \pmod{p}$$ for all odd integers $a$ greater than one .

You can run this test here. I have tested this claim for many random values of $p$ , $\alpha$ and $\beta$ and there were no counterexamples .

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    $\begingroup$ Are you including $0$ in the natural numbers? $\endgroup$ – Carl Schildkraut Jul 27 '20 at 4:53
  • $\begingroup$ @CarlSchildkraut Yes. $\endgroup$ – Peđa Terzić Jul 27 '20 at 5:20
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We use the identity $$P_n^{(\alpha,\beta)}(x)=\sum_{s=0}^n \binom{n+\alpha}{n-s}\binom{n+\beta}{s}\left(\frac{x-1}2\right)^s\left(\frac{x+1}2\right)^{n-s},$$ and assume that $p\geq 3$. For $s=0$, the sum is simply $(x+1)/2$ modulo $p$ (by Fermat's little theorem), and for $s=p$, the sum is $(x-1)/2$, so those two terms sum to $x$. It suffices to show that $$\sum_{s=1}^{p-1} \binom{p+\alpha}{p-s}\binom{p+\beta}{s}\left(\frac{x-1}2\right)^s\left(\frac{x+1}2\right)^{p-s}\equiv 0\bmod p$$ for all integer $x$. We claim, in fact, that each of these terms are multiples of $p$. By Lucas's theorem, for $0<s<p$, $$\binom{p+\alpha}{p-s}\equiv \binom{1}{0}\binom{\alpha}{p-s}\bmod p,$$ and $$\binom{p+\beta}{s}\equiv \binom{1}{0}\binom{\beta}{s}\bmod p.$$ However, since $\alpha+\beta<s+(p-s)$, either $\alpha<p-s$ or $\beta<s$, so one of these is $0\bmod p$.

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