# Compositeness testing using Jacobi polynomials

Mathematics Asked by Peđa Terzić on July 30, 2020

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 $$alpha + beta , then $$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 .

We use the identity $$P_n^{(alpha,beta)}(x)=sum_{s=0}^n binom{n+alpha}{n-s}binom{n+beta}{s}left(frac{x-1}2right)^sleft(frac{x+1}2right)^{n-s},$$ and assume that $$pgeq 3$$. 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 $$sum_{s=1}^{p-1} binom{p+alpha}{p-s}binom{p+beta}{s}left(frac{x-1}2right)^sleft(frac{x+1}2right)^{p-s}equiv 0bmod p$$ for all integer $$x$$. We claim, in fact, that each of these terms are multiples of $$p$$. By Lucas's theorem, for $$0, $$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, either $$alpha or $$beta, so one of these is $$0bmod p$$.

Answered by Carl Schildkraut on July 30, 2020

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