Jacobi polynomials and primality testing Can you provide a proof or a counterexample for the claim given below ?
Inspired by Agrawal's conjecture in this paper I have formulated the following claim :

Let $n$ be an odd natural number greater than one . Let $r$ be the smallest odd prime number such that $r \nmid n$ and $n^2 \not\equiv 1 \pmod r$ . Let $P_n^{(\alpha,\beta)}(x)$ be Jacobi polynomial such that $\alpha$ , $ \beta$ are natural numbers and $\alpha +\beta < n$  , then $n$ is a prime number if and only if $P_n^{(\alpha,\beta)}(x) \equiv x^n \pmod {x^r-1,n}$ .

You can run this test here .
I have tested this claim for many random values of $n$ , $\alpha$ and $\beta$ and there were no counterexamples .
Mathematica implementation of test :
(* n>a+b *)
n=139;
a=14;b=22;
r=3;
While[Mod[n,r]==0 || PowerMod[n,2,r]==1,r=NextPrime[r]];
If[PolynomialMod[PolynomialRemainder[JacobiP[n,a,b,x],x^r-1,x],n]-PolynomialRemainder[x^n,x^r-1,x]===0,Print["prime"],Print["composite"]];

 A: This is a partial answer.
This answer proves that if $n$ is a prime number, then $P_n^{(\alpha,\beta)}(x) \equiv x^n \pmod {x^r-1,n}$.
Proof : 
Assuming that $x$ is real and using the following expression
$$P_n^{(\alpha,\beta)}(x)=\sum_{s=0}^{n}\binom{n+\alpha}{n-s}\binom{n+\beta}{s}\left(\frac{x-1}{2}\right)^s\left(\frac{x+1}{2}\right)^{n-s}$$
we have
$$\begin{align}&2^n\left(P_n^{(\alpha,\beta)}(x)-x^n\right)
\\\\&=-2^nx^n+\sum_{s=0}^{n}\binom{n+\alpha}{n-s}\binom{n+\beta}{s}(x-1)^s(x+1)^{n-s}
\\\\&=-2^nx^n+\binom{n+\alpha}{n}(x+1)^{n}+\binom{n+\beta}{n}(x-1)^n
\\&\qquad \qquad+\sum_{s=1}^{n-1}\binom{n+\alpha}{n-s}\binom{n+\beta}{s}(x-1)^s(x+1)^{n-s}
\\\\&=-2^nx^n+\binom{n+\alpha}{n}\sum_{k=0}^{n}\binom{n}{k}x^{n-k}+\binom{n+\beta}{n}\sum_{k=0}^{n}\binom nk(-1)^{k}\cdot x^{n-k}
\\&\qquad \qquad+\sum_{s=1}^{n-1}\binom{n+\alpha}{n-s}\binom{n+\beta}{s}(x-1)^s(x+1)^{n-s}
\\\\&=\left(-2^n+\binom{n+\alpha}{n}+\binom{n+\beta}{n}\right)x^n+\left(\binom{n+\alpha}{n}-\binom{n+\beta}{n}\right)
\\&\qquad\qquad +\binom{n+\alpha}{n}\sum_{k=1}^{n-1}\binom{n}{k}x^{n-k}+\binom{n+\beta}{n}\sum_{k=1}^{n-1}\binom nk(-1)^{k}\cdot x^{n-k}
\\&\qquad \qquad+\sum_{s=1}^{n-1}\binom{n+\alpha}{n-s}\binom{n+\beta}{s}(x-1)^s(x+1)^{n-s}\end{align}$$
Here, we use the following facts :


*

*By Fermat's little theorem, $2^n\equiv 2\pmod n$. Also, since $(n+\alpha)(n+\alpha-1)\cdots (n+1)\equiv \alpha !\pmod n$, we get $\binom{n+\alpha}{n}=\binom{n+\alpha}{\alpha}=\frac{(n+\alpha)(n+\alpha-1)\cdots (n+1)}{\alpha !}\equiv 1\pmod n$. Similarly, we get $\binom{n+\beta}{n}\equiv 1\pmod n$. Therefore, we have $-2^n+\binom{n+\alpha}{n}+\binom{n+\beta}{n}\equiv -2+1+1\equiv 0\pmod n$.

*Since $\binom{n+\alpha}{n}\equiv 1\pmod n$ and $\binom{n+\beta}{n}\equiv 1\pmod n$, we get $\binom{n+\alpha}{n}-\binom{n+\beta}{n}\equiv 1-1\equiv 0\pmod n$.

*$\binom{n}{k}\equiv 0\pmod n$ for each $k$ such that $1\le k\le n-1$.

*Suppose that $\alpha\ge n-s$ and $\beta\ge s$. Then, we get $\alpha+\beta\ge n$ which contradicts $\alpha+\beta\lt n$. So, we have $\alpha\lt n-s$ or $\beta\lt s$. If $\alpha\lt n-s$ with $1\le s\le n-1$, then the numerator of $\left(\binom{n+\alpha}{n-s}=\right)\frac{(n+\alpha)(n+\alpha-1)\cdots (\alpha+s+1)}{(n-s)!}$ is divisible by $n$, but the denominator isn't. So, $\binom{n+\alpha}{n-s}$ is divisible by $n$.Similarly, if $\beta\lt s$, then $\binom{n+\beta}{s}$ is divisible by $n$. As a result, we see that, for each $s$ such that $1\le s\le n-1$, $\binom{n+\alpha}{n-s}\binom{n+\beta}{s}$ is divisible by $n$.
Therefore, we see that there is a polynomial $f$ with integer coefficients such that 
$$2^n\left(P_n^{(\alpha,\beta)}(x)-x^n\right)=nf$$
from which
$$P_n^{(\alpha,\beta)}(x)=x^n+(x^r-1)\times 0+n\times \frac{1}{2^n}f,$$
follows.
It follows from this and $\gcd(n,2^n)=1$ that
$$P_n^{(\alpha,\beta)}(x) \equiv x^n \pmod {x^r-1,n}$$
