Quote:
Originally Posted by carpetpool
Bottom line, factor a^(p1)1 into irreducible polynomials I, and one of these polynomials when evaluated at a will be a multiple of p, if a is not 0 (mod p).

For any prime number \(p\), in the ring of polynomials with coefficients in the integers modulo \(p\) we have
\[ X^{p1}1=(X1)(X2)(X3)\ldots (X(p1)). \]