Thread: Covering sets for a^n-1 View Single Post
2017-12-28, 12:48   #2
Nick

Dec 2012
The Netherlands

32008 Posts

Quote:
 Originally Posted by carpetpool Bottom line, factor a^(p-1)-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
$X^{p-1}-1=(X-1)(X-2)(X-3)\ldots (X-(p-1)).$