If a number x has several divisors (integers), then if a, b and c are three prime divisors, (a^m)(b^n)(c^p)=x, where m, n and p are integers>0. When x=1, a=1, b=-1, can we find c≠a,b?
(1^m)((-1)^n)(c^p)=1, 1^m=1, so we are left with ((-1)^n)(c^p)=1. If n is even, c^p=1; if n is odd c^p=-1. But we have seen that (-1)^n=-1 when n is odd and (-1)^n=1 when n is even, so c must be either 1 or -1, that is, c=a or c=b, so c, as a new divisor, cannot exist. That leaves 1 and -1 as the only divisors of 1.