# Prime number problem

• Feb 12th 2009, 06:36 AM
slevvio
Prime number problem
Let a be a non zero integer and p prime.

Show p| $a^{2}$ => p|a.

This is deeply perplexing me as all I can work out is gcd(p, $a^2$) = p. Because p is prime, p has only 1 and p as its divisors. p divides a^2 so p is a divisor of a^2. p>1 so gcd(p,a^2) is p. If it was equal to 1 this would make my job easier but I can't work this out.

any help would be appeciated :)
• Feb 12th 2009, 07:34 AM
Laurent
Quote:

Originally Posted by slevvio
Let a be a non zero integer and p prime.

Show p| $a^{2}$ => p|a.

This is deeply perplexing me as all I can work out is gcd(p, $a^2$) = p. Because p is prime, p has only 1 and p as its divisors. p divides a^2 so p is a divisor of a^2. p>1 so gcd(p,a^2) is p. If it was equal to 1 this would make my job easier but I can't work this out.

any help would be appeciated :)

Suppose $p|a^2$ and $p$ is prime. If $p\not|a$ then, since $p$ is prime, $\gcd(a,p)=1$. By Gauss' theorem, we deduce from $p|a\cdot a$ and $\gcd(a,p)=1$ that $p|a$, which is a contradiction.