# Thread:

1. ## 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

2. 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.
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