1. ## Prime number problem

Let a be a non zero integer and p prime.

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

This is deeply perplexing me as all I can work out is gcd(p,$\displaystyle a^2$) = 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|$\displaystyle a^{2}$ => p|a.

This is deeply perplexing me as all I can work out is gcd(p,$\displaystyle a^2$) = 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 $\displaystyle p|a^2$ and $\displaystyle p$ is prime. If $\displaystyle p\not|a$ then, since $\displaystyle p$ is prime, $\displaystyle \gcd(a,p)=1$. By Gauss' theorem, we deduce from $\displaystyle p|a\cdot a$ and $\displaystyle \gcd(a,p)=1$ that $\displaystyle p|a$, which is a contradiction.