Prove that for all $a\in \mathbb{Z}$ and all primes $p$, $p^2$ does not divide $a^2-p$

Question

What would be a method to start, or some can prove useful theorem for this problem

Prove that for all $a\in \mathbb{Z}$ and all primes $p$, $p^2$ does not divide $a^2-p$

Answer 1

$p^{2}\mid a^{2}-p\Rightarrow p\mid a^{2}-p\Rightarrow p\mid a^{2}\Rightarrow p\mid a$. The last implication because $p$ is prime.

So $a=kp$ resulting in:

$p^{2}\mid k^{2}p^{2}-p\Rightarrow p\mid k^{2}p-1\Rightarrow p\mid1$.

A contradiction is found.

Answer 2

Suppose $p^2 \mid a^2 - p$. Then $p \mid a^2 - p$, thus $a^2$ is divisible by $p$. This means that $a$ is divisible by $p$ (since $p$ is prime), so $p^2 \mid a^2$, yielding $p^2 \mid (a^2-p) - a^2 = -p$. This contradiction shows that $a^2-p$ is not divisible by $p$.

Answer 3

Consider two cases. When a is divisible by p and when it's not. When it's not, a^2 and p are coprime. When it is, expand it as q*something and proceed.