Prove if p is a prime number that does not divide a, then $a^{p^2}$ congruent to $a^p\mod p^2$

Question

I tried to do proof by contradiction, so I started to do use real arbitrary numbers. I decided I use p for prime still, but a being a multiple of p and by doing this, the statement is still true. I tried it with p=3, a =6 or p=2 and a =6. Anyways to solve it by contradiction or should I use another method

Answer 1

As by Fermat's little theorem $a^p\equiv a\pmod p$ for any integer $a$

$(a^p)^p\equiv( a)^p$

So, we don't need $p\nmid a$

Answer 2

You have

$$a^{p^2} - a^p =a^p(a^{p^2-p} -1) =$$ $$ a^p(a^{\phi(p^2)}-1)\equiv a^p(1-1) \pmod{p^2},$$

where Euler's theorem is used in the last step.

Answer 3

Recall $\ b\equiv a \pmod{\!m}\,\Rightarrow\,b^{\large m}\equiv a^{\large m} \pmod{\!m^{\large 2}}.\,$ Put $\,b=a^{\large p},\ m = p$