Proof for divisions that in include prime number.

Question

How do I prove, that if $m^2$ can be divided $p$ (where $m$ is a whole number and $p$ is a prime number) then also m can be divided by $p$?

Answer 1

Suppose not, then $\gcd(m,p)=1$, so $am+bp=1$ for suitable integers $a,b$. But then $1^2=(am+bp)^2=a^2m^2+2abmp+b^2p^2$ is a multiple of $p$, contradiction.

Answer 2

Hint: $$m=p_1\cdot p_2\cdot p_3\cdots p_n$$

$$m^2=p_1^2\cdot p_2^2\cdot p_3^2\cdots p_n^2$$

Where $p_i$ is prime

If $$m^2\equiv 0 \mod p$$ Then $$p=p_i$$ where $ 1\le i \le n $

I hope you can take it from here.