Suppose $n\in\mathbb{Z}$. If $3 \nmid n^2$, then $3 \nmid n$.

Question

• Suppose $n\in\mathbb{Z}$. If $3 \nmid n^2$, then $3 \nmid n$.

Assume $3|n$, that is $n=3m$ for some $m\in\mathbb{Z}$, so $n^2=(3m)^2=9m^2=3(3m^2)$. Let $3m^2=x$, then we have $n^2=(3m)^2=9m^2=3(3m^2)=3x$, it means that $3|n^2$, so we are done.

Can you check my proof? Thankss...

Answer 1

Also, for any prime $p$ and positive integer $a$, if $p^a | n$ then $n = p^am$ so $n^2 = p^{2a}m^2 = p^a(p^{a}m^2) $ so $p^{a} | n^2$.

Therefore, if $p^{a} \not\mid n^2$ then $p^{a} \not\mid n$.

Also see this question of mine for a further generalization:

If $n \mid a^2 $, what is the largest $m$ for which $m \mid a$?