# If $p\mid a^2$ then $p^2\mid a^2$. Show that it is true or false.

For natural numbers $$a$$ and $$p$$ with $$p$$ prime, if $$p$$ divides $$a^{2}$$ then $$p^{2}$$ also divides $$a^{2}$$.

My understanding to the is if $$p\mid a$$, then $$\gcd(a,p)$$ should not be equal to $$1$$ and so as $$\gcd(a^2,p)$$ and I do not know about $$p^2\mid a^2$$.

• Do you know the properties of a prime number? – Mhenni Benghorbal Sep 9 '14 at 5:57
• canonical representation!!!!!!!!!! – Bumblebee Sep 9 '14 at 5:59
• Is it group theory? – Bumblebee Sep 9 '14 at 6:00
• @Nilan, I was just thinking the same. It's number theory of course, but possibly the OP needs it for an application to group theory. – David Sep 9 '14 at 6:04

Hint. A fundamental property of primes is: if $p\mid ab$ then $p\mid a$ or $p\mid b$ (or both).

What do you get if you apply this to the statement $p\mid a^2$?

Can you then finish the problem?

• @Lynnie You are asked by David to apply the property to the statement. Not to repeat the statement :). – drhab Sep 9 '14 at 6:02
• But $p\mid a^2$ is given, it doesn't help just to write it down twice ;-) [Thanks @drhab for making that comment just as I was writing!] Can you use the property I quoted in my first line to find some more specific information? – David Sep 9 '14 at 6:03
• Well note if b|a and c|a then cb|a^2, im pretty sure – Kamster Sep 9 '14 at 6:05
• @user159813 That's also important. Lynnie, can you now put all the pieces together? – David Sep 9 '14 at 6:07
• Yes thanks ya'll it clear now. – Lynnie Sep 9 '14 at 6:09

If you know the Fundamental Theorem of Arithmetic, this is very simple. Write $$a=\prod_{i=1}^np_i^{\alpha_i},$$ for primes $$p_i$$ and positive integers $$\alpha_i$$, so that $$a^2=\prod_{i=1}^np_i^{2\alpha_i}.$$ Since $$p\mid a^2$$, this implies that there exists $$k$$ with $$p_k=p$$. But then, $$a^2$$ has a factor of $$p^{2\alpha_k}\geq p^2$$, and we're done. $$\blacksquare$$