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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 $p^2\mid a^2$.

Please need clarification.Thank you for your help.

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  • $\begingroup$ Do you know the properties of a prime number? $\endgroup$ – Mhenni Benghorbal Sep 9 '14 at 5:57
  • $\begingroup$ canonical representation!!!!!!!!!! $\endgroup$ – Bumblebee Sep 9 '14 at 5:59
  • $\begingroup$ Is it group theory? $\endgroup$ – Bumblebee Sep 9 '14 at 6:00
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    $\begingroup$ @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. $\endgroup$ – David Sep 9 '14 at 6:04
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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?

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  • $\begingroup$ @Lynnie You are asked by David to apply the property to the statement. Not to repeat the statement :). $\endgroup$ – drhab Sep 9 '14 at 6:02
  • $\begingroup$ 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? $\endgroup$ – David Sep 9 '14 at 6:03
  • $\begingroup$ Well note if b|a and c|a then cb|a^2, im pretty sure $\endgroup$ – Kamster Sep 9 '14 at 6:05
  • $\begingroup$ @user159813 That's also important. Lynnie, can you now put all the pieces together? $\endgroup$ – David Sep 9 '14 at 6:07
  • $\begingroup$ Yes thanks ya'll it clear now. $\endgroup$ – Lynnie Sep 9 '14 at 6:09
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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. $\blacksquare$

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