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  • 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...

JMoravitz
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    Yes, it is fine. – Mark Mar 04 '19 at 21:19
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    This is correct and a good example of a proof by contrapositive. I might have used fewer words to do it, but it is important to write the proof with your audience in mind. In your case, you would be writing the proof for a teacher who is teaching you the fundamentals of proof writing and this is one of your early examples for proofs with divisibility so the extra steps are worthwhile to show explicitly. – JMoravitz Mar 04 '19 at 21:20
  • It might be worthwhile to note that you are are proving the (equivalent) contra-positive statement. – Thomas Andrews Mar 04 '19 at 21:22
  • Thanks for comments... –  Mar 04 '19 at 21:25
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    Correct, however it could be argued for the sake of rigour that you should finally assume that 3 doesn't divide $n^2$ and assume that 3 does indeed divide n, leading to a contradiction. – Matthew Mar 04 '19 at 21:25

1 Answers1

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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 $m \mid a$?

marty cohen
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  • Primality plays no role in this trivial direction of that inference, i.e. $,d\mid n,\Rightarrow, d\mid n^2,$ is true for all $d.$ $\ \ $ – Bill Dubuque Mar 04 '19 at 21:47