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Prove that for any integer $n$, if $b^2$ divides $n$, then $b$ divides $n$.

Trying to figure out this proof. The proof I'm looking at is written as $n$ = any integer, if $25|n \implies 5|n$. I've been trying to figure this for days and have been running around in circles. Would appreciate a general proof for this.

$n$ = any integer, if $(b^2)|n \implies b|n$.

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    $\begingroup$ Hint: $b~\mid~ b^2$. $\endgroup$ – Christopher Carl Heckman Feb 10 '16 at 5:24
  • $\begingroup$ You could also try breaking $n,b$ down into their prime factorizations. $\endgroup$ – basket Feb 10 '16 at 5:26
  • $\begingroup$ Yes, but that's too much work. Use my hint, and a property of $\mid$. $\endgroup$ – Christopher Carl Heckman Feb 10 '16 at 5:27
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    $\begingroup$ $$b^2\mid n\implies n=b^2k=bbk=b\ell\implies b\mid n.$$ $\endgroup$ – Pixel Feb 10 '16 at 9:23
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Use the definition: If $a \mid b \Rightarrow$ there is an integer $k$ such that $b=ak$.

So if $b^2 \mid n \Rightarrow$ there is an integer $k$ such that $n=b^2k= bbk$. Now the product of two integers is also an integer, setting $c = bk$, what can you conclude?

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  • $\begingroup$ This helped so much! Thank you $\endgroup$ – Shan Solo Feb 10 '16 at 6:51
  • $\begingroup$ @ShanSolo no problem. $\endgroup$ – user13451345 Feb 10 '16 at 7:00
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Since an answer has been given ...

The $\mid$ relation is transitive; if $x \mid y$ and $y\mid z$, then $x\mid z$. Let $x=b$, $y=b^2$, and $z=n$. Done.

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