# Prove that for all integers $a$ and $b$, if $a$ divides $b$, then $a^{2}$ divides $b^{2}$.

I just need to know that if $a$ divides $b$, where $a$ and $b$ are integers, does $a^{2}$ divide $b^{2}$?

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If $a$ divides $b$, then $b=ka$ for some integer $k$, so $b^2=k^2a^2$ where $k^2$ is an integer.

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How do you prove that k^2 is an integer? Is that by closure? – amster27 Feb 22 '13 at 1:07
@amster27: If you are concerned about whether $k^{2}$ is an integer, then you should be equally concerned about whether both $a^{2}$ and $b^{2}$ are integers. :) – Haskell Curry Feb 22 '13 at 1:23
Thanks, I'm just starting to write proofs in a Mathematical Reasoning class, and I just feel so lost. But these explanations helped. – amster27 Feb 22 '13 at 1:28
@amster: Note that $+_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ and $\times_{\mathbb{Z}}: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. Therefore, adding two integers or multiplying two integers yields an integer. – Haskell Curry Feb 22 '13 at 1:44

Hint $\rm\ \ a\mid b\ \Rightarrow\ \dfrac{b}a\in \Bbb Z\ \Rightarrow\ \dfrac{b^2}{a^2} = \left(\dfrac{b}a\right)^2\!\in \Bbb Z^2\subseteq \Bbb Z\:\Rightarrow\: a^2\mid b^2\$

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That's closer to a proof than a hint. – 1015 Feb 22 '13 at 1:29
The argument is elegant, but I think that the OP wants to stay within $\mathbb{Z}$ and does not wish to jump into $\mathbb{Q}$, as he seems to be very interested in the axioms governing the properties of $\mathbb{Z}$. – Haskell Curry Feb 22 '13 at 1:29
@Haskell Possibly, but there was no hint of any such constraint in the question. – Math Gems Feb 22 '13 at 1:46