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If $a$ and $b $ are two integers, and $a \mid b$, then gcd$(a^2,b^2)$ =

I think the answer is $a^2$. Is it correct?

user109886
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2 Answers2

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Certainly $a^2\mid b^2$, and $n \nmid a^2$ for $n>a^2$.

Ian Coley
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If $a\mid b$, then $b = ka$ for some integer $k$.

Then we aim to find $\gcd(a^2, (ka)^2)=\gcd(a^2, k^2a^2)$. So, yes, $a^2$ is the greatest common divisor.

apnorton
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