<s>[INST] If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime If $a$ and $b$ are coprime, prove that $a^2+b^2$ and $a^2b^2$ are coprime.
Answer:
We have to prove $\gcd(a^2+b^2,a^2b^2)=1$, $\gcd(a,b)=1$. Then $\gcd(a^2,b^2)=1$. Now we have $\gcd(a^2,a^2+b^2)=1$ and $\gcd(b^2,a^2+b^2)=1$ by $\gcd(a,b)=\gcd(p\pm q)$.
Then I am lost.
 [/INST] Let $p$ be a common prime factor of $a^2 + b^2$ and $a^2 b^2$, and recall that for any integer $n$, $p | n \iff p | n^2$. Thus $p | a^2 b^2$ implies $p | ab$, so $p$ must divide both of $a^2 + b^2 \pm 2ab = (a\pm b)^2$; i.e., both $a+b$ and $a-b$ must be multiples of $p$. Thus, both $a$ and $b$ are multiples of $p/2$. If $p$ is odd, then $a$ and $b$ have $p$ as a common factor, so we're done. If $p = 2$, then $p | a^2 b^2$ forces at least one of $a$ and $b$ to be even; as $a^2 + b^2$ is even, both $a$ and $b$ must be even. QED.
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