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I'm trying to solve this problem. I should be able to do it using simple divisibility properties but I don't know how.

Let a and b be integers such that they are coprime. Prove that $\gcd(a^2b^3,a+b)=1$

For instance... I thought that the gcd divides both $a^2b^3$ and $a+b$ so it must divide a sum of them. I've tried going this way but it's not clear to me where it should lead me. Any hint will be welcomed. Thanks.

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    $\begingroup$ Hint: suppose otherwise and let $p$ be a prime dividing the gcd. Show that $p$ divides either $a$ or $b$ and deduce that it divides both. $\endgroup$ – lulu Dec 5 '16 at 13:58
  • $\begingroup$ This question is probably not a duplicate, but it feels like a duplicate. $\endgroup$ – Bob Happ Dec 5 '16 at 22:04
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Suppose that $p$ is a prime number such that $p|a^2b^3$ then $p|a$ or $p|b$. Let's say $p|a$. If $p|(a+b)$ then we should have $p|b$ what is impossible because $a,b$ are coprimes.

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Primes that divide $a^2b^3$ are the primes that divide $a$ and $b$, put together. None of them divide $a+b$, since $a$ and $b$ are coprime.

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By Euclid $\ (a,a\!+\!b) = \overbrace{(a,b)}^{1} = (b,a\!+\!b)\ $. Since $\,a,b\,$ are coprime to $\,a\!+\!b\,$ so is their product $\,a^j b^k,$ by Euclid's Lemma.

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