2
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ 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
    Commented Dec 5, 2016 at 13:58
  • $\begingroup$ This question is probably not a duplicate, but it feels like a duplicate. $\endgroup$
    – Bob Happ
    Commented Dec 5, 2016 at 22:04

3 Answers 3

Reset to default
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .