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.
Answer
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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