# Thread: divisibility problem?

1. ## divisibility problem?

Let a, b be positive integers such that a^5 = b^7. Prove there exists some n in natural numbers such that a = n^7 and b = n^5.

Any help would be appreciated - I'm not really sure how to approach this question or even what it is asking. I suspect it relates to divisibility or prime factors but I don't really know how to solve it. Thanks.

2. Originally Posted by kimberu
Let a, b be positive integers such that a^5 = b^7. Prove there exists some n in natural numbers such that a = n^7 and b = n^5.

Any help would be appreciated - I'm not really sure how to approach this question or even what it is asking. I suspect it relates to divisibility or prime factors but I don't really know how to solve it. Thanks.

Hint: Let $p$ be a prime dividing $a$ , and suppose $n$ is its maximal power that divides a, i.e.: $p^n\mid a\,,\,\,p^{n+1}\nmid a$ ; of course, $p\mid b$ as well, of course, since it divides

the LHS and thus also the RHS, say $p^m\mid b\,,\,\,p^{m+1}\nmid b$ , but then:

$p^{5n}\mid a^5=b^7$ ,and also $p^{7m}\mid b^7=a^5$ $\Longrightarrow p^{5n}\mid p^{7m}\,\,\,and\,\,\,p^{7m}\mid p^5\Longrightarrow 5n=7m$ , so...

Tonio