# divisibility problem?

• Feb 23rd 2010, 09:40 PM
kimberu
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.
• Feb 24th 2010, 04:18 AM
tonio
Quote:

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

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

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

Tonio