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Math Help - divisibility problem?

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by kimberu View Post
    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
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