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Let $a,b$ be positive integers. Suppose $(a^n-1)\mid(b^n-1)$ $\forall n\in\mathbb{Z}^+$. Can we conclude that $b$ is a power of $a$?

Zerinf Zhang
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  • what has been tried ? –  Aug 26 '17 at 15:18
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    I can prove that a prime factor of $b$ must be a factor of $a$. So the statement holds for the case $a$ is a prime number. – Zerinf Zhang Aug 26 '17 at 15:20
  • In my opinion is quite obviously true also the inverse: if $b$ is a power of $a$ then $(a^n-1)\mid(b^n-1)$ – Raffaele Aug 26 '17 at 16:27
  • @Zerinf Zhang ; See each of the following:

    https://math.stackexchange.com/questions/1461203/if-an-1-divides-bn-1-too-often-then-b-ak

    https://math.stackexchange.com/questions/1919550/if-fracan-1bn-1-is-a-natural-number-for-every-n-then-a-bm?noredirect=1&lq=1

    https://math.stackexchange.com/questions/417340/do-there-exist-two-primes-pq-such-that-pn-1-mid-qn-1-for-infinitely-many/1466790#1466790

     – Davood Aug 26 '17 at 18:08
  • @Famke - I haven't looked very carefully, but at first sight it's not clear to me that this is a duplicate. – Pierre-Yves Gaillard Aug 26 '17 at 18:18
  • @ Pierre-Yves Gaillard ; as Elaqqad has been mentioned; in page 27 of the following there is an answer for this problem:
    http://www.emis.de/journals/JTNB/2005-1/pages423-435.pdf     – Davood Aug 26 '17 at 18:22
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    @Famke - Thanks! (I wasn't notified of your comment, perhaps because you put a space between @ and my name.) – Pierre-Yves Gaillard Aug 26 '17 at 19:18
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    @Pierre-Yves Gaillard ; yes, you are right. – Davood Aug 26 '17 at 19:20

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