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The problem is to find all positive integers $a$ and $b$ such that $a^2(2^a-a^3)+1=7^b$.

I found a=10, and my intuition tells me there are no more solutions. I've also shown that $a=42k+10$ for some nonnegative integer $k$, but I can't prove anymore than this. (It could help to know that it's from the problems section of a book, so it should have a fairly nice solution.)

Thanks for your help!

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  • $\begingroup$ I see $b$ has to be a multiple of $4$. Other than that, no idea. $\endgroup$ – Mike Feb 22 '14 at 20:34
  • $\begingroup$ And a must be even. $\endgroup$ – eyphka Feb 22 '14 at 22:01
  • $\begingroup$ For any positive integer a, LHS is negative and with a positive b, RHS is positive and I don't see how you could solve this?Is the equation by any chance, $(a^{3}-2^{a})$? $\endgroup$ – Satish Ramanathan Feb 22 '14 at 22:31
  • $\begingroup$ @Arkan Could you please tell us what book this is from? Thanks! $\endgroup$ – Matthew Conroy Feb 23 '14 at 4:41
  • $\begingroup$ @satishramanathan $2^a>a^3$ for all $a\geq10$. $\endgroup$ – user11977 Feb 23 '14 at 8:29
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The OP claims to have solved this some time ago (see comment under main question). I'm just adding this answer so that this thread doesn't show up in a "No Answers" search/filter.

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