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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!

Singhal
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Is Ne
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  • I see $b$ has to be a multiple of $4$. Other than that, no idea. – Mike Feb 22 '14 at 20:34
  • And a must be even. – eyphka Feb 22 '14 at 22:01
  • 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, $(a^{3}-2^{a})$? – Satish Ramanathan Feb 22 '14 at 22:31
  • @Arkan Could you please tell us what book this is from? Thanks! – Matthew Conroy Feb 23 '14 at 04:41
  • @satishramanathan $2^a>a^3$ for all $a\geq10$. – user11977 Feb 23 '14 at 08:29
  • More or less similar to the [Ramanujan-Nagell equation](http://en.wikipedia.org/wiki/Ramanujan-Nagell_equation). – Lucian Feb 23 '14 at 11:40
  • Which theory or context surrounded the exercise? If it's in a chapter on, say, Pell's equations or any other specific subject, could you please mention that? It would help searching in the right direction. – Bart Michels Mar 03 '14 at 19:55
  • @Arkan Megraoui Can you tell me the title of that book? – lsr314 Apr 13 '14 at 14:37
  • Well, we have $$ a^2(2^a-a^3)=7^b-1=6(7^{b-1}+7^{b-2}+\dotsb+1), $$ and hence $2 \mid a$, say $a=2c$ for an integer $c \ge 1$. Now \begin{align} (2c)^2(2^{2c}-(2c)^3) &= 6(7^{b-1}+7^{b-2}+\dotsb+1) \\ 2c^2(4^c-8c^3) &= 3(7^{b-1}+7^{b-2}+\dotsb+1). \end{align} First assume $3 \mid c$, say $c=3d$ for an integer $d \ge 1$. Then \begin{align} 2(3d)^2(4^{3d}-8(3d)^3) &= 3(7^{b-1}+7^{b-2}+\dotsb+1) \\ 6d^2(4^{3d}-8(3d)^3) &= 7^{b-1}+7^{b-2}+\dotsb+1. \end{align} After working that case out, assume $3 \nmid c$, and work it out. Does that help? – Kieren MacMillan Aug 19 '14 at 00:34
  • Ah, okay... Maybe answer your own question, then, so that it doesn't show up in the "Unanswered" list? – Kieren MacMillan Aug 19 '14 at 13:18

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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.

Kieren MacMillan
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