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Solve in integers the equation: $a^{5} +1 = 2b^{5}$

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  • $\begingroup$ Is it asked to find all? At least we have already a rational solution:1+1=2. But, to find all integer solutions is a little harder...And I will give it a try. $\endgroup$
    – awllower
    Jan 9, 2013 at 18:48
  • $\begingroup$ I think it's just $a=\pm1.$ $\endgroup$
    – Charles
    Jan 9, 2013 at 19:39
  • $\begingroup$ See also; by the same author, that is me: math.stackexchange.com/questions/272454/… $\endgroup$
    – user55514
    Jan 9, 2013 at 22:51
  • $\begingroup$ I think there are not integer solutions but the trivial (a,b) =(1,1) but could not prove it for now. $\endgroup$
    – user55514
    Jan 9, 2013 at 22:58
  • $\begingroup$ In fact i had not time yet to try to solve this particular one. I have been working in showing there are no solutions but the trivial one to the general Diophantine a^(2n) +1 = 2b^(2n) (even exponents). $\endgroup$
    – user55514
    Jan 9, 2013 at 23:05

1 Answer 1

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There are no non-trivial solutions in positive integers to $a^p + 1 = 2b^p$ for any odd prime p.

This is a consequence of Denes Conjecture which states that if p is an odd prime and three natural non-zero integers $x^p, y^p$ and $z^p$ lie in an arithmetic progression, then x = y = z. Although referred to as a conjecture, this has been proved by Darmon & Merel. For further information and sources see section 9 and the references in this article by Ribet. Also useful is the English translation of item 13 in Ribet's references: Hellegouarch Y,, Invitation to the Mathematics of Fermat-Wiles, English Translation 2002, Academic Press, pp 342-3.

Clearly if $a^p + 1 = 2b^p$, then $1(= 1^p), b^p$ and $a^p$ are in arithmetic progression. Hence Denes Conjecture implies that $a = b = 1$, and so the only positive solution is $(a, b) = (1, 1)$.

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  • $\begingroup$ Thank you, Adam. This is a nice general result and proves also r^3 + s^3 = 2*t^3 has no non trivial solutions (i have just read the Ribet paper) which was proven by conventional and elemental means by Euler. I am just wondering now two things: 1) Would it be very difficult to extend the proof to r^(2*n+1) + s^(2*n+1) = t^(2*n+1) has no non trivial solutions ? 2) How to prove a^5 -1 = 2*b^5 has no non integral solutions ? $\endgroup$
    – user55514
    Jan 17, 2013 at 18:09
  • $\begingroup$ I read the easy-to-read Ribet paper just for the statement that there was indeed a proof; not for the proof itself i would not have been able to understand. Now how to prove a generalization of the equation, that a^(2*n+1) +1= 2*b^(2*n+1) has no non trivial solutions for any positive whole n ? $\endgroup$
    – user55514
    Jan 17, 2013 at 18:21
  • $\begingroup$ The equation in the fist comment is : r^(2*n+1) + s^(2*n+1) = 2*t^(2*n+1) has no non trivial solutions $\endgroup$
    – user55514
    Jan 17, 2013 at 18:23
  • $\begingroup$ For a classical and elemental proof by Euler that x^3 +- y^3 = 2*z^3 (the case p=3) has no non trivial integer solutions: books.google.es/… $\endgroup$
    – user55514
    Jan 17, 2013 at 23:01

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