# Solve a Diophantine equation with 2 variables and odd degree 5

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

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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. – awllower Jan 9 '13 at 18:48
I think it's just $a=\pm1.$ – Charles Jan 9 '13 at 19:39
See also; by the same author, that is me: math.stackexchange.com/questions/272454/… – user55514 Jan 9 '13 at 22:51
I think there are not integer solutions but the trivial (a,b) =(1,1) but could not prove it for now. – user55514 Jan 9 '13 at 22:58
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). – user55514 Jan 9 '13 at 23:05

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 $(a, b) = (1, 1)$.