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

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

-
Welcome to MSE! Is this homework? If so, please tag it as such. What have you tried on your own and where are you confused? Can you please add those details so the MSE Community can provide answers? Thanks and regards! –  Amzoti Jan 9 '13 at 18:33
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

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