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See also the already solved question:

Solve a Diophantine equation with 2 variables and odd degree 5

Prove that there are no non trivial integer solutions to the equation $a^{5} -1 = 2b^{5}$

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why do you care?? – jspecter Jan 17 at 20:18
jspecter said : "why do you care??" I care because i want to understand. If nobody cared, what would happen ? we have to care for better. – user55514 Jan 23 at 7:45

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Let $a$ and $b$ satisfy $a^5 - 1 = 2b^5.$ Then $A = -a$ and $B = -b$ satisfies $A^5 + 1 = 2B^5.$ So by your previous question...

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Thank you jspecter. I had not been able to see that simple fact because i did not consider going from one equation to the other. If (a,b) is a solution of a^5 +1 = 2*b^5 then (-a,-b) is a solution of a^5 -1 = 2*b^5. But i do not see in such a clear way that if one has only trivial solutions, the other one must also have only the (-a,-b) trivial solutions and no more. – user55514 Jan 17 at 22:18
I have edited my answer to the original question as I had overlooked that Denes' Conjecture relates only to positive integers. It seems therefore to remain possible that there are non-trivial solutions to $a^5 + 1 = 2b^5$ with a, b negative. I am not sure therefore that the above answer resolves $a^5 - 1 = 2b^5$, although it is a correct inference if it is assumed that $a^5 + 1 = 2b^5$ has no non-trivial solutions in positive or negative integers. – Adam Bailey Jan 18 at 9:41
Thanks again; Adam. Personally i believe we need deeper mathematical considerations to prove this equation has no non trivial solutions. If (a,b) is a solution of a^5 +1 = 2*b^5 then (-a,-b) is a solution of a^5 -1 = 2*b^5. But we cannot deduce from this fact, all solutions will be negative. And we also cannot deduce that if the first equation has no non trivial solutions, then the second will also have no solutions, if we do not have a theorem (or else can prove it) that clearly says so, applied to this case. – user55514 Jan 18 at 11:00
I would like to recall that though equations a^5 +1 = 2*b^5 and a^5 -1 = 2*b^5 look very similar the first one belongs to the general equation x^n +y^n = 2*z^n which was proven to have no solutions for any n (it suffices to prove it for any n = p, prime) by first P.Denés for p<= 29; then by k. Ribet for all p = 1 mod 4 and finally by H.Darmon and L.Merell for all p = 3 mod 4 using the same kind of methods used by Wiles to prove the Fermat "not-enough-margin-in-the-book-to-write-it" statement. But the case x^n -y^n = 2*z^n to which belongs the equation of this question is different. – user55514 Jan 19 at 16:33
Here is the link to K. Ribet paper;many thanks to Adam Bailey who let me know it. math.berkeley.edu/~ribet/Articles/icfs.pdf – user55514 Jan 19 at 16:38

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