All natural solutions of $2x^2-1=y^{15}$ How can I find all positive integers $x$ and $y$ such that $2x^2-1=y^{15}$?
PS. See here.
 A: I was working on Byron's suggestion before he made it but it took me a while because I'm not a real number theorist.  And, I used Sage.  I consider $2y^2 = x^3 + 1$ and I want to put this in a form where the coefficient of $y^2$ and $x^3$ are 1 so I multiply both sides by 1/2 first and then use the transform $(x,y) \mapsto (X/2,Y/4)$ to get $Y^2 = X^3 + 8$.  Now, Sage can find all integral points and they are $(-2, 0)$, $(1, \pm3)$, $(2, \pm4)$, and $(46, \pm312)$.  This is good enough because note that any integer solution to the original $2y^2 = x^3 + 1$ will be mapped to an integer solution here.  So, now you just need to consider these 7 solutions.  Our map is clearly invertible by $(X, Y) \mapsto (2x, 4y)$, so we have a one-to-one correspondence between points and everything makes sense.  All we need to do is map those 7 points backward to see what they were on $2y^2 = x^3 + 1$.  They were $(-1, 0)$, $(1/2, \pm 3/4)$, $(1, \pm1)$, and $(23, \pm78)$.  Thus, we have all integer solutions to $2y^2 = x^3 + 1$, 3 are obvious and the other two, $(23, \pm78)$ can not correspond to solutions of $2y^2 = x^{15} + 1$ as Byron explained.
By the way, the elliptic curve has rank 1 which means it has an infinite number of solutions over the rationals.  But, whether or not any of these correspond to nontrivial solutions of the original $2y^2 = x^{15} + 1$, I have no idea.
A: There are no solutions other then $(x,y)=(1,1)$
See http://rmmc.asu.edu/abstracts/rmj/vol31-2/lucapag1.pdf
A: $(1,1)$ is available by inspection.  Otherwise, if you write $2x^2=y^{15}+1=(y+1)(y^2-y+1)(y^4-y^3+y^2-y+1)(y^8+y^7-y^5-y^4-y^3+y+1)$ (Thanks, Alpha) you can look at where the factors of $2x^2$ can come from.  As the last three terms are odd, $y+1$ must have an odd numbers of factors of $2$. $y^4-y^3+y^2-y+1=(y^3-2y^2+3y-4)(y+1)+5$, so these two can share a 5, but no other prime.  $y^2-y+1=(y-2)(y+1)+3$ so these can share a factor $3$ but no other prime.  No other pairs can have common factors.  Aside from $2, 3, 5$, all primes must divide each factor an even number of times and can only divide one of them.  This gives you a lot of information on how the factors of $2x^2$ can be distributed.  I think it is pretty unlikely there are any more solutions
