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How can I find all positive integers $x$ and $y$ such that $2x^2-1=y^{15}$?

PS. See here.

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@quanta: you seem to have dropped the $-1$. – Qiaochu Yuan Apr 4 '11 at 18:34
@Qiaochu, I don't understand, what was my mistake? – quanta Apr 4 '11 at 18:59
@quanta: you seem to be solving $2x^2 = y^{15}$. This equation is very easy to solve. – Qiaochu Yuan Apr 4 '11 at 19:01
Do you have a good reason to suspect that mathematics is ready for this question? If I write down a random Diophantine equation then there's a chance that it's an unsolved problem. If you've just made this question up then there's a chance that it's an unsolved problem. But if it's from a book or an old Olympiad question or something, then there will be a relatively low-level solution. What is the source of the question? – Kevin Buzzard Apr 4 '11 at 20:28
I suggest looking at the simpler equation $2x^2-1=y^3$. Apart from the trivial solution $(x,y)=(1,1)$ there is also $(x,y)=(78,23)$. Since 23 is not a fifth power, we don't get a solution to the original equation. I suspect that a real number theorist could make short work of $2x^2-1=y^3$. – Byron Schmuland Apr 4 '11 at 20:54
up vote 4 down vote accepted

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.

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Thanks for this! – Byron Schmuland Apr 4 '11 at 21:13

There are no solutions other then $(x,y)=(1,1)$

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Thanks very much. Those with access can find the result in Cohn, J. H. E. The Diophantine equations $x^3=Ny^2\pm 1$. Quart. J. Math. Oxford Ser. (2) 42 (1991), no. 165, 27–30. – Byron Schmuland Apr 5 '11 at 14:48

$(1,1)$ 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

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