# Nontrivial Rational solutions to $y^2=4 x^n + 1$

Are there any nontrivial rational solutions to the following equation: $$y^2=4 x^n + 1,$$ where $n>2$?

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$x=0$, $y=\pm1$ come to mind. – alex.jordan Mar 9 '12 at 1:26
Thank you alex.jordan. I changed the title to include the word nontrivial. – Craig Feinstein Mar 9 '12 at 1:33
This is a great question. It shows lots of research, effort, and motivation. – The Chaz 2.0 Mar 9 '12 at 2:31
We are looking at family of Elliptic curves (for $n>2$) For n=3, it is discussed here math.lsu.edu/~hoffman/papers/elmod.pdf – Kirthi Raman Mar 11 '12 at 13:47
If $n>4$ it's hyperelliptic, not elliptic, isn't it? – Hurkyl Mar 11 '12 at 15:21
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Building on Théophile's answer, say $k=a/b$ and $x=c/d$ in reduced form, so that $a(a+b)d^n = b^2c^n$.

$c$ is coprime to $d$, and $a$, $b$, $a+b$ are pairwise coprime, so we get $a(a+b) = c^n$ and $b^2 = d^n$. Furthermore, $a$ and $a+b$ have to be $n$th powers, so there exists two coprime numbers $e,f$ such that : $a = e^n, a+b = f^n, b^2 = d^n, c = ef$.

If $n$ is odd, then $d$ has to be a square : $d = g^2, b = g^n$, and $e^n+g^n=f^n$, which is impossible by the Fermat-Wiles theorem.

If $n$ is even, then $n=2m, b=d^m$, and $(e^2)^m + d^m = (f^2)^m$. Again, this is impossible for $m>2$, which leaves the case $n=4$.

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 The first equation should be $a(a+b)d^n=b^2c^n$. – Craig Feinstein Mar 11 '12 at 4:40 @Craig : thanks, so this is not finished yet. I don't think there are solutions, but it's becoming ugly. – mercio Mar 11 '12 at 13:04 This is the elliptic curve $v^2 = u^3 - u$, by the (invertible, rational) change of variable $x = u/v$ and $y = (2u^3 - v^2)/v^2$. Unfortunately, I do not have handy a CAS I can actually ask for the rank and torsion of it. But, I think, $u$ and $u^2 - 1$ both have to be squares? Should be easy to work out the only solutions. (don't forget the point at infinity!) – Hurkyl Mar 11 '12 at 17:20 Thank you mercio! Great answer! – Craig Feinstein Mar 12 '12 at 19:45

Let's dispense with integer solutions. In that case, clearly $y$ is odd, so $y = 2k+1$ for some $k$. This gives $4k^2+4k+1 = 4x^n+1$.

Therefore $k(k+1) = x^n$. No prime $p$ that divides $x^n$ can divide both $k$ and $k+1$, so the latter are both perfect $n$th powers, which is absurd.

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 OP asks for rational solutions, not integer ones. – alex.jordan Mar 9 '12 at 1:30 He asked for rational solutions, didn't he? So y need not be an integer at all. – Mike Mar 9 '12 at 1:32 Thanks for noticing that. I've rewritten my answer. – Théophile Mar 9 '12 at 1:42 Is it possible to dispense with all nontrivial rational solutions? – Craig Feinstein Mar 9 '12 at 1:58