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