Without applying Fermat's Last Theorem, how can one show that the hyperelliptic curve $y^2 = x^{p} + \frac{1}{4}$ has only one positive rational solution $(x,y) = (0, \frac{1}{2})$ for ever prime $p \geq 5$ ?
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$\begingroup$ May you kindly forgive me for the use of the ''real analysis'' tag. Since i'm quite new on this platform, my rating is not yet sufficient for me to use the number theory, Diophantine analysis or algebraic number theory tags. Thank you. $\endgroup$– user83236Commented Dec 5, 2015 at 10:19
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$\begingroup$ Observe that by factoring $y^2 - \frac{1}{4} = (y-\frac{1}{2})(y + \frac{1}{2} )= x^p$, it follows that $y-\frac{1}{2} = u^p$ and $y + \frac{1}{2} = v^p$ hence $v^p - u^p =1$. By FLT we know that this has only one rational solution $(u, v) = (0,1)$ hence in the original question $(x, y) = (0, \frac{1}{2})$. But is there any other way to arrive at this result apart from this one, maybe via the theory of (hyper)elliptic curves ? $\endgroup$– user83236Commented Dec 5, 2015 at 10:28
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4$\begingroup$ I have added the tags. But normally you should have been able to do it, AFAIK all tags can be chosen freely, up to 5. $\endgroup$– WolfgangCommented Dec 5, 2015 at 11:05
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4$\begingroup$ I think it's reasonable to downvote someone who asks a question which they know to be equivalent to FLT and then ask for a proof which doesn't use FLT. $\endgroup$– ericCommented Dec 5, 2015 at 18:09
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1$\begingroup$ @Fedor: the direction that you suggest might not be known to the OP is far easier than the one they told us they knew :-) Furthermore, neither implication is at the level of this site. This is definitely stackexchange material. $\endgroup$– ericCommented Dec 5, 2015 at 23:04
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1 Answer
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It is equivalent to FLT. Indeed, if $a^p+1=b^p$ for positive rational $a,b$, we have $x^p:=(ab)^p=a^p(a^p+1)=(a^p+1/2)^2-1/4:=y^2-1/4$.
Opposite implication (moving from the comments): if $x^p=y^2-1/4=(y-1/2)(y+1/2)$, $x\ne 0$, denote $y-1/2=a/m$ for coprime non-zero integers $a,m$. Then also $a+m\ne 0$, $y+1/2=(a+m)/m$ and $x^p=a(a+m)/m^2$. Since $a,a+m,m$ are mutually coprime and $p$ is odd, we get that they should be all perfect $p$-th powers, $a=A^p$, $m=B^p$, $a+m=C^p$, $A^p+B^p=C^p$.
answered Dec 5, 2015 at 11:14
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$\begingroup$ I do not, but it is given by OP. $\endgroup$ Commented Dec 5, 2015 at 12:26
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$\begingroup$ You mean equivalent to the prime case of FLT. $\endgroup$ Commented Dec 5, 2015 at 12:27
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1$\begingroup$ Well, FLT is a priori equivalent to its prime case. $\endgroup$ Commented Dec 5, 2015 at 12:30
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1$\begingroup$ "denote y−1/2=a/m for coprime positive integers a,m": this is not possible if $y < 1/2$. $\endgroup$– joroCommented Dec 5, 2015 at 16:36