2
$\begingroup$

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 $p \geq 5$ ?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user83236 Dec 5 '15 at 10:19
  • $\begingroup$ 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 ? $\endgroup$ – user83236 Dec 5 '15 at 10:28
  • 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. $\endgroup$ – Wolfgang Dec 5 '15 at 11:05
  • 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. $\endgroup$ – eric Dec 5 '15 at 18:09
  • 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. $\endgroup$ – eric Dec 5 '15 at 23:04
12
$\begingroup$

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, $a=A^p$, $m=B^p$, $a+m=C^p$, $A^p+B^p=C^p$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ You don't need $p$ to be prime? $\endgroup$ – joro Dec 5 '15 at 12:05
  • $\begingroup$ I do not, but it is given by OP. $\endgroup$ – Fedor Petrov Dec 5 '15 at 12:26
  • $\begingroup$ You mean equivalent to the prime case of FLT. $\endgroup$ – user236182 Dec 5 '15 at 12:27
  • 1
    $\begingroup$ Well, FLT is a priori equivalent to its prime case. $\endgroup$ – Fedor Petrov Dec 5 '15 at 12:30
  • 1
    $\begingroup$ "denote y−1/2=a/m for coprime positive integers a,m": this is not possible if $y < 1/2$. $\endgroup$ – joro Dec 5 '15 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy