Timeline for Can you find squares in this class?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 20, 2015 at 18:29 | comment | added | Michael Stoll | @NoamD.Elkies: I know, but I when I first tried, I used a bound of $10^5$ (and then I got the point by pulling it back from $E_p$). "Quite large" was meant relative to the solutions for smaller $p$. | |
| Mar 20, 2015 at 16:28 | vote | accept | OriginalBBB | ||
| Mar 20, 2015 at 4:16 | comment | added | Noam D. Elkies | @Michael Stoll For $p=997$, your ratpoints program doesn't take long to find the minimal solution $(l,m) = (130792, 148329)$. | |
| Mar 19, 2015 at 16:46 | comment | added | Michael Stoll | @JeremyRouse: I'm not so sure about this. One should get a criterion for when the rank is positive, but it is conceivable that there is some $p$ such that $E_p$ has positive rank and non-trivial Sha, with a non-trivial element represented by the curve in question. | |
| Mar 19, 2015 at 16:31 | comment | added | Jeremy Rouse | Thanks for the correction jmc. If one wanted to study this problem in more detail, one could find a criterion analogous to Tunnell's for the congruent number problem. | |
| Mar 19, 2015 at 16:10 | history | edited | Jeremy Rouse | CC BY-SA 3.0 |
added 6 characters in body
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| Mar 19, 2015 at 15:07 | comment | added | Michael Stoll | Another comment: The condition $p \equiv \pm 1 \bmod 12$ is only necessary, but not sufficient for $p$-adic solubility. In fact, the only primes $\equiv 1 \bmod 12$ below 1000 for which the equation is $p$-adically soluble are $p = 37,61,157,193,313,349,373,397,433,577,601,613,661,673,769,853,877,937,997$. Combining this with the rank condition, only $37, 193, 349, 373, 433, 601, 673, 877, 997$ remain. Of these, all actually have solutions (for $p=997$, the smallest solution seems to be quite large). | |
| Mar 19, 2015 at 14:50 | comment | added | Michael Stoll | The argument for $p = 61$ generalizes. Let me write $E_p$ for what is $E$ in the answer. Then if $E_p(\mathbb Q)$ has rank zero, the original equation has no solutions (since it then represents either not even an element of Ш or, if it does, a nontrivial one). This happens for $p = 13,61,73,97,109,157,181,229,241,277,313,337,397,409,421,457,541,577,613,661,709,733,757,769,829,853,937,\ldots$. | |
| Mar 19, 2015 at 12:44 | comment | added | jmc | Awesome answer! I think though, that $p$ should be $\ell$ for some $p$ in the sentence “if $p = 61$ … for all primes $p$”. | |
| Mar 19, 2015 at 12:40 | history | answered | Jeremy Rouse | CC BY-SA 3.0 |