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Timeline for Can you find squares in this class?

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Mar 20, 2015 at 16:28 vote accept OriginalBBB
Mar 19, 2015 at 12:40 answer added Jeremy Rouse timeline score: 6
Mar 18, 2015 at 17:04 comment added OriginalBBB @JeremyRouse That is great news for me (and definitevely an answer to my question - you should add it as an answer), could you please point me to some reference about the informations you used in your deduction?
Mar 18, 2015 at 16:52 comment added Jeremy Rouse If there is a solution, then $p \equiv 1 \pmod{12}$. You are asking about rational points on the hyperelliptic curve $y^{2} = p(x^{4} + 6x^{2} - 3)$. There are no $3$-adic solutions if $p \equiv 2 \pmod{3}$, and there are no $p$-adic solutions unless $p \equiv \pm 1 \pmod{12}$. It follows that all solutions have $p \equiv 1 \pmod{12}$. There are also solutions if $p = 193, 349, 373, 433, 601, ...$.
Mar 18, 2015 at 16:00 history edited OriginalBBB CC BY-SA 3.0
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Mar 18, 2015 at 15:58 comment added OriginalBBB Thanks @jmc, are there no solutions between the primes 37 and 8929? Futhermore, all the primes you found are congruent to 1 mod 12, so the class of primes congruent to 7 mod (12) seems to be outside of it (a guilible mathematician would say). Any ideas?
Mar 18, 2015 at 15:27 comment added jmc And more tuples $(l,m,p)$: (7, 8, 8929) (8, 11, 6637) (10, 9, 38917) (11, 8, 48817) (13, 10, 99961) (14, 9, 113989) (16, 7, 133597) (17, 6, 142057)
Mar 18, 2015 at 15:23 comment added jmc Taking $(l,m) = (8,1)$ gives $11^{2} \cdot 37$. So combine that with $p = 37$, and you get a square.
Mar 18, 2015 at 15:21 comment added jmc I quickly tested whether $l^{4} + 6l^{2}m^{2} - 3m^{4}$ is square-free for $l,m \in \{1,\ldots,50\}$, and there are 1491 tuples where it is not.
Mar 18, 2015 at 14:40 history edited OriginalBBB CC BY-SA 3.0
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Mar 18, 2015 at 14:25 history edited OriginalBBB CC BY-SA 3.0
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Mar 18, 2015 at 14:24 review First posts
Mar 18, 2015 at 14:29
Mar 18, 2015 at 14:19 history asked OriginalBBB CC BY-SA 3.0