Timeline for A specific Diophantine equation related to the congruent number question
Current License: CC BY-SA 4.0
9 events
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| Jul 24, 2021 at 13:50 | comment | added | Jeremy Rouse | I confess that it was unclear to me if you were still interested in the question of when $2x^{2} + y^{2} + 8(2z+1)^{2} = n$ has no solutions, because it is not as closely connected with the congruent number problem as you had originally thought. | |
| Jul 24, 2021 at 13:04 | comment | added | roydiptajit | @JeremyRouse can you give any direction to the problem, as of how to solve the diophantine problem? | |
| Jul 17, 2021 at 18:03 | comment | added | roydiptajit | Okay thanks.. for correcting me. I was making a silly mistake.. | |
| Jul 17, 2021 at 17:04 | comment | added | Jeremy Rouse | The map $(\alpha,\beta,\gamma) \mapsto (\alpha,\beta,2\gamma)$ is a bijection between the set of solutions to $2x^{2}+y^{2}+32z^{2}=n$ and the set of solutions to $2x^{2}+y^{2}+8z^{2}=n$ with $z$ even. (You say that one solution $(\alpha,\beta,\gamma)$ gives rise to two solutions $(\alpha,\beta,\pm 2 \gamma)$, but that's not quite right because $(\alpha,\beta,-\gamma)$ also maps to the same pair.) | |
| Jul 17, 2021 at 16:43 | comment | added | roydiptajit | @JeremyRouse If $z$ is odd then how the twice relation is satisfying which I have stated in the 1st paragraph? One solution of the later is getting mapped to exactly two solution of the former all with even $z$.. The definition I have obtained from from Neal Koblitz's book. There tunnel's theorem was stated like this.. | |
| Jul 17, 2021 at 14:57 | comment | added | Jeremy Rouse | One of the things you say about $n$ being congruent is not correct. The correct interpretation of Tunnell's theorem is that $n$ is congruent if and only if $n$ has the same number of representations in the form $2x^{2} + y^{2} + 8z^{2}$ with $z$ even and with $z$ odd. For example, $n = 41$ is congruent because of the $16$ representations of $41$ in the form $2x^{2}+y^{2}+8z^{2}$, exactly $8$ of them have $z$ odd. (The ones with $z$ odd are $2 \cdot (\pm 2)^{2} + (\pm 5)^{2} + 8 \cdot (\pm 1)^{2}$.) | |
| Jul 17, 2021 at 9:25 | history | edited | roydiptajit |
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| Jul 17, 2021 at 9:05 | review | First posts | |||
| Jul 17, 2021 at 16:41 | |||||
| Jul 17, 2021 at 9:00 | history | asked | roydiptajit | CC BY-SA 4.0 |