Timeline for Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 5, 2013 at 22:10 | comment | added | Peter Mueller | @Kieren: I suggest that you restrict your many updates to real progress. Also, instead of asking if or erroneously claiming that certain things work, you better first check the details for yourself. By the way, you never said why you are so much interested in this one specific equation. | |
| Oct 5, 2013 at 18:16 | comment | added | Kieren MacMillan | It's a convoluted argument that I'm working on simplifying. It starts by showing that $\gcd((s^4-1)/r,r)= \gcd((r^4-1)/s,s)=1$, and then using that fact to trigger a chain of constraints. | |
| Oct 5, 2013 at 16:01 | comment | added | Noam D. Elkies | What's stopping some prime factor(s) of $r$ from dividing $2t^2+2t+1$ with multiplicity $\geq 2$? | |
| Oct 5, 2013 at 14:32 | comment | added | Kieren MacMillan | Yes, I believe I can prove $r^2 \mid (t^2+t)$, so $t=0$. | |
| Oct 5, 2013 at 13:04 | comment | added | Kieren MacMillan | From ($\star$), we see that for any prime $p \mid r$, we have $p^2$ dividing exactly one of the factors on the right-hand side (because they're pairwise relatively prime). This, in concert with the previous divisibility restrictions, forces $t$ out of the possible range of solutions. At least, I believe so — I'm double-checking my proof now… | |
| Oct 5, 2013 at 3:37 | comment | added | Noam D. Elkies | "can be shown" how? It seems unlikely that this kind of elementary manipulation can solve the problem. | |
| S Oct 4, 2013 at 16:39 | history | answered | Kieren MacMillan | CC BY-SA 3.0 | |
| S Oct 4, 2013 at 16:39 | history | made wiki | Post Made Community Wiki by Kieren MacMillan |