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
34 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2014 at 2:32 | vote | accept | Kieren MacMillan | ||
| May 12, 2014 at 2:32 | vote | accept | Kieren MacMillan | ||
| May 12, 2014 at 2:32 | |||||
| May 11, 2014 at 5:44 | answer | added | Noam D. Elkies | timeline score: 7 | |
| Oct 5, 2013 at 20:01 | comment | added | Kieren MacMillan | @PeterMueller: In light of your Pell equation observation, is it valid to claim that $$ \frac{r^2+2rs-s^2}{2rs} $$ must be a convergent to $\sqrt{2}$, i.e., one of $3/2, 17/12, 99/70, \dotsc$? | |
| Oct 5, 2013 at 19:08 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
corrected bounds on r/s
|
| Oct 5, 2013 at 18:49 | comment | added | Kieren MacMillan | @PeterMueller: My mistake. I can prove $4/3 < r/s < 3/2$ when $r > 3$. | |
| Oct 4, 2013 at 17:13 | comment | added | Barry Cipra | @NickS, I think you mean the equation reduces to $Re(z^4(1+i))=1$, not $Im$, after setting $z=r+is$. | |
| Oct 4, 2013 at 16:39 | answer | added | Kieren MacMillan | timeline score: 1 | |
| Oct 4, 2013 at 13:06 | comment | added | Kieren MacMillan | @Peter: Excellent! I know I can get some upper and lower bounds, e.g. $3r > 3s \ge 2r$ is trivial. The question is how close I can bring them. In any case, I'm encouraged. | |
| Oct 4, 2013 at 7:42 | comment | added | Peter Mueller | @Kieren: If you really can prove the inequalities between $r$ and $s$, then you are done. Note that by the symmetries of your equation (generated by $(r,s)\mapsto(-r,-s)$ and $(r,s)\mapsto(-s,r)$) you may assume $r,s>0$. Set $x=r/s$. You claim that $5/4\le x\le 4/3$. On the other hand, for $s$ big $x$ is close to a positive root of $y^4+y^3-6y^2-4y+1$, and these roots are $0.19891\dots$ and $1.49660\dots$, clearly outside the allowed range. So your inequalities imply a tight upper bound on $s$. | |
| Oct 3, 2013 at 17:21 | comment | added | Kieren MacMillan | @NickS: Could you (despite your initial claim that you don't know) point me in a direction how that might that help? It sounds very intriguing. | |
| Oct 3, 2013 at 16:22 | review | Close votes | |||
| Oct 3, 2013 at 18:14 | |||||
| Oct 3, 2013 at 16:11 | comment | added | Kieren MacMillan | Sorry! I didn't know that. I was just trying to follow the "add new information to the original post" rule. I will try to save the edits for single big chunks, much less often. Thanks for the tip. | |
| Oct 3, 2013 at 16:05 | comment | added | Felipe Voloch | Every time you edit your question, you bump it up to the front page. Doing that too many times is not nice. | |
| Oct 3, 2013 at 15:42 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
new bounds obtained; Post Made Community Wiki
|
| Oct 3, 2013 at 15:02 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
added congruence calculations with $4 \le r \le 13000$
|
| Oct 2, 2013 at 1:07 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
changed wording of last edit
|
| Oct 2, 2013 at 0:34 | comment | added | Nick S | I don't know if this helps but with $z=r+is \in \mathbb Z[i]$ the equation reduces to $Im(z^4(1+i))=1$. | |
| Oct 2, 2013 at 0:21 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
added divisibility restrictions
|
| Oct 1, 2013 at 14:38 | comment | added | percusse | $$\begin{pmatrix}1\\r\\r^2\\r^3\\r^4\end{pmatrix}^T\begin{pmatrix}-1 &0&0&0&1\\0&0&0&-4&0\\0&0&-6&0&0\\0&4&0&0&0\\1&0&0&0&0\end{pmatrix}\begin{pmatrix}1\\s\\s^2\\s^3\\s^4\end{pmatrix}=0$$ can be used to derive such nontrivial identities by matrix operations though doesn't help much in terms of finding the solutions. You might try some integer programming tricks by extending the outer vectors and looking for a binary matrix for the quadratic equation. Maybe there is some structure. | |
| Sep 30, 2013 at 22:25 | answer | added | Peter Mueller | timeline score: 9 | |
| Sep 30, 2013 at 21:29 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
eliminated possibly false conclusion from previous edit
|
| Sep 30, 2013 at 20:59 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
added 103 characters in body
|
| Sep 30, 2013 at 20:41 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
added congruence modulo (r-s)(r+s) comment
|
| Sep 30, 2013 at 18:51 | comment | added | Peter Mueller | @Kieren: I believe the question is more difficult than it looks at a first glance. The fact that there are non-trivial solutions makes it unlikely that some easy ad hoc argument works. One other observation I'm not sure if it helps: Your equation is $s^4h(r/s)=1$, where $h(x)$ has the cyclic group of order $4$ as Galois group, as can be seen by the fact that $x\mapsto (x-1)/(x+1)$ maps roots of $h(x)$ to roots. | |
| Sep 30, 2013 at 18:33 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
added new factorization discovery
|
| Sep 30, 2013 at 15:38 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
clarified "r \pm s" observation
|
| Sep 30, 2013 at 15:21 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
added Pell equation observation
|
| Sep 30, 2013 at 14:23 | comment | added | Kieren MacMillan | Ooo! @PeterMueller, that could be very helpful — thanks! | |
| Sep 30, 2013 at 14:14 | comment | added | Peter Mueller | The equation is equivalent to $(r^2+2rs-s^2)^2-2(2rs)^2=1$, don't know if this relation to the Pell equation $x^2-2y^2=1$ is helpful. | |
| Sep 30, 2013 at 14:03 | comment | added | Alicia Garcia-Raboso | Notice that the $\mathbb{Z}/2\mathbb{Z}$ symmetry $(r,s) \mapsto (-r,-s)$ comes in fact from a $\mathbb{Z}/4\mathbb{Z}$ symmetry $(r,s) \mapsto (s,-r)$. | |
| Sep 30, 2013 at 13:02 | comment | added | Felipe Voloch | I don't know about "elementary solutions" but Pari/GP via the command thue (but check the docs as it needs initialization) reports that the solution set is as you expect. | |
| Sep 30, 2013 at 12:58 | history | edited | Kieren MacMillan | CC BY-SA 3.0 |
added 72 characters in body
|
| Sep 30, 2013 at 12:51 | history | asked | Kieren MacMillan | CC BY-SA 3.0 |