Timeline for Is 36 a sum of 4 rational fourth powers?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 1 at 22:22 | vote | accept | Bogdan Grechuk | ||
| May 31 at 15:43 | answer | added | Bogdan Grechuk | timeline score: 23 | |
| May 30 at 9:07 | comment | added | Achim Krause | At odd $p$, the situation is easier, since the exponent $4$ is coprime to $p$. In this case, any solution $x^4+y^4+z^4+t^4=36$ mod $p$ where one of the $x,y,z,t$ is coprime to p (this is automatic for all primes except $3$, but for $3$ it's easy to check that you have such solutions, e.g. $(1,1,1,0)$) can be deformed to a solution in $\mathbb{Z}_p$ by the ordinary version of Hensel's lemma. | |
| May 30 at 9:03 | comment | added | Achim Krause | Yes. Specifically, if you have a solution $x^4 + y^4 + z^4 + t^4 = 4$ mod $16$, then one of $x,y,z,t$ needs to be odd, say $x$. Now by a version of Hensel's lemma, it is possible to deform $x$ in $\mathbb{Z}_2$ to an $\widetilde{x}=x + 4h$ with $\widetilde{x}^4 + y^4 + z^4 + t^4 = 36$ in $\mathbb{Z}_2$. | |
| May 30 at 8:47 | comment | added | R.P. | @BogdanGrechuk To construct a lift of a solution mod p, we just need to be able to extract a fourth root, which we can do by Hensel's lemma, if p > 2. | |
| May 30 at 7:06 | comment | added | Bogdan Grechuk | Thank you. Chevalley-Warning indeed implies solutions mod $p$. But by what theorem "At p=2 it suffices to find a solution mod 16, and at all other primes we are looking for a solution mod p"? | |
| May 29 at 20:37 | comment | added | Achim Krause | Ah, yes, I guess so! | |
| May 29 at 19:50 | comment | added | R.P. | @AchimKrause Wouldn't Chevalley-Warning also apply in this case? | |
| May 29 at 18:10 | comment | added | Achim Krause | Ah, here's an argument: Let $P\subseteq \mathbb{F}_p$ be the subset of $4$-th powers. Then $|P|=(p-1)/4 +1$ or $|P|=(p-1)/2+1$ depending on whether $p=1$ or $3$ mod $4$. By the Cauchy-Davenport inequality, it follows that $P+P+P+P$ contains all $p$ elements. So the equation has a solution mod every odd prime (and with a bit of extra manual work at $p=2$ and $p=3$, a solution in $\mathbb{Z}_p$ for any $p$.) | |
| May 29 at 15:48 | comment | added | Bogdan Grechuk | No, I do not know how to check the local solubility for all $p$ at once, so I just checked it for small primes. | |
| May 29 at 15:28 | comment | added | Achim Krause | Do you already know that $x^4+y^4+z^4+t^4=36$ admits a $p$-adic solution for all p? You write "modulo small primes", but not "modulo all primes". (At $p=2$ it suffices to find a solution mod 16, which is easy to do, and at all other primes we are looking for a solution mod $p$) | |
| May 29 at 14:31 | history | asked | Bogdan Grechuk | CC BY-SA 4.0 |