2

What are the solutions of equation

$$x^2+y^2+z^2+w^2=0,$$

in the $p$-adic integers? I think that for $p = 2$ it has only the trivial solution, but for $p$ odd there are nontrivial solutions.

zacarias
  • 3,158
  • 1
    I think in this, by modding by $p$ we know that by Chevalley's theorem, this has a solution over $GF(p)$, then apply the general Hensel's lemma, for $p$ odd. – LASV Dec 03 '13 at 22:09
  • Correct for $p=2$. In $2$-adics the equation holds $\bmod4$ only if all terms are even or all odd, and except for $w=x=y=z=0$ all even has to imply a not-all-even solution by descent on terminal zeroes base $2$. But the all odd case gives a sum $\not\equiv0\bmod8$. – Oscar Lanzi Jul 13 '21 at 19:44
  • For $4n+1$ primes there must be non-trivial solutions because $-1$ will be $p$-adically square and so we can render $y^2=-x^2, w^2=-z^2$. – Oscar Lanzi Jan 05 '23 at 16:57
  • Related: https://math.stackexchange.com/q/149520/96384 – Torsten Schoeneberg Feb 29 '24 at 17:21

2 Answers2

3

Well, there if $x_0^2\equiv a\pmod p$ then there is a solution to $x^2=a$ in $p$-adic solutions. So taking an integer solution $x_0^2+y_0^2+z_0^2+w_0^2=p$, with $x_0\neq 0$, we can find a $p$-adic solution to $x^2+y_0^2+z_0^2+w_0^2=p$.

We can certainly find infinitely many different answers by mutliplying $(x,y_0,z_0,w_0)$ by any $p$-adic number, but the real question is how many projective solutions.

Thomas Andrews
  • 177,126
  • 1
    Not quite enough, the derivative of the function has to be non-zero for Hensel's lifting to work, so $2(x_0 + y_0 + z_0 + w_0)$ has to be nonzero for your solution mod $p.$ – Igor Rivin Dec 03 '13 at 22:13
  • 2
    @Igor: Just $2x_0$: the others are constants with respect to $x$. –  Dec 03 '13 at 22:15
  • 1
    Yes you need $x_0\not\equiv 0\pmod p$, but at least one of the $x_0,y_0,z_0,w_0$ is non-zero and hence non-zero modulo $p$. @IgorRivin – Thomas Andrews Dec 03 '13 at 22:17
  • 1
    Ah, OK, that's slightly different from what I thought... – Igor Rivin Dec 03 '13 at 22:18
3

This answer may be disproportionate to the context of the question, but, if by chance one knows about local behavior of quaternion algebras... one might recognize that quaternary quadratic form as being the norm-form of the (usual rational form of) the Hamiltonian quaternions. The only ramified places (at which that quaternion algebra remains a division algebra) are $2$ and $\mathbb R$. Thus, at all odd $p$ that form is isotropic, and at $2$ (and $\mathbb R$, obviously) it is anisotropic.

(Yes, certainly, Hensel's lemma is used in setting up the usual properties of division algebras over local fields, as in Weil's "Basic Number Theory". Also, my on-line notes at http://www.math.umn.edu/~garrett/m/algebra/algebras.pdf do things similar to Weil, ...)

paul garrett
  • 52,465