I recently came across the following indefinite quadratic forms: \begin{align} f_1 & = x^2 + 2y^2 - z^2 \\ f_2 & = x^2 + y^2 + z^2 - w^2 - u^2 \\ f_3 & = x^2 + y^2 + z^2 + 6 w^2 - t^2 - u^2 - v^2 \\ f_4 & = x^2 + y^2 + z^2 + w^2 + 2r^2 - s^2 - t^2 - u^2 - v^2 \end{align} over the non-negative integers. What general methods are available to determine if these quadratic forms are universal or represent certain classes of integers satisfying some congruence constraints, etc.? In particular, what does the theory of genera and the Hasse Principle say about these indefinite forms? I'm aware of the 15 and 290 Theorems.
Asked
Active
Viewed 318 times
4
-
2Your question about general methods is interesting. But the specific examples you give all succumb very easily upon noting that $x^2-y^2$ represents every integer not of the form $4k+2$. – Erick Wong Feb 17 '14 at 19:49
-
All of these forms allow you to describe any number. The question is only in what form to do it. Can for this use and the Pell equation. http://math.stackexchange.com/questions/74931/integral-solutions-of-x2y21-z2/789972#789972 – individ Apr 28 '16 at 15:53
1 Answers
4
Well, $$ x^2 + 2 y^2 - C z^2 $$ is universal if and only if $C > 0,$ $C$ is odd, and every prime factor $p$ of $C$ (if $C > 1$) satisfies $p \equiv 1,3 \pmod 8.$ This is Problem 2 on page 161 of L.E.Dickson, Modern Elementary Theory of Numbers (1939).
Such problems get rapidly easier with more variables. With 5 or more, a form can only be isotropic in every $p$-adic field $\mathbb Q_p.$ As long as you stick with "diagonal" forms as above, all $a_i x_i^2$ and no $a_{ij} x_i x_j,$ congruence constraints, if any, will be quite visible.
I should emphasize how very different indefinite forms are from definite ones, as far as questions of universality. The 15 and 290 theorems have very little to do with this. For one thing, there are infinitely many ternary universal forms; I displayed one infinite family above.
This is quite strong, from page 160 in Dickson METN, Theorem 113: $$ a x^2 + b y^2 + c z^2 $$ is universal if and only if (0) $abc$ is odd or twice odd, (1) $a,b,c$ are not all of like sign and all are nonzero, (2) $a,b,c$ are relatively prime in pairs, (3) $-bc,-ca,-ab$ are quadratic residues for every prime factor of $a,b,c$ respectively.
The big change from Legendre's Theorem on page 156 is allowing square factors in the individual entries. I like the presentation on page 80 of Cassels, he makes it clear that each prime factor matters. Here is an example i learned as a case of an exercise in Lam's 2005 book; $$ 85 x^2 + 81 y^2 - z^2 $$ is universal. Now, 5 and 17 are not quadratic residues mod 3, but 85 is a residue. Similarly, 3 is not a residue mod 5 or mod 17, but 81 is, in fact it is a square. Finally, with $c=-1,$ there are no conditions, no prime factors.
