# Rational roots to quadratic forms in 4 variables

Hi,

I am interested in the following question. Let $F(x_1, x_2, x_3, x_4)$ be a quadratic form in four variables with integer coefficients. Let $B > 0$ be a parameter. Define $N_1(F,B)$ to be the number of rational solutions to the equation $F(x_1, x_2, x_3, x_4) = 0$, such that $|x_i| \leq B$ for $i = 1,2,3,4$. I am looking for an upper bound of the form $$\displaystyle N_1(F,B) \ll_\epsilon B^{\theta + \epsilon},$$ where the constant $\theta < 2$. Is there any such results known? One can assume that $F$ is absolutely irreducible and if necessary, assume that $F$ is non-singular.

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I have an answer for $F=x_1^2+x_2^2+x_3^2+x_4^2$. :-) And I guess that your $\theta$ should highly depend on the signature of $F$, so you probably have some extra condition (and motivation?) in mind. –  Wadim Zudilin Jul 2 '11 at 17:35
The way you have described this, with no bounds on the denominators, as soon as there is any nontrivial solution there are infinitely many. –  Will Jagy Jul 2 '11 at 20:23
The problem title and body say rational but with if $|x_i|$ means absolute value then indeed the number could well be infinite. So did you mean integer solutions or some height on $\mathbb{Q}$ such as $|\frac{p}{q}|$ being the larger of the absolute values of $p$ and $q$? Of course that would allow more than $B^3$ solutions with $|x_i| \leq B$ –  Aaron Meyerowitz Jul 2 '11 at 21:36

I'm afraid that, even if we assume (as suggested in A.Meyerowitz's comment) that the intention is integer solutions of $F(x_1,x_2,x_3,x_4) = 0$, the number of solutions with $\max_i |x_i| \leq B$ will grow at least as $B^2$ as long as there's any nonzero solution, and sometimes the growth will even be a bit faster.

The usual heuristics suggest that if there's no local obstruction (such as there is for W.Zudilim's example of $X_1^2 + X_2^2 + X_3^2 + X_4^2$, with no nontrivial zeros even over $\bf R$) then the number solutions up to height $B$ should grow as $B^2$: there are about $B^4$ candidates, and for each one of them $F(x_1,x_2,x_3,x_4)$ has size at most $B^2$, so if we imagine they're randomly distributed then about $B^4 / B^2 = B^2$ values should be zero.

It is not hard to prove that for some choices of quadratic form $F$, even nonsingular ones, the count is $\gg B^2$. Namely suppose $F(x_1,x_2,x_3,x_4)$ has the form $Q(x_1,x_2) - Q(x_3,x_4)$ for some quadratic $Q$. Then the line $(x_1,x_2,x_1,x_2)$ already gives $B^2$ zeros. Geometrically, the two rulings of the quadric $Q=0$ in ${\bf P}^3$ are defined over ${\bf Q}$, and any line — not just the trivial $\{(x_1:x_2:x_1:x_2)\}$ — will give some positive multiple of $B^2$. Using all the lines in the ruling one finds that in fact the correct growth rate is $B^2 \log B$.

If I remember right it is known that for an unobstructed smooth quadric in ${\bf P}^3$ whose rulings are not rational but Galois conjugate over ${\bf Q}$, the counting function $N_1(F,B)$ is asymptotically proportional to $B^2$ as the heuristics suggested.

This is all consistent with Manin's conjecture: the quadrics with rational rulings are precisely those for which the rational subgroup of the Néron-Severi group has rank 2 rather than 1, which accounts for the extra factor of $\log B$.

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To rephrase Elkies's answer in elementary terms, here is an approach to derive an asymptotic formula for the number $N(T)$ of integer solutions to $x_1 x_3 - x_2 x_4 = 0$, with $x_1^2 + x_2^2 + x_3^2 + x_4^2 \leq T^2$.

Let $A = \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix}$. Then the above equation is $\det A = 0$. If $\det A = 0$ for an integer matrix $A$, then there exists an integer vector $v$ with $A v = 0$. Now given an integer vector $v$, let $N_v(T)$ denote the number of integer matrices $A$ of norm at most $T$ such that $A v = 0$. This is a linear problem, one is counting the number of integer points in the two dimensional subspace $$R_v = \{ A \in M_2(\mathbf{R}) \mid Av = 0 \}.$$ Now it is easy to see that the integer lattice $M_2(\mathbf{Z})$ intersects $R_v$ in a lattice of covolume $\|v\|^2$. Therefore, $N_v(T)$ is asymptotic to $\pi T^2/\|v\|^2$, and so,

$$N(T) \sim \sum_{\stackrel{ v \in \mathbf{Z}^2}{\|v \| \le T}} N_v(T) \sim \sum_{\stackrel{ v \in \mathbf{Z}^2}{\|v \| \le T}} \frac{\pi T^2}{\|v\|^2} \sim \pi T^2 \log T.$$

(One still has to justify why the correct place to cut off the sum over $v$ is around $\|v \| = T$, and also to justify the interchange of limits. This is done e.g. in the following paper:

Katznelson, Y. R. Singular matrices and a uniform bound for congruence groups of $\mathrm{SL}_n(\mathbf{Z})$. Duke Math. J. 69 (1993), no. 1, 121–136.)

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...and then doing this in the sublattice with $x_1 \equiv x_4$ and $x_2 \equiv x_3 \bmod 2$ also lets us count solutions of $a^2 + b^2 = c^2 + d^2 \leq H^2$ (with $H = T/2$). –  Noam D. Elkies Jul 3 '11 at 16:18

I don't think (I'm assuming the correct problem is over the integers) you can get $\theta < 2.$ I did experiments with $$x_1^2 - x_2^2 - x_3^2 - x_4^2 = 0$$ and so

         B                 N
1                13
10               601
100             54433
1000           5390401
10000         538790017


I'm just not seeing $N / B^2$ necessarily having limit 0. If the limit really is 0, that still need not give you an exponent below 2. The result to remember is the asymptotic count of numbers up to some positive $x$ that are representable as the sum of two squares. There is a known constant $C \approx 0.7642$ such that the count is $$\sim \frac{Cx}{\sqrt {\log x}}$$
A full proof is in a Dover reprint, both volumes together of Topics in Number Theory by William J. LeVeque, section 7-5 in Volume 2. Also see link. It would appear there is a chance of getting fairly precise asymptotics for $$x_1^2 + x_2^2 - x_3^2 - x_4^2 = 0$$ although your uniform bound of $B$ on the individual variables is inconvenient for that problem. Your bound is convenient for the problem I tabulated, so maybe that one can be finished up.

EDIT: taking $R(n)$ as the number of ways of writing an integer $n$ as the sum of three integral squares, for an odd prime $p$ we have $$R(p^2 n) = (p + 1 - (-n|p) ) \; \; R(n) - \; \; p \; R( n / p^2) .$$ With $R(1) = 6,$ given any odd squarefree $m,$ we have $$R(m^2) \geq 6m.$$ As a result, the number of solutions to $$x_1^2 - x_2^2 - x_3^2 - x_4^2 = 0$$ with $|x_1| \leq B$ is asymptotically at least a constant times $B^2.$

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It's somewhat easier to get asymptotics for $x_1^2 + x_2^2 - x_3^2 - x_4^2$ by writing it as $ad-bc$ with $a,d = x_1 \pm x_3$ and $b,c = x_4 \pm x_2$. The asymptotic constant may be annoying to figure out, but it's not too hard to give upper and lower bounds proportional to $B^2 \log B$. See my answer from ~35 minutes ago for more information. –  Noam D. Elkies Jul 3 '11 at 3:50
Thanks, Prof. Elkies. I've been trying to finish some bounds on the form I tabulated. Not getting much, except that for an odd prime $p$ the number of representations of $p^2$ as the sum of three squares (both primitive and imprimitive) is at least $6p.$ So the number of solutions with all $x_j \leq B$ is at least $\frac{3 B^2}{\log B},$ so at least the quest for an exponent below 2 is futile. –  Will Jagy Jul 3 '11 at 4:39
Geometrically $x^2_1=x^2_2+x^2_3+x^2_4$ is the twist of ${\bf P}_1 \times {\bf P}_1$ where the two factors become conjugate over ${\bf Q}(i)$, i.e. the projective line over ${\bf Q}(i)$. Unwinding this yields the following recipe: let $w,z$ be Gaussian integers of the same parity, and let $x_1,x_2=\frac12(|w|^2 \pm |z|^2)$ and $x_3,x_4$ be the real and imaginary parts of $wz$. In effect we're identifying the Riemann sphere (over ${\bf Q}(i)$) with the Euclidean sphere (with rational coordinates) :-) This should give $cB^2$. –  Noam D. Elkies Jul 3 '11 at 15:20
...and primitive solutions should correspond to $w,z$ that are either relatively prime in ${\bf Z}[i]$ or have $\gcd(w,z)=1+i$ but are not congruent mod 2. –  Noam D. Elkies Jul 3 '11 at 15:49
Thanks again, Noam. –  Will Jagy Jul 3 '11 at 20:35