Rational solutions of quadratic forms

Question

Is there an algorithm or a method that one can use to determine whether an equation of the form $(\text{E})$: $$ax^2+by^2+cz^2+dt^2=0$$ has a solution $(x,y,z,t)$ in whole numbers. In other words, given the tuple $(a,b,c,d)\in\mathbb{Z}^4$, how to determine whether there is a non-trivial tuple $(x,y,z,t)\in\mathbb{Z}^4$ that satisfies $(\text{E})$?

In case when one of $a,b,c$ or $d$ is zero, one can use Legendre's theorem on ternary quadratic forms. This paper shows the full process of how it's done. Is there anything similar for the case of four variables?

Answer 1

Clearly, it is enough to decide whether their is rational solutions or not (multiply by the square of a common denominator to have integer solutions).

I assume that $a,b,c,d$ are all nonzero (otherwise, you are reduced to known cases).

Multiplying by $a$ and replacing $x$ by $ax$, one may assume WLOG that $a=1$, so we are reduced to the equation $x^2+by^2+cz^2+dz^2=0$. (*)

This step is not necessary, but it allows to write down simpler conditions.

Hasse Minkowski says that (*) has a nonzero rational solution if and only if it has a nonzero solution over $\mathbb{R}$ and $\mathbb{Q}_p$ for all $p$.

1. Having a solution over the reals is equivalent to say that $b,c,d$ are not all $>0$.

2. For the $p$-adic case, it depends on the determinant and local Hasse invariants of the rational quadratic form $x^2+by^2+cz^2+dt^2$.

Here, the determinant is the square class of $bcd$, and if $p$ is prime , the local Hasse invariant is $(b,cd)_p(c,d)_p$.

I will below how to define $( , )_p$.

If $r,s$ are two non zero rationals, write $r=p^\alpha u, s=p^\beta v, p\nmid u, p\nmid v$.

Then $(r,s)_p=(-1)^{\alpha\beta\cdot\frac{p-1}{2}}\left(\dfrac{u}{p}\right)^\beta \left(\dfrac{v}{p}\right)^\alpha$, if $p\neq 2$, where $\left(\dfrac{\phantom{a}}{p}\right)$ is the Legendre symbol, and $\displaystyle (r,s)_2=(-1)^{\frac{u-1}{2}\cdot \frac{v-1}{2}+\alpha\frac{v^2-1}{8}+\beta\frac{u^2-1}{8}}$

Note for later use that $(-1,-1)_p=1$ if $p\neq 2$ and $(-1,-1)_2=-1$.

Note also that if $p\neq 2$, and the $p$-adic valuations of $r,s$ are both zero, $(r,s)_p=1.$

Recall also the following fact.

Fact. Let $r=p^\alpha u, p\nmid u$. Then $r$ is a square in $\mathbb{Q}_p$ if and only if :

• $\alpha$ is even and $u$ is a square modulo $p$ if $p\neq 2$ (which can be decided using Legendre symbol)

• $\alpha$ is even and $u\equiv 1 \ [8]$ if $p=2$.

If we translate Thm 6 of Chapter IV $\S$ 2 of Serre's "A course in arithmetic", we get that (*) has a nonzero solution over $\mathbb{Q}_p$ if and only if one of the following cases hold:

Case 1. $bcd$ is not a square modulo in $\mathbb{Q}_p^\times$

Case 2. $bcd$ is a square in $\mathbb{Q}_p^\times$ and $(b,cd)_p=(c,d)_p$ if $p\neq 2$

Case 3. $bcd$ is a square in $\mathbb{Q}_2^\times$ and $(b,cd)_2=-(c,d)_2$

All these conditions amounts to computations of finitely many Legendre symbols, because if $p\nmid bcd$ and $p\neq 2$, then either $bcd$ is not a square in $\mathbb{Q}_p^\times $, or $bcd$ is a square in $\mathbb{Q}_p^\times$ but in this case both symbols $(b,cd)_p$ and $(c,d)_p$ are equal to $1$ !!

Hence for odd prime which are not divisors of $bcd$, you are automatically in one of the first two cases.

So you only have to test if you are in one of the three previous cases only for $p\mid 2bcd $. Note that if the $p$-adic valuation of $bcd$ is odd, $bcd$ is automatically not a square, so you can reduce to prime numbers $p$ such that $p\mid 2bcd$ and $v_p(bcd)$ is even !

Hence, you have an algorithm to decide the existence of a nontrivial solution over $\mathbb{Q}$ (note that the Legendre symbol coincide with the Jacobi symbol, so you can compute it without factoring you integers). However, it won't give you an explicit solution.

Example. Consider $3x^2+3\cdot 5y^2+7z^2-2\cdot 7\cdot 23 t^2=0$.

This is equivalent to consider $x^2+3^2 \cdot 5y^2+3\cdot 7z^2-2\cdot 3 \cdot 7\cdot 23 t^2=0$.

Here $b=3^2 \cdot 5, c=3\cdot 7, d=-2\cdot 3\cdot 7 \cdot 23, cd=-2\cdot 3^2\cdot 7^2\cdot 23$ , and $bcd=-2\cdot 3^4\cdot 5\cdot 7^2\cdot 23$.

Since $b,c,d$ are not all positive, we have solutions over the reals.

For the $p$-adic case, we just have to check two cases: $p=3,7$ since they are the only prime divisors of $bcd$ with an even valuation.

For $p=3$, we need to check that $-2\cdot 5\cdot 23$ is a square modulo $3$. But $-2\cdot 5\cdot 23=1 \ [3]$, which is a square.

Now since $( r, s)_p$ only depends on the square classes of $r$ and $s$, we have $(b,cd)_3= (5,-2\cdot 23)_3=(-1)^{\frac{5-1}{2}\frac{-46-1}{2}}=1$ and $(c,d)_3=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_3 =(-1)^{\frac{3-1}{2}}\left(\dfrac{7}{3}\right) \left(\dfrac{-2\cdot 7\cdot 23}{3}\right)=-1\cdot 1\cdot -1=1.$

Hence we are in Case 2.

For $p=7$, we need to check that $-2\cdot 5\cdot 23$ is a square modulo $7$. But $-2\cdot 5\cdot 23=1 \ [7]$, which is a square.

Since $7\nmid 5,2$ and $3$, $(b,cd)_7=1$. Now $(c,d)_7=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_7=(-1)^{\frac{7-1}{2}}\left(\dfrac{3}{7}\right) \left(\dfrac{-2\cdot 3\cdot 23}{7}\right)=-1\cdot -1\cdot 1=1.$

Hence we are in Case 2.

All in all, the original equation must have a non trivial solution.

This is indeed the case since $3\cdot 2^2+3\cdot 5\cdot 3^2+7\cdot 5^2-2\cdot 7\cdot 23\cdot 1^2=0$.

Answer 2

I actually realized some time after my comment above that, because we're trying to hit zero, at least one member of $(a,b,c,d) < 0$. If we assume WLOG that $a \geq b \geq c \geq d$, then at least $d$ is negative. If we set $x = t$ and $a+d = m$, then we can simplify to

$$mx^2 + by^2 + cx^2 = 0$$

But now one of those coefficients still has to be negative if we want to hit $0$. Either $|a| < |d|$ and $m$ is negative, or $b$ and/or $c$ are negative. Again, WLOG we can assume $c$ is negative. Then setting $y = z$ and $b+c-n$, we get:

$$mx^2 + ny^2 = 0$$

And either $m$ or $n$ has to be negative, so we really have

$$mx^2 = ny^2$$

I have to imagine there are solutions for this that don't have $m = n$ and $x = y$. But I'm not sure if there's a generalized form.

Edit: Oh, right. Either $m = n, x = y$, or both $m$ and $n$ must be squares for there to be integer solutions, because (for instance) $2x^2 = 5y^2$ can't have an integer solution, since both $x$ and $y$ have prime factorizations where all the exponents are $2$.