Rational solutions of quadratic forms 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?
 A: 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 $2$.
