I think this must be well-known (and probably not hard to prove either), but I cannot find a reference: for a (positive) rational number $a$, the number of integral solutions to the equation $$ x^2+y^2-az^2=0, $$ with $|x|,|y|,|z|<T$ is $C(a) T \log T$, where $C(a)$ is a constant depending only on $a$. I would very much appreciate a reference which also includes a proof.
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asked Nov 16, 2015 at 10:23
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3$\begingroup$ The answer below gives a nice reference. But this particular case is pretty easy. Clearly one can assume that $a$ is an integer and a sum of two squares. Then you are asking for $\sum_{z \le T} r(az^2)$ where $r(n)$ is the number of ways of writing $n$ as a sum of two squares. This is essentially summing a multiplicative function, which is $3$ on primes $p$ that are $1\pmod 4$ (and not dividing $a$), and $1$ on the primes $p$ that are $3\pmod 4$. The usual argument via Dirichlet series now gives the asymptotic. $\endgroup$– LuciaCommented Nov 16, 2015 at 15:22
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You need some further assumptions on $a$ to guarantee the existence of a solution to this equation, as there might not be solutions modulo some prime, say. The theorem of Hasse and Minkowski tells you exactly when solutions exist.
But anyway, the result you want is Theorem 8 in the paper: Heath-Brown - A New Form of the Circle Method, and its Application to Quadratic Form.
You can find this here: http://eprints.maths.ox.ac.uk/144/1/circle.pdf (Wayback Machine)
answered Nov 16, 2015 at 10:47