How many solutions are there to $n_1^2-n_2^2-n_3^2+n_4^2=k$?

Question

How many solutions are there to $n_1^2-n_2^2-n_3^2+n_4^2=k$ where $n_i\in [1,N]$ and $n_i,k\in \mathbb Z$?

For $k=0$ by paucity we know it should be $\ll N^2 \log N$ but what about different $k$?

Answer 1

Let us allow $n_i\in[-N,N]$ for simplicity; this will not change the order of magnitude of the number of solutions. So let us consider $$R(k,N):=\#\left\{(n_1,n_2,n_3,n_4)\in\{-N,\dotsc,N\}^4:n_1^2-n_2^2-n_3^2+n_4^2=k\right\}.$$ Let us also assume that $k$ is odd, and $N\gg\sqrt{k}$ with a sufficiently large implied constant.

If $r(n)$ denotes the number of ways $n$ can be written as a sum of two squares, then clearly $$\sum_{m\leq N^2-k}r(m)r(m+k)\leq R(k,N)\leq\sum_{m\leq 2N^2}r(m)r(m+k).$$ Hence it follows from Theorem 12.5 in Iwaniec: Spectral methods of automorphic forms (2nd ed.) that $$R(k,N)\asymp \Bigl(\sum_{d\mid k}d^{-1}\Bigr)N^2.$$ The same result would also follow from Theorem 4 in Heath-Brown: A new form of the circle method, and its application to quadratic forms, J. reine angew. Math. 481 (1996), 149-206, upon calculating the singular series $\sigma(x_1^2-x_2^2-x_3^2+x_4^2,k)$ there.

Surely, with more work, an asymptotic formula can also be derived, and the result should be extendable to even $k$ as well.

Answer 2

GH from MO's answered your question in full. Some possible points of interest:

1. Estermann was the first to evaluate $\sum_{m \le N} r(m)r(m+k)$ in "An asymptotic formula in the theory of numbers" (Proc. Lond. Math. Soc. (2) 34, 280-292 (1932)). In fact his estimate is uniform in $k\le N$ (possibly even), at least as it's stated in Lemma 2 of Hooley's "On the intervals between numbers that are sums of two squares" (Acta Math. 127, 279-297 (1971)).

2. One might ask about solutions to $P(n_1)-P(n_2)-P(n_3)+P(n_4)=k$ for $n_i\in [1,N]$ and $P$ a polynomial of degree $\ge 3$. For $k\neq 0$ the number of solutions is known to be $\ll N^{2-c_{\deg P}}$ by McGrath's recent preprint "On the asymmetric additive energy of polynomials" (2022).

3. The case $k=0$ was resolved earlier. We have the diagonal solutions $(n_1,n_4)=(n_2,n_3)$ and $(n_1,n_4)=(n_3,n_2)$ which contribute $2N^2-N$, and the number of the rest of the solutions is known to be $\ll N^{2-c}$ for an absolute $c$. This was first studied, for $P(x)=x^d$, by Hooley in "On another sieve method and the numbers that are a sum of two h-th powers" (Proc. Lond. Math. Soc., III. Ser. 43, 73-109 (1981)), and works of Wooley, of Heath-Brown and of Browning resolved this in full, see Browning's "The polynomial sieve and equal sums of like polynomials" (Int. Math. Res. Not. 2015, No. 7, 1987-2019 (2015)).