Number of Solutions in Z/2027Z

More generally, let $f(n)$ denote the number of solutions to $a_1^2 - b_1^2 + a_2^2 - b_2^2 + \dots + a_n^2 - b_n^2 \equiv 0 \pmod{p},$ where $p$ is a odd prime.

For $n = 1$ , $a_1^2 - b_1^2 \equiv 0 \pmod{p}$ If $b_1 \equiv 0 \pmod{p}$ , then $a_1 \equiv 0 \pmod{p}$ . Otherwise, for each nonzero residue $b_1$ , there are two solutions $a_1$ , so $f(1) = 1 + 2(p - 1) = 2p - 1$ .

Now, assume $n \ge 2$ . Then $a_n^2 - b_n^2 \equiv -(a_1^2 - b_1^2 + \dots + a_{n - 1}^2 - b_{n - 1}^2) \pmod{p}.$ By definition, there are $f(n - 1)$ solutions where the right-hand side is 0 modulo $p$ . For each such solution, there are $2p - 1$ solutions in $a_n$ and $b_n$ , so there are $(2p - 1) f(n - 1)$ solutions in this case.

Otherwise, there are $p^{2n - 2} - f(n - 1)$ solutions where the right-hand side is nonzero modulo $p$ ; let the right-hand side be $r$ . Then $(a_n + b_n)(a_n - b_n) \equiv r \pmod{p}.$ If we set $c \equiv a_n - b_n$ , then $a_n + b_n \equiv rc^{-1} \pmod{p}$ . This system has a unique solution in $a_n$ and $b_n$ .

There are $p - 1$ possible values of $c$ , so there are $(p - 1)[p^{2n - 2} - f(n - 1)]$ solutions in this case. Hence, $\begin{aligned} f(n) &= (2p - 1) f(n - 1) + (p - 1)[p^{2n - 2} - f(n - 1)] \\ &= p f(n - 1) + (p - 1) p^{2n - 2}. \end{aligned}$

This gives $f(2) = p^3 + p^2 - p$ and $f(3) = p^5 + p^3 - p^2$ . In particular, for $p = 2027$ , the answer is $2027^5 + 2027^3 - 2027^2 = 34219120973043861.$

Observe that $p = 2027$ is a prime number.

If $f_n$ is the number of tuples such that $\sum_{i = 1}^3 a_i^2 - b_i^2 = n$ , then $F(x) = \sum_{n} f_n x^n = \sum_{a_i, b_i , i} x^{a_1^2 + a_2^2 + a_3^2 - b_1^2 - b_2^2 - b_3^2} = \left ( \left ( \sum_{a = 0}^{p-1} x^{a^2} \right ) \left ( \sum_{b = 0}^{p-1} x^{-b^2} \right ) \right )^3.$ If $N$ is the desired number, then $pN = F(1) + ... + F(\zeta^{p-1})$ where we define $\zeta = e^{\frac{2\pi}{p}i }$ . Observe that $\left ( \sum_{a = 0}^{p-1} \zeta^{ka^2} \right ) \left ( \sum_{b = 0}^{p-1} \zeta^{-kb^2} \right ) = \sum_{a, b} \zeta^{k(a^2 - b^2)}.$ Now, since the map $(a,b) \to (a+b, a-b)$ is bijective in $(\mathbb{Z}/p \mathbb{Z})^2$ , this is equal to $\sum_{x,y} \zeta^{kxy} = \begin{cases} p \quad k \neq 0 \\ p^2 \quad k = 0 \end{cases}$ So, $pN = p^6 + p^3(p-1) \implies N = p^5 + p^2(p-1)$ and we may conclude.