2
$\begingroup$

Given a natural number $n$ and an element $k \in \mathbb{Z}_n$, how many solutions are there in $\mathbb{Z}_n$ to the equation $x^2+y^2 =k$? That is, I'm wondering whether there is a mod-$n$ version of the sum of squares function discussed in this post.

As an example, to write 1 as a sum of squares modulo 7, one has permutations of $0^2+1^2$ but also $2^2+2^2$ and $2^2+5^2$ and $5^2+5^2$.

$\endgroup$
  • $\begingroup$ As stated, there are infinitely many if there are any. Or do you want to restrict to $0 \le x,y < k$? Also, why $k \le n-1$ rather than the opposite? $\endgroup$ – Robert Israel Jul 19 '18 at 17:21
  • $\begingroup$ You are looking for the Igusa zeta function. For example, for $n=1$ and $k = p$ one has $p - (-1/p)$ solutions. $\endgroup$ – TKe Jul 19 '18 at 17:47
1
$\begingroup$

A formula for the quantity you are considering is given is computed by elementary means in https://arxiv.org/abs/1404.4214 .

In fact, they consider the same counting problem for general quadratic congruences of the form

$$ a_1 x_1^2 + \dots + a_k x_k^2 \equiv b \mod n .$$

$\endgroup$
1
$\begingroup$

Here is a quick way to compute it in the case of an odd sphere (so even number of variables) and $p$ is an odd prime. Note that for a general odd number its easy to reduce to this case by the Chinese reminder theorem and an easy lifting from $p$ to $p^m$. The equation $x^2+y^2=a$ is equivalent to the equation $N_{k[i]/k}(\alpha)=a$ where $k=\mathbb{F}_p$ is the field with $p$-elements. If $p\equiv 3 mod(4)$ then this equation has exactly $p+1$ solutions if $a \ne 0$ and one solution if $a=0$ because the Norm map $Nm:k[i]^\times\to k^\times$ is a surjective homomorphism.

If we denote by $f_\ell(a)$ the function that count the solution with $\ell$ variables, then byt clearly we have

$$f_{2\ell}(a)=\sum_{a_1+...+a_\ell = a}f_2(a_1)f_2(a_2)...f_2(a_\ell)$$, because we can count conditionally on the value of $x_{2i-1}^2+x_{2i}^2$ for every $1\le i\le \ell$ and sum the results. In other words, $f_{2\ell}$ is the $\ell$-fold convolution $f_2*f_2*...*f_2$. Let $I$ denote the constant function with value 1 on $k$, and $\delta_0$ the function which is $1$ at $0$ and $0$ otherwise. The calculation we did for $f_2$ gives exactly that $f_2=(p+1)I -p\delta_0$. Using the linearity of the convolution and the convolutions $I*I=pI$, $\delta_0*f=f$ for every $f$ we get that $$((p+1)I - p\delta_0)^{*\ell}=\sum_{k=0}^{\ell-1} {\ell \choose k} (-1)^{\ell - k}(p+1)^kp^{k-1}p^{\ell-k}I+(-1)^\ell p^\ell \delta_0=$$

$$=p^{\ell-1} \sum_{k=1}^{\ell} {\ell \choose k} (-1)^{\ell - k}(p+1)^kI+(-1)^\ell p^\ell\delta_0=p^{\ell-1}(p^\ell-(-1)^\ell)I+(-1)^\ell p^\ell \delta_0$$

and this (if I don't have any computation mistake!) gives you the answer in the case $p\equiv3 mod 4$. A similar argument works for $p\equiv 1 mod 4$ with different $f_2$ (here the algebra $k[i]$ splits into $k\oplus k$)

For even spheres, its slightly more complicated since eventually you have to convolve with the function $f_1$ which is the sum of $I$ and a quadratic character, but this is not a big difference when you notice that if $\chi$ is such a character then $\chi*I=0$ and $(I+\chi)*(I+\chi)=I+2\chi+\chi*\chi=f_2$ so you have all the convolution formulas you need.

$\endgroup$
  • 1
    $\begingroup$ When $p \equiv 1$ (mod 4), the equation can be replaced with $x^2-y^2=k$ because $-1$ is a square, and this equation has $p-1$ solutions for non-zero $k$. $\endgroup$ – Aravind Jul 20 '18 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.