# On the product $\prod_{1\leq a<b\leq \frac{p-1}{2}}\,\left(a^2+b^2\right)$, where $p$ is prime

Let $p$ be an odd prime natural number. Define $$M(p):=\prod_{1\leq a<b\leq \frac{p-1}{2}}\,\left(a^2+b^2\right)\,.$$ Prove that, over the field $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$ (i.e., in modulo $p$), $$M(p)= \left\{ \begin{array}{ll} 0&\text{if }p\equiv1\pmod{4}\,,\\ +1&\text{if }p\equiv3\pmod{16}\text{ or }p\equiv15\pmod{16}\,,\\ -1&\text{if }p\equiv7\pmod{16}\text{ or }p\equiv11\pmod{16}\,. \end{array} \right.$$ The first equality (i.e., that $M(p)=0$ for $p\equiv 1\pmod{4}$) is trivial. For now (see this thread), we know that $M(p)=\pm1$ for $p\equiv 3\pmod{4}$. Under my answer in that thread, jpvee found that the claim above holds for prime natural numbers $p$ such that $p\leq 1000$. Does anybody know whether the claim is true, and if it so, how to prove the claim?

• Quite a lot of the expressions there hint at quadratic reciprocity and the Legendre symbol... – TheSimpliFire Jul 3 '18 at 8:45

• A good find! Nice to learn that I can still make serious headway into a Schweitzer Problem :-) – Jyrki Lahtonen Jul 3 '18 at 10:54
• @JyrkiLahtonen You are being very nice here :) Though i found the answer, to be honest, i don't understand a thing whats written :) – crskhr Jul 3 '18 at 11:00

Let's assume that $p\equiv3\pmod4$ is a prime. As you already did in the linked thread, we can use the field $K=\Bbb{F}_p[i]=\{a+bi\mid a,b\in\Bbb{F}_p\}$.

By the properties of the norm map $N:K^*\to\Bbb{F}_p^*, N(a+bi)=a^2+b^2$, we know that every element of $F_p^*$ is the image of exactly $p+1$ elements. This is because the norm map is a surjective homomorphism between cyclic groups of respective orders $p^2-1$ and $p-1$.

To get a handle on the constraints on $a,b\in\Bbb{F}_p^2\setminus\{(0,0)\}$ we need to do the following. We have the dihedral group $G=D_4$ of order eight acting as obvious symmetries of the norm form: $(a,b)\mapsto (\pm a,\pm b)$ or $(a,b)\mapsto (\pm b,\pm a)$. It is clear that each orbit of $G$ on $K^*$ contains a unique pair $(a,b)$ such that $0\le a\le b\le(p-1)/2$.

Here the orbits of $(a,b)$ such that $0<a<b\le(p-1)/2$ have full size $8$, but the points $(a,0),0<a\le (p-1)/2$, and $(a,a),0<a\le(p-1)/2$ have orbits of size four only as their stabilizers in $G$ have order two.

So if we write $N_1(x)$ for the number of solutions of $x=a^2+b^2,0<a<b\le(p-1)/2$, $N_2(x)$ for the number of solutions of $x=a^2,0<a\le(p-1)/2$, and $N_3(x)$ for the number of solutions of $x=a^2+a^2$, then we have, for all $x\in\Bbb{F}_p^*$, the identity $$p+1=8N_1(x)+4N_2(x)+4N_3(x).$$ Clearly $N_2(x)=1$ or $0$ according to whether $x$ is a quadratic residue or not. Similarly $N_3(x)=1$ or $0$ according to whether $x$ is two times a quadratic residue or not.

I am fairly sure that the formula follows from these observations and the facts that

• the product of all the elements of $\Bbb{F}_p^*$ is $-1$,
• the product of all the quadratic reside of $\Bbb{F}_p^*$ is equal to $+1$,
• the product of all the elements that are twice a square is equal to $2^{(p-1)/2}=\left(\dfrac2p\right)$ where the Legendre symbol is $+1$ when $p\equiv 7\pmod8$ and $-1$ when $p\equiv3\pmod8$.

The way this shows up in your product is that the factor $x$ occurs in it $$N_1(x)=\frac{p+1}8-\frac{N_2(x)+N_3(x)}2$$ times. If two is a quadratic non-residue, then $N_2(x)+N_3(x)=1$ for all $x$ and $N_1(x)=(p-3)/8$ for all $x$ meaning that your product is $(-1)^{(p-3)/8}$. If two is a quadratic residue, then $N_2(x)+N_3(x)=2$ when $x$ is a QR and zero otherwise. In this case your product is $(-1)^{(p+1)/8}$ because the extra factor from quadratic residues is equal to one.