For every prime $p$ there exists integer numbers $x,y,z$ such that $(x^2+1)(y^2+1)(z^2+1)\equiv1\pmod p$

Question

For every prime $p>3$ there exists integer numbers $x,y,z$ such that $(x^2+1)(y^2+1)(z^2+1) \equiv 1 \pmod{p}$ and $x \not\equiv y \pmod{p},y \not\equiv z \pmod{p}, z \not\equiv x \pmod{p}$.

Is this statement true ?

Answer 1

The answer is yes when $p\equiv1$, $3$, or $7\pmod8$ even if one insists that $x=0$.

If $p\equiv1,3\pmod8$ then there exists $y$ such that $y^2\equiv-2\pmod p$; then take $z=-y$, so that $(y^2+1)(z^2+1)\equiv(-1)(-1)=1\pmod p$.

If $p\equiv7\pmod8$, there are $(p-1)/2$ distinct residue classes modulo $p$ of the form $t^2+1$ with $p\nmid t$. However, these residue classes do not include $0$ (since $p\equiv3\pmod4$) or $1$ (since $p\nmid t$) or $-1$ (since $p\not\equiv1,3\pmod8$). The remaining $p-3$ residue classes are partitioned into $(p-3)/2$ pairs consisting of an element and its inverse; and by the pigeonhole principle, one of those pairs must have both its elements represented by the form $t^2+1$ modulo $p$. Take $y$ and $z$ to be values achieving those two inverses, so that $(y^2+1)(z^2+1)\equiv1\pmod p$.

When $p\equiv5\pmod8$ the answer is still yes, although we need $x$ to be flexible again. By the above observations, the only way we can't solve the congruence already with $x=0$ is if the $(p-3)/2$ residue classes of the form $t^2+1$ not congruent to $1$ or $0$ or $-1$ are comprised of exactly one element from each pair of multiplicative inverses modulo $p$. [Presumably this can't happen, but let's continue.] Let $R$ and $N$ partition these $(p-3)/2$ residue classes of the form $t^2+1$ into quadratic residues and quadratic nonresidues, respectively. Since $(p-3)/2$ is odd when $p\equiv5\pmod8$, one of $R$ and $N$ has more elements than the other.

If $\#R>\#N$, then take some element of $N$ and multiply it by every element of $R$, yielding $\#R$ nonresidues; one of them must not be of the form $t^2+1$, hence is the inverse of such a residue class, and $x$ can be chosen to make the triple product equal to $1$. ($N$ can't be empty, since some residue will be followed by a nonresidue modulo $p$.)

If $\#N>\#R$, then choose some element of $N$ and multiply it by every element of $N$, yielding $\#N$ quadratic residues; as above, one of them must not be of the form $t^2+1$, and we are done. Except: what if $\#N=\#R+1$ and the single extra quadratic residue is $-1\pmod p$? This can be avoided, because $\#N = (p-1)/4$ is odd when $p\equiv5\pmod8$, and so it must contain exactly one element from some pair of residue classes whose product is $-1\pmod p$; choose that one to be the fixed element of $n$. But, flaw detected: two residue classes are square roots of $-1\pmod p$ and thus don't form pairs. This actually happens for $p=5$. Can we argue around this for larger $p$?