Modular arithmetic problem from Chinese remainder Theorem

Question

Take $p,q$ to be coprimes with $p<q<2p$. If $w,x,y,z$ are unknowns such that $w,x,y,z<\min(p,q)$ and we know $(wy+xz)\bmod (p+q)$, $(wz+xy)\bmod (p+q)$ $wy\bmod q$ and $xz\bmod p$ then are $w,x,y,z$ uniquely determined upto some symmetry?

Answer 1

No; switching $x$ and $z$, or $w$ and $y$, yields another solution. That is, given $(w,x,y,z)$, the 4-tuples $(w,z,y,x)$, $(y,x,w,z)$ and $(y,z,w,x)$ satisfy the same congruences.

Concretely, take $p=3$, $q=5$ and $(w,x,y,z)=(1,1,2,2)$. Then $wy+xz\equiv4\pmod{8}$ and $wy\equiv2\pmod{3}$ and $xz\equiv2\pmod{5}$. But the same congruences hold for $(w,x,y,z)=(2,2,1,1)$.

This shows that almost every choice of $p,q,w,x,y,z$ is a counterexample, which makes me wonder whether you have even thought about it at all.

Answer 2

Even ignoring the defects in the question, a "rescued" version of the question is hopeless. There are primes and distinct quadruples (even ignoring sign and swap symmetries) that have the same residues you call out.

For $p=5,q=7$, a very easy computer search finds many distinct satisfying pairs of quadruples (where "$(u)_v$" means $u \pmod{v}$): $$\begin{align} (wy+&xz)_{p+q},& (w&y)_{q},& (x&z)_{p},& &(w,x,y,z) \\ &2& &1& &1& &(1,1,1,1),(2,2,4,3) \\ &4& &1& &3& &(1,1,1,3),(2,2,4,4) \\ &3& &2& &1& &(1,1,2,1),(3,2,3,3) \\ &4& &2& &2& &(1,1,2,2),(4,3,4,4) \\ &5& &2& &3& &(1,1,2,3),(3,2,3,4) \\ &5& &1& &4& &(1,2,1,2),(2,3,4,3) \\ &6& &2& &4& &(1,2,2,2),(3,3,3,3) \\ &8& &2& &1& &(1,2,2,3),(4,4,4,4) \\ &10& &2& &1& &(3,1,3,1),(4,2,4,3) \\ &0& &2& &3& &(3,1,3,3),(4,2,4,4) \\ &1& &2& &4& &(3,2,3,2),(4,3,4,3) \end{align} $$ (I think this is all of them. I may have omitted too may pairs as symmetrically redundant, but this doesn't change the content of this answer.)