(Determine the polynomials $P(x, y, z, w)$ and $Q(x, y, z, w)$ satisfying:

$$
(x y+z+w)^{2}-\left(x^{2}-2 z\right)\left(y^{2}-2 w\right)=(P(x, y, z, w))^{2}-\left(x^{2}-2 z\right)(Q(x, y, z, w))^{2}
$$
We will reduce the number of variables by establishing equality in front of the coefficients of $y$. The left-hand side expression being of degree 2 in $y$, the polynomials $P$ and $Q$ are at least of degree 1 in $y$ because there must be a non-zero coefficient in front of $y^{2}$. Suppose they are of degree $d$ strictly greater than 1 and denote $c_{P}(x, z, w)$ and $c_{Q}(x, z, w)$ the coefficients in front of $y^{d}$ of $P$ and $Q$ respectively. Then, by expanding, the coefficient in front of $y^{2 d}$ of the right-hand side expression is of the form $c_{P}(x, z, w)^{2}-\left(x^{2}-2 z\right) c_{Q}(x, z, w)^{2}$, which cannot be the zero polynomial because $\left(x^{2}-2 z\right)$ is not a square (we can also look at the degree in $z$, which should be even on the left but odd on the right. We deduce that $P$ and $Q$ are of degree 1 in $y$. If we write $P$ in the form $A(x, z, w) y+R(x, z, w)$ and $Q$ in the form $B(x, z, w)+S(x, z, w)$, we can compare the degrees in $w, z$ to find that $A$ and $B$ depend only on $x$, then by looking at $A(x)^{2}-\left(x^{2}-2 z\right) B(x)^{2}$ as an equation in $z$, we find $A(x)^{2}=x^{2}$ and $B(x)^{2}=1$.

By looking for the coefficients in front of $y$ and the evaluation for $y=0$ again, we find that $R=x^{2}-z+w$ and $S=x$ in the case $A=x$ and $B=1$, so finally $P= \pm\left(x y+x^{2}-z+w\right)$ and $Q= \pm(x+y)$.