Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$? Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$?
 A: I suppose you mean : nonzero $p,q$.
The answer is no. In fact, there do not even exist nonzero univariate polynomials
$A,B$ such that $(*) : A(x)^2=B(x)^2(x^2+1)$. This is because the ring
${\mathbb K}[X]$ has the unique factorization property, and
$X^2+1$ is square-free in it.
Thus if $(A,B)$ is a solution of $(*)$, we must have
$A(i)=A(-i)=0$, so $X^2+1$ divides $A$ in ${\mathbb K}[X]$, and then
$(B,\frac{A}{X^2+1})$ is another solution of $(*)$. Iterating,
$(\frac{A}{X^2+1},\frac{B}{X^2+1})$ is yet another solution
of $(*)$. We then have a contradiction if we take
$A,B$ minimizing ${\sf deg}(A)+{\sf deg}(B)$.
A: The answer is no.
If $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$, then $p(X,Y)^2 = q(X,Y)^2 (X^2 + Y^2)$ in $\mathbb R[X,Y]$. 
Since $X^2+Y^2$ is irreducible (hence prime) in $\mathbb R[X,Y]$ we get $p(X,Y)=(X^2+Y^2)p_1(X,Y)$, and therefore $(X^2+Y^2)p_1(X,Y)^2=q(X,Y)^2$. Using the same argument we have $q(X,Y)=(X^2+Y^2)q_1(X,Y)$, and thus we get $p_1(X,Y)^2 = q_1(X,Y)^2 (X^2 + Y^2)$. Obviously, $\deg_X p_1<\deg_Xp$ and $\deg_X q_1<\deg_Xq$. Continuing the procedure we obtain a relation of one of the above forms, say $p(X,Y)^2 = q(X,Y)^2 (X^2 + Y^2)$ with $\deg_Xp=1$ and $\deg_Xq=0$. Can you continue from here?
Remark. A slight generalization: in a UFD is not possible to have a relation of the form $a^2=b^2p$ with $a,b$ non-zero and $p$ prime.
