Find all pairs of functions such that $f(x) +g(y) = x^2y^2$ Find all pairs of functions $f,g\colon \mathbb{R} \to \mathbb{R}$ such that $f(x) + g(y) = x^2y^2$.  If there are no such pairs then explain why?
I believe the answer is that there are no pairs because there is no way to add a function of $x$ to a function of $y$ and get a product of the two.  
 A: Here is a hint. Whenever you see some sort of functional equation like this, it is a good idea to plug in some easy values of $x,y$ to see what happens. For instance, what happens when you plug in $x=0$ and vary $y$? How about $x=1$? Now what about fixing a $y$ value and varying $x$?
A: Let $L(x,y) = f(x)+g(y)$ and $R(x,y) = x^2y^2$. 
The easiest way to see no $f(x)$ and $g(y)$ can make $L(x,y) = R(x,y)$ identically is 
$$\frac{\partial^2}{\partial x\partial y}L(x,y) = 0 \ne 4xy = \frac{\partial^2}{\partial x\partial y}R(x,y)$$
Of course, this argument require $f(x)$ and $g(y)$ to be differentiable. To remove
this restriction, we can "integrate" both sides formally and convert above
contradiction to one that involve only finite differences. 
Notice
$$\begin{align}
  L(0,0) - L(0,1) &- L(1,0) + L(1,1) \\
& = (f(0)+g(0)) - (f(0)+g(1)) - (f(1)+g(0)) + (f(1)+g(1)) = 0\\
  R(0,0) - R(0,1) &- R(1,0) + R(1,1) \\
&=  0^20^2 - 0^21^2 - 1^2 0^2 + 1^21^2 =  1
\end{align}
$$
$L(x,y)$ need to differ from $R(x,y)$ at at least one of following four points
$(0,0), (0,1), (1,0), (1,1)$. Otherwise, we will be faced with the contradiction that $0 = 1$. 
As a consequence, it is impossible to rewrite $x^2y^2$ in the form $f(x) + g(y)$.
