$F(x)+G(y)= e^{x+y}?$

Question

Are there functions $F(x)$, $G(y)$, such that $F(x)+G(y)=e^{x+y}$ , where $x,y$ are real numbers? I have been trying all elementary functions, and have no clues on what else I could do.

Answer 1

Here is another solution. Applying $\dfrac{\partial^2}{\partial x \partial y}$ to the both sides of $F(x)+G(y)=e^{x+y}$, we have $0=e^{x+y}$. This never holds. Therefore the functions you want do not exist.

Answer 2

No such $F$ and $G$ exist.

The fact that $F(0) + G(y) = e^{y}$ for all real $y$ would imply $G(y) = e^{y} - F(0)$.

Similarly, $F(x) = e^{x} - G(0)$ for all real $x$. No matter how $F(0)$ and $G(0)$ are chosen, you're sunk.

Answer 3

If that were the case, you would have for all $y$, $$F(1) - F(0)=F(1)+G(y) -(F(0)+G(y)) = e^{1+y}-e^y=e^y(e-1),$$ so $e^{y}(e-1)$ would be constant.

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Motivation: Initially I thought of taking a partial derivative, but with more generality, it suffices to take a difference $f(0+1,y) -f(0,y)$.