# Factoring x + y

I'm trying to find functions $f(x)$ and $g(y)$ such that $$f(x)\cdot g(y) = x + y$$

I can't seem to find a single solution to this problem. Anything I try becomes of the form $f(x,y) \cdot g(y) or f(x) \cdot g(x,y)$

Here is my work so far:

$$f(x)g(y) = x + y$$

$$f'(x)g(y) + f(x)g'(y)\frac{dy}{dx} = 1 +\frac{dy}{dx}$$

$$f'(x)g(y) - 1 = \frac{dy}{dx} - f(x)g'(y)\frac{dy}{dx}$$

$$f'(x)g(y) - 1 = \frac{dy}{dx} (1 - f(x)g'(y))$$

$$\frac{f'(x)g(y) - 1}{1 - f(x)g'(y)} = \frac{dy}{dx}$$

But since $g(y)$ doesn't have a known or restricted order, I have no idea what $f(x)$ would be:

I am considering taking higher derivatives and then using substitutions (ex: $g(y) = (x+y)/f(x))$ but I'm not sure if that will work.

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You have that $$\tag 1 g(0)f(0)=0$$ while $$f(1)g(0)=1$$ $$f(0)g(1)=1$$ The first equations says that either $g(0)=0$ or $f(0)=0$, or both. But this contradicts the last two equations.

ADD Note that what I wrote holds also when $x=-y$, and when $x=0$, $y=\alpha$ a constant, and vice-versa. In fact, the last observation means your function cannot be defined for any value of $x$ or $y$. Indeed, we have that for any $\alpha,\beta\in\Bbb R$

$$f(0)g(\alpha)=\alpha$$ $$f(\beta)g(0)=\beta$$

But beacuse of $(1)$, the last relations are impossible, so we cannot define either $f$ or $g$ for any $x\in\Bbb R$.

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The last line can be simplified to "multiplying the last two relations you get $0=1$." :) –  N. S. May 2 at 19:40
But can the following possible? –  frogeyedpeas May 2 at 19:41
@N.S. "Simplified"? Is it really that complicated? =) –  Pedro Tamaroff May 2 at 19:41
@frogeyedpeas The point of my answer is that no such decomposition can exist. –  Pedro Tamaroff May 2 at 19:41
15 seconds of posting delay is worth commenting about, which is why I'm doing it too! And it's taking me 30+ seconds to do so, too. :) –  Kaz May 2 at 23:20
This would mean that $\frac{a+y}{b+y}$ would have to be a constant function of $y$ for all $a,b$. But $\frac{a+y}{b+y}=1-\frac{b-a}{b+y}$ is only constant when $a=b$.