factor $1/(x\pm y)$ as $f(x)g(y)$

Question

Is it possible to rewrite $$\frac{1}{x\pm y}$$ as multiplication of an expression containing only $x$ and another containing only $y$ ($x$ and $y$ are real independent variables), i.e. $$\frac{1}{x\pm y}\stackrel{?}{=}f(x)g(y)$$ If it is possible, what are $f(x)$ and $g(y)$?

Answer 1

No, this is not possible.

For one, you have a problem when taking $x,y$ such that $x+y=0$, since in that case, $\frac{1}{x+y}$ is not defined, while $f(x)\cdot g(y)$ is a well defined number.

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But let's avoid the region around $0$. Say, for example, that $x$ and $y$ are both positive. That means $x+y>0$ and you are in the clear.

Fix $y=1$ and observe your expression. It becomes $$\frac1{x+1}=f(x)g(1)$$ which means that $f(x)=\frac{1}{g(1)(x+1)}$ for all values of $x$. Similarly, you can show that $g(y)=\frac{1}{f(1)(y+1)}$ for all values of $y$. Put the two expressions together, and you get $$f(x)g(y)=\frac{1}{(x+1)(y+1)f(1)g(1)} = \frac{1}{x+y}$$ and this rewrites to $$f(1)g(1)=\frac{x+y}{(x+1)(y+1)}.$$ The right hand side of the equation is a non-constant function (just take $(1,1)$ and $(2,2)$ as examples of input values that do not return the same value), while the left hand side is a constant, which leads to a contradiction.

Answer 2

Sure it's possible. Take $A=\frac{a}{a+b}$ and take $B=0$. Then it should be trivial to show that $$\frac Aa + \frac Bb = \frac{\frac{a}{a+b}}{a} + \frac0b = \frac{1}{a+b}.$$

In fact, taking any value of $B$, you can set $A=a\cdot\left(\frac{1}{a+b} - \frac Bb\right)$ and get the equality $$\frac{A}{a} + \frac Bb = \frac1{a+b}.$$

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However, if you demand that $a,b, A, B$ are all integers, then the answer is no. A simple counterexample can be found by taking $a=b=1$. Then, $\frac{1}{a+b}=\frac12$, while $\frac Aa + \frac Bb = \frac A1 + \frac B1 = A + B$, and since $A$ and $B$ are integers, this means $\frac Aa + \frac Bb$ must also be an integer.

Since $\frac12$ is not an integer, it's clear that $\frac Aa + \frac Bb$ cannot be equal to $\frac1{a+b}$.