# Writing $1-e^{-xy}$ as a square.

Is it possible to write $1-e^{-xy} = r(x)r(y)$ for some function $r$ where $x,y$ are positive real numbers. I was just wondering to try to express that quantity like that. I tried solving the equation by Brut force but was not able to make any impact. Any suggestion will be helpful.

• Hint: consider $r(x)r(1)$ and $r(x)r(2)$; their quotient should be constant.... (Indeed, this addresses the more general problem $1-e^{-xy} = r(x)s(y)$.) – Greg Martin Jun 20 '18 at 6:04

That would imply $1-e^{-x^2}=r(x)^2$ and then $r(x)=\sqrt{1-e^{-x^2}}$. Therefore $1-e^{-xy}=\sqrt{(1-e^{-x^2})(1-e^{-y^2})}$. But this is false: try $x=1$ and $y=2$.
Do partial differenciation on the both sides with $x$, and then assume that let $x=y$, then integrate on both sides with $x$ with limits $[o,x]$ as we know $r[o]$ so we can get the $r[x]$.