Find $f:\mathbb{R}\to\mathbb{R}$ such that $f(xy+x+y)=f(xy)+f(x)+f(y)$

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function that satisfies $$ f(xy+x+y)=f(xy)+f(x)+f(y) $$ Find $f$ and prove that $$ f(x+y)=f(x)+f(y) $$

Answer 1

With $x=y=0$ we find $$f(0)=0.$$ Then with $y=-x$ we find $$f(x)=-f(-x)$$ for all $x\in\mathbb R$

Given $u,v\in\mathbb R$ with $u+v\ne -2$, let $x=\frac{u+v}2$ and $y=\frac{v-u}{u+v+2}$. Then $$ f(v)=f(xy+x+y)=f(xy)+f(x)+f(v)$$ and $$ f(u)=f(-xy+x-y)=-f(xy)+f(x)-f(y)$$ so that $$ f(u)+f(v)=2f(\tfrac{u+v}2).$$ This also holds if $u+v=-2$, for then $-u-v\ne-2$ and $$f(u)+f(v)=-(f(-u)+f(-v))=-2f(-\tfrac{u+v}2)=2f(\tfrac{u+v}2).$$ As also $$f(u+v)=f(u+v)+f(0)=2f(\tfrac{u+v}2)$$ we conclude $$ f(u+v)=f(u)+f(v)$$ for all $u,v\in\mathbb R$.