> [!NOTE] Lemma (IMO shortlist, 1979) > For all functions $f:\mathbb{R}\to \mathbb{R},$ such that for all $x,y\in \mathbb{R},$ $f(xy+x+y)=f(xy)+f(x)+f(y)\tag{1}$then $f(x+y)=f(x)+f(y)$that is, $f$ is additive. ###### Proof Plugging $x=y=0$ gives $f(0)=0.$ Also $y=-x$ gives $f(-x)=-f(x).$ Plugging in $y=1$ we get $f(2 x+1)=2 f(x)+f(1)$ hence for all $u,v\in \mathbb{R},$ $\begin{align} f(2(u+v+u v)+1)&=2 f(u+v+u v)+f(1) \\ &=2 f(u v)+2 f(u)+2 f(v)+f(1) \end{align}$On the other hand, plugging in $x=u$ and $y=2 v+1$ in $(1)$ gives $\begin{align} f(2(u+v+u v)+1)&=f(u+(2 v+1)+u(2 v+1)) \\ &=f(u)+2 f(v)+f(1)+f(2 u v+u) \end{align}$ Hence it follows that $ f(2 u v+u)=2 f(u v)+f(u)\tag{2}$ Plugging in $v=-1 / 2$ we get $0=2 f(-u / 2)+f(u)=-2 f(u / 2)+f(u)$. Hence, $f(u)=2 f(u / 2)$ and consequently $f(2 x)=2 f(x)$ for all real $x.$ Now $(2)$ reduces to $f(2 u v+u)=f(2 u v)+f(u)$. Plugging in $u=y$ and $x=2 u v$, we obtain $f(x)+f(y)=f(x+y)$ for all nonzero reals $x$ and $y$. Since $f(0)=0$, it trivially holds that $f(x+y)=f(x)+f(y)$ when one of $x$ and $y$ is $0$.