Indija TS 2012

Put $x=y=0$; we get $f(0)=3f(0)$ so that $f(0)=0$. Taking $y=-1$, we get $f(-1)=f(x)+f(-1)+f(-x)$ so that $f(-x)=-f(x)$ for all $x\in \mathbb{R}$. Taking $y=1$, we get $f(2x+1)=2f(x)+1$, for all $x\in \mathbb{R}$. Replace $x$ by $u+v+uv$ in this, we get $$f(2(u+v+uv)+1)=2f(u+v+uv)+f(1)=2f(u)+2f(v)+2f(uv)+f(1),(1)$$

for all $u,v\in \mathbb{R}$. Taking $x=u$, $y=2v+1$ in the original equation, we get $$f(u+2v+1+2uv+u)=f(u)+f(2v+1)+f(2uv+u),$$ and this reduces to $$f(2(u+v+uv)+1)=f(u)+f(2v+1)+f(2uv+u),$$ (2) for all $u,v\in \mathbb{R}$. Comparing (1) and (2) and using $f(2x+1)=2f(x)+1$, we get $$f(2uv+u)=f(u)+2f(uv),$$ (3) for all $u,v\in \mathbb{R}$. Put $v=-1/2$ in (3) to get $$0=f(0)=f(u)+2f(-u/2)=f(u)-2f(u/2).$$ This shows that $f(u/2)=f(u)/2$ or $f(2u)=2f(u)$ for all $u\in \mathbb{R}$. Thus $$f(2uv+u)=f(2uv)+f(u),$$ for all $u,v\in \mathbb{R}$. Given any $x\ne 0$ and $y$, we can find $u,v$ such that $2uv=y$ and $u=x$. Thus $f(x+y)=f(x)+f(y)$ for all $x\ne 0$ and $y$. However this also valid for $x=0$. Thus $f(x+y)=f(x)+f(y)$ for all $x,y\in \mathbb{R}$.