$y=-1$ gives $f(f(-1))=(f(x)+\frac{1}{2})(f(-1)+\frac{1}{2})$. If $f(-1)$ is not equal to $-\frac{1}{2}$, then $f(x)$ must be constant for all $x$. If $f$ is a constant $k$, then the functional equation is equivalent to $k=(k+\frac{1}{2})(k+\frac{1}{2})$ $k=k^2+k+\frac{1}{4}$ $0=k^2+\frac{1}{4}$, which is impossible, contradiction! Therefore, we must have $f(-1)=-\frac{1}{2}$. $y=-1$ gives then $f(f(-1))=0$, so $f(-\frac{1}{2})=0$. $(x,y)=(0,-\frac{1}{2})$ gives $f(0)=(f(0)+\frac{1}{2})(f(-\frac{1}{2})+\frac{1}{2})=\frac{1}{2}(f(0)+\frac{1}{2})$ $2f(0)=f(0)+\frac{1}{2}$ $f(0)=\frac{1}{2}$ If there is a $k \ne -1$ such that $f(k)=-\frac{1}{2}$, then $y=k$ gives $f((1+k)x-\frac{1}{2})=0$, which by varying $x$ causes $f(x)=0$ for all $x$, contradicting there being no non-constant solution. Therefore, $-1$ is the unique $x$ s.t. $f(x)=-\frac{1}{2}$. $y=0$ gives $f(x+\frac{1}{2})=f(x)+\frac{1}{2}$, which implies $f(x)-\frac{1}{2}=f(x-\frac{1}{2})$. $x=-1$ gives $f(f(y)-y-1)=0$, so $-\frac{1}{2}=f(f(y)-y-1)-\frac{1}{2}=f(f(y)-y-\frac{3}{2})$. This implies that $f(y)-y-\frac{3}{2}=-1$, since $-1$ is the unique $x$ s.t. $f(x)=-\frac{1}{2}$. Therefore, $f(y)=y+\frac{3}{2}-1=y+\frac{1}{2}$. Plugging this solution back in, $x+y+xy+1=(x+1)(y+1)$ which is true. Therefore, $f(x)=x+\frac{1}{2}$ is the only solution. Cheers, Rofler

amparvardi wrote: Find all functions $f : \mathbb R \to \mathbb R$ such that \[f\left(x+xy+f(y)\right)= \left( f(x)+\frac 12 \right) \left( f(y)+\frac 12 \right) \qquad \forall x,y \in \mathbb R.\] Let $P(x,y)$ be the assertion $f(x+xy+f(y))=\left(f(x)+\frac 12\right)\left(f(y)+\frac 12\right)$ 1) $f(x+y)=f(x)+f(y)-\frac 12$ $\forall x,y$ and $f(x)=x+\frac 12$ $\forall x\in\mathbb Q$ ============================================================= $P(x,-1)$ $\implies$ $f(f(-1))=\left(f(x)+\frac 12\right)\left(f(-1)+\frac 12\right)$ If $f(-1)\ne -\frac 12$, this implies $f(x)=c$ constant but, plugging this in original equation, we get $c=(c+\frac 12)^2$ and so no solution. So $f(-1)=-\frac 12$ and $f(f(-1))=0$ and so $f(-\frac 12)=0$ $P(0,-\frac 12)$ $\implies$ $f(0)=\frac 12\left(f(0)+\frac 12\right)$ and so $f(0)=\frac 12$ $P(x,0)$ $\implies$ $f(x+\frac 12)=f(x)+\frac 12$ $P(x+\frac 12,y)$ $\implies$ $f(x+xy+f(y)+\frac y2)+\frac 12=\left(f(x)+1\right)\left(f(y)+\frac 12\right)$ Subtracting $P(x,y)$ from this equation, we get : $f(x+xy+f(y)+\frac y2)=f(x+xy+f(y))+\frac 12f(y)-\frac 14$ If $y\ne -1$, and setting then in this equation $x=\frac{u-f(y)}{y+1}$, this becomes $f(u+\frac y2)=f(u)+\frac 12f(y)-\frac 14$, still true for $y=-1$ Setting $u=0$ in this equality, we get $f(\frac y2)=\frac 12f(y)+\frac 14$ and so the equality becomes : $f(u+\frac y2)=f(u)+f(\frac y2y)-\frac 12$ and so $f(x+y)=f(x)+f(y)-\frac 12$ So $g(x)=f(x)-\frac 12$ is such that $g(x+y)=g(x)+g(y)$ and so $g(x)=xg(1)$ $\forall x\in\mathbb Q$ So $f(x)=ax+\frac 12$ $\forall x\in\mathbb Q$ and, setting $x=-1$ in this equality, we get $a=1$ Q.E.D 2) $f(x)$ is injective ============ If $f(y_1)=f(y_2)$ with $T=y_1-y_2\ne 0$ : $f(x+y_1)=f(x)+f(y_1)-\frac 1=f(x)+f(y_2)-\frac 12=f(x+y_2)$ and so $f(x+T)=f(x)$ $P(x+T,y)$ $\implies$ $f(x+xy+f(y)+Ty)=f(x+xy+f(y))$ If $y\ne -1$, and setting then in this equation $x=\frac{u-f(y)}{y+1}$, this becomes $f(u+Ty)=f(u)$, still true for $y=-1$ And this implies that $f(x)=c$ is constant, which is not a solution. So such $y_1,y_2$ dont exist and $f(x)$ is injective. Q.E.D 3) $f(x)=x+\frac 12$ $\forall x$ ==================== Comparing $P(x,0)$ and $P(0,x)$, we get $f(x+\frac 12)=f(f(x))$ and injectivity implies then $f(x)=x+\frac 12$ Q.E.D And since this mandatory value indeed is a solution, we get the result : $\boxed{f(x)=x+\frac 12}$ $\forall x$ *edited* : too late

I have a solution! $P(x,-1)$ $\rightarrow $ $f(f(-1))=(f(x)+\frac{1}{2})(f(-1)+\frac{1}{2}). $ If $f(-1)\not= -\frac{1}{2} $ then $f(x)=const $. Let $f(x)=c$, so $c=(c+0,5)^2$ contradiction. Hence, $f(-1)=-\frac{1}{2},$ $f(-\frac{1}{2})=0$ and $f(0)=\frac{1}{2}.$ Put on original equation of $P(x,0)$, we have $f(x+\frac{1}{2})=f(x)+\frac{1}{2}.$ If $f(a)=-\frac{1}{2}$ then $a=-1.$ Let $y\not=-1$, then $P(\frac{-0,5-f(y)}{1+y}, y)$ $ \rightarrow $ $f(y)=y+\frac{1}{2}.$ it's true at $y=-1.$