GeorgeRP wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(2x + y + f(x + y)) + f(xy) = y f(x) \]for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $f(2x+y+f(x+y))+f(xy)=yf(x)$ 1) If $f(0)\ne 0$, then $\boxed{\text{S1 : }f(x)=2-2x\quad\forall x}$

2) If $f(0)=0$, then $\boxed{\text{S2 : }f(x)=0\quad\forall x}$

A shifted version of IMO 2015/5.

lol i didnt even notice the shifting to 2015/5 (take $g(x) = f(x) + x$, the equation for $g$ is 2015 IMO 5) while doing the problem The only solutions are $f(x) = -2x + 2$ and $f \equiv 0$, which work. Let $P(x,y)$ be the given assertion. $P(0,x): f(x + f(x)) = (x-1) f(0)$. Case 1: $f(0) \ne 0$ . Note that varying $x$ in $P(0,x)$ gives $f$ is surjective. Now picking $x$ where $f(x) = 0$ in $P(0,x)$ gives $(x-1) f(0) = 0\implies x = 1$. Hence $f(1) = 0$ and $f$ is injective at $0$. $P(x,1): f(2x + 1 + f(x+1)) = 0$. Thus, $2x + 1 + f(x+1) = 1$, so $f(x+1) = -2x\implies f(x) = -2(x-1)= -2x + 2$. $\square$ Case 2: $f(0) = 0$. $P(x,-x): f(x) + f(-x^2) = -xf(x)$, so $f(-x^2) = f(x)(-x - 1)$. Setting $x = 1$ here gives that $f(-1) = -2f(1)$. Setting $x = -1$ here gives that $f(-1) = 0$. Hence $f(1) = 0$ also. Now comparing $P(x,-x)$ with $P(-x,x)$ gives that\[ f(x) (-x - 1) = f(-x) (x - 1) \implies f(-x) (x - 1) + f(x) (x + 1) = 0 \ \ \ \ \ \ (1) \]for all reals $x$. Claim: If $f(x) = 0$, then $f(-x) = 0$. Proof: If $x = 1$, this is already proven. If not, simply note that by plugging $x$ in $(1)$ we get $f(-x) (x - 1) = 0$ if $f(x) = 0$, so $f(-x) = 0$ also. $\square$ Claim: If $f(x - 1) = f(x-2) = 0$, then $f(x) = 0$. Proof: $P(-1,x): f(x - 2 + f(x-1)) + f(-x) = 0$. The LHS becomes $f(-x)$, so $f(-x) = 0\implies f(x) = 0$. $\square$ $P(x-1,1): f(f(x) + 2x - 1) = 0$. $P(x,0): f(f(x) + 2x) = 0$. Hence by the claim, we have $f(f(x) + 2x + 1) = 0$. $P(x+1, -1): f(f(x) + 2x + 1) + f(-(x+1)) = -f(x+1)$, so $f(-(x+1)) = -f(x+1)$ for all reals $x$, implying that $f$ is odd. Now, $(1)$ becomes $-f(x) (x - 1) + f(x) (x + 1) = 0$, so $2f(x) = 0\implies f \equiv 0$.