f(1)=0 (obviously) f(x^2+(x+1)f(x))=(1-x)f(1)=0 Let's prove that f(x) is a one-one-mapping when x=0 f(a)=0 x<-a: f(a^2+yf(a))=(1-a)f(y-a) f(a^2)=(1-a)f(y-a) because f(a^2), (1-a) is a variable, a=1 (f(x)=0 can be a solution too) x^2+(x+1)f(x)=1 f(x)=1-x f(x)=0 or 1-x .

hanulyeongsam wrote: x^2+(x+1)f(x)=1 f(x)=1-x. Not exactly. $x^2+(x+1)f(x)=1$ $\implies$ $f(x)=1-x$ only if $x\ne -1$ It remains to study the case $x=-1$ (not very difficult, but mandatory)

Denote $P(x,y)$ the assertion of the given F.E. $P(1,x)$ gives $f(1+xf(1))=0$, so if $f(1) \ne 0$ then we get $f$ constant (and thus $0$ everywhere) so now suppose $f$ non-constant and thus $f(1)=0$, now by $P(x,x+1)$ we get $f(x^2+(x+1)f(x))=0$, now note that if $f(c)=0$ then $f(c^2)=0$ from here and then from $P(c,x)$ we get that $(1-c)f(x-c)=0$, if there was some $c \ne 1$ then $f$ constant so now we get $f$ injective at $0$ which means $x^2+(x+1)f(x)=1$ or just $f(x)=1-x$ for all $x \ne -1$, to finish set $P \left(x, \frac{-1-x^2}{1-x} \right)$ for all $|x| \ne 1$ to get $f(-1)=(1-x)f \left( \frac{-1-x}{1-x} \right)$, which is just $f(-1)=(1-x) \cdot \frac{2}{1-x}=2$, therefore $f(x)=0$ or $f(x)=1-x$ both work thus we are done .

The only solutions are $f\equiv 0$ and $f(x) = 1 - x$, which work. Now we show they are the only solutions. Clearly the only constant solution is $0$. Now assume $f$ is not constant. Let $P(x,y)$ be the given assertion. $P(1, 0): f(1) = 0$. Claim: $f$ is surjective. Proof: Let $r$ satisfy $f(r) \ne 0$ (since $f$ isn't constant). $P(x, x+r)$ gives that $(1 - x) f(r)$ is in the image of $f$ and this is clearly surjective over reals as $x$ varies. $\square$ $P(-x,0): f(x^2) = (1 + x) f(x)$. Claim: $f$ is injective at $0$. Proof: Suppose otherwise and there existed $c\ne 1$ with $f(c) = 0$. $P(-c,0)$ gives $f(c^2) = 0$. $P(c, x + c): 0 = (1-c) f(x) $. Since $1 - c \ne 0$, we have $f$ is constant at $0$, a contradiction. $\square$ $P(x, x + 1): f(x^2 + (x+1) f(x)) = 0 \implies x^2 + (x+1) f(x) = 1$. Note that if $x\ne -1$, we get that $(x+1) f(x) = 1 - x^2 = (1-x)(1+x)$, so dividing both sides by $x + 1$ gives $f(x) = 1 - x$. Thus, it remains to show that $f(-1) = 2$. However, note that if this wasn't the case, then $2$ wouldn't be in the image, a contradiction to the fact that $f$ is surjective. Hence $f(-1) = 2$ and $f(x) = 1 - x$ for all reals $x$.

The only functions that work are $\boxed{f(x)=0}$ and $\boxed{f(x)=1-x}$ which can both be directly verified. Let $P(x,y)$ be the given assertion. Claim: $f(1)=0$ The assertion $P(1,0)$ suffices. Claim: If $f(a)=0$ then either $a=1$ or $f(x)=0$ for all $x$ Assume $a\neq 1$. Consider the assertion $P(a,y+a)$ then $f(y)=\frac{f(a^2)}{1-a}$, then $f$ as constant which combined with the previous claim finishes. Now assume that $f(1)$ is the only zero of $f$. Then $P(x,x+1)$ gives that $f(x^2+(x+1)f(x))=0$ so we must have $$x^2+(x+1)f(x)=1$$Thus $f(x)=1-x$ for all $x\neq -1$. This can be resolved easily by the assertion $P(2,1)$.