Let $P(x,y)$ be the assertion. $P(x,0) \rightarrow f(0)=1$. $P(0,y) \rightarrow f(yf(y))=y^2+f(y)$. Since $f$ is surjective, replace $f(y)$ with $x$, yielding $f(xy)=x+y^2$. $P(x,1) \rightarrow f(x)=x+1$, which works, so we are done.

Taco12 wrote: Let $P(x,y)$ be the assertion. $P(x,0) \rightarrow f(0)=1$. $P(0,y) \rightarrow f(yf(y))=y^2+f(y)$. Since $f$ is surjective, replace $f(y)$ with $x$, yielding $f(xy)=x+y^2$. $P(x,1) \rightarrow f(x)=x+1$, which works, so we are done. 1. $f$'s surjectivity is not trivial. 2. In the last line, you abuse the notation $P(x,y)$, though I presume it means $f(xy)=x+y^2$. 3. When you replace $f(y)$ with $x$ you are no longer able to let $y=1$ and let $x$ vary, since fixing $y$ fixed $x$. Try $P(1,-1)$.

Let's nail this. I claim either $f(x)=1-x$ for all $x$ or $f(x)=x+1$ for all $x$. Denote by $P(x,y)$ the given assertion. Note that $f\not\equiv 0$, hence $P(x,0)$ yields $f(0)=1$. Next, $P(1,-1)$ yields $f(1)f(-1)=0$ as $f(0)=1$. Case 1:$f(1)=0$. Assume $f(1)=0$. $P(x,1-x)$ yields $f(x) = (1-x)^2 + f(x)f(1-x)$. Inserting $1-x$ in place of $x$ in here, we get $f(1-x) = x^2 +f(1-x)f(x)$. Subtracting, we deduce $f(x)-f(1-x) = 1-2x$. Finally, letting $f(1-x)=f(x)-(1-2x)$ in the first equation, we get \[ f(x) = (1-x)^2 + f(x)\bigl(f(x)-(1-2x)\bigr) \iff f(x)^2 +2(x-1)f(x) + (x-1)^2 = 0 \iff \bigl(f(x)+(x-1)\bigr)^2=0. \]Thus $f(x) = 1-x$ for all $x$. Case 2: $f(-1)=0$. In this case, do the exact same analysis to obtain $f(x)=x+1$ for all $x$.

i got f(x) = (1-x)^2 ....

$f(x+yf(x+y))=y^2+f(x)f(y)$ Let $P$ be assertion of FE 1.$P(0,0)$ $f(0)=f(0)^2$ Case1: $f(0)=0$ Then , $P(x,0)$ $\implies$ $f(x)=f(x)f(0)$ contradicition $f$ isn't constant. So,$f(0)=1$ Let's prove : $f$ is injective at $0$ Take $q$ such that $q \ne 0$ and $f(q)=1$ $P(0,q)$ $\implies$ $q=0$ . 2.$P(1,-1)$ $f(1)f(-1)=0$ Case2'1: $f(1)=0$ $P(x,1)$ $f(x+f(x+1))=1=f(0)$ $x+f(x+1)=0$ $f(x)=1-x$ Case2'2: $f(-1)=0$ $P(x,-1)$ $f(x-f(x-1))=f(0)=1$ $f(x)=1+x$ Remark: there isn't pointwise trap

$y=0$ gives $f(x)=f(x)f(0)$,it's easy to see that $f(0)=1$ is the only right case. $y=-x$ gives $1-x^2=f(x)f(-x)$,$x=1$ gives: $f(1)f(-1)=0$. First we'll show that $f(t)=1$ if and only if $t=0$. Put $x=x,y=t-x$ and $x=t-x,y=x$ and $f(t)=1$ which proves this. Case 1:$f(1)=0.$ $y=1$ gives $f(x+f(x+1))=1$ which is our first answer $f(x)=-x+1$. $f(-1)=0$ case is similar to $f(1)=0$.

easy one