A little idea: if $y=1$, we can prove the surjectivity and the injectivity, it is right?

ThE-dArK-lOrD wrote: Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that $$f(x)+f(yf(x)+f(y))=f(x+2f(y))+xy$$for all $x,y\in \mathbb{R}$. Proposed by Adrian Beker Let $P(x,y)$ denote the equation. First,we will prove that $f$ is injective. Assume the contrary that there exists $p\neq q$ such that $f(p)=f(q),$ then comparing $P(p,y)$ and $P(q,y)$ gives $f(p+2f(y))-f(q+2f(y))=(q-p)y$. However, this means that $(q-p)p=f(p+2f(p))-f(q+2f(p))=f(p+2f(q))-f(q+2f(q))=(q-p)q,$ a contradiction. Hence $f$ is injective. Hence, if $f(0)=0,$ then $P(0,y)$ implies that $f(f(y))=f(2f(y))\implies f\equiv 0,$ clearly not possible. Back to the main problem, we have $P(-f(0),0)\implies f(c)=0$ where $c=-f(0)\neq 0.$ $P(c,c)\implies f(0)=c^2\implies f(0)=1\implies f(-1)=0.$ $P(x,-1) \implies f(-f(x))=-x.$ Thus, $f$ is surjective. $P(x,0)\implies f(x)+f(1)=f(x+2)$ $P(x+2,y)-P(x,y)\implies f(1)+f(yf(x)+f(y)+yf(1))-f(yf(x)+f(y))=f(1)+2y\implies f(yf(x)+f(y)+yf(1))-f(yf(x)+f(y))=2y.$ However, as $f$ is surjective, this means that $f(x+yf(1))-f(x)=2y$ for all $y\neq 0.$ (This is also true for $y=0,$ though.) Thus, $f(yf(1))=2y+f(0)=2y+1$ implies that $f$ is a linear function. Since $f(0)=1$ and $f(-1)=0,$ we have $\boxed{f(x)\equiv x+1}$ which clearly is the solution.

Let $P(x,y)$ denote the given equation We will prove that $f(x) = x+1$ is the only solution and it clearly works. $P(-f(0),0) \Rightarrow f(-f(0))=0$ $P(-f(0),-f(0)) \Rightarrow f(0) = f(0)^2 \Rightarrow f(0) \in \{0,1\}$ If $\boxed{f(0)=0}$ $P(0,y) \Rightarrow f(f(y))=f(2f(y))$

$\Rightarrow f(y) = 0$ for all $y$ which is not possible If $\boxed{f(0)=1}$ $P(x,0) \Rightarrow f(x)+f(1)=f(x+2)$ Since $f(0)=1$ and $f(-1)=0$ $P(x,-1) \Rightarrow f(-f(x))=-x$ so $f$ is bijective. $P(0,y) \Rightarrow 1+f(y+f(y))=f(2f(y))$

So $f(k) = k+1$ for every $k \in \mathbb{Z}$ and $f(x+2)=f(x)+2$ $P(x,2) \Rightarrow f(2f(x)+3)+f(x)=f(x+6)+2x=f(x)+2x+6$ so $f(2f(x)+3)=2x+6$ $\Rightarrow f(2f(x)-1)=2x+2$ From this one and $P(-1,x)$ we get that $f(f(x))=x+2$ $f(f(x))+x=f(2f(x)-1)$ for $x = -f(-y)$ gives that $f(-y)+f(2y-1)=f(y)$ Since $f(f(x))+f(-f(x))=2$ and $f$ is bijective we have that $f(x)+f(-x)=2$ So $f(2x-1)=2f(x)-2$ $1+f(f(y)+y)=f(2f(y))=f(f(2x-1))+2=2x+3 \Rightarrow f(f(x)+x)=2x+2=f(f(2x)) \Rightarrow f(x)+x=f(2x)$ From $P(0,y) \Rightarrow 2y+3=1+f(y+f(y))=f(2f(y))=f(f(y))+f(y)=f(y)+y+2$ So $f(y)=y+1$ for all $y$ $\in \mathbb{R}$

$P(-f(0),0) \implies f(-f(0))=0. P(-f(0),-f(0)) \implies f(0)^2=f(0)$. $i)$ $f(0)=0:$ $P(0,x) \implies f(f(x))=f(2f(x))$. $P(f(x),y), P(2f(x),y) \implies f(f(x)+2f(y))+f(x)f(y)=f(2f(x)+2f(y))+2f(x)y, y \rightarrow f(y)$ and $2f(y) \implies f(x)f(y)=0 \implies f(x)=0$ for all $x$ which is not a solution. $ii)$ $f(0)=1:$ $f(-1)=0, P(x,-1) \implies f(-f(x))=-x \implies f$ is injective and $P(-1,x) \implies f(f(x))=f(2f(x)-1)-x$. $x \rightarrow -f(x) \implies f(-x)=f(-2x-1)+f(x) \implies f(x)=f(2x-1)+f(-x) \implies f(2x-1)+f(-2x-1)=0 \implies *f(x-2)+f(-x)=0$ for all $x$. $x \rightarrow -f(x) \implies f(f(x)-2)=x, x \rightarrow f(x)-2 \implies f(x-2)=f(x)-2 \implies f(x)+f(-x)=2$. $f(x-2)=f(x)-2 \implies **f(x)+4=f(x+4)$. $P(0,x) \implies 1+f(x+f(x))=f(2f(x))$ and $x \rightarrow -f(x) \implies 1+f(-x-f(x))=f(-2x)$. Summing these we get $4=f(-2x)+f(2f(x))$. $$f(-2x)+f(2x+2)^*=f(-2x)-f(-2x-4)^{**}=4=f(-2x)+f(2f(x)) \implies f(2f(x))=f(2x+2) \implies f(x)=x+1$$and it fits.

Assume $f(a)=f(b)$ for some $a,b$. $P(a,x)-P(b,x)\Rightarrow f(a+2f(x))+ax=f(b+2f(x))+bx$, setting $x=a$ and $x=b$ gives $a^2-ab=ab-b^2$ and hence $a=b$. If $f(0)=0$, then $P(0,x)\Rightarrow f(x)=0$ which doesn't fit. Then $f(0)\ne0$ and: $P(-f(0),0)\Rightarrow f(-f(0))=0$ $P(-1,-f(0))\Rightarrow f(-f(-1)f(0))=f(0)\Rightarrow -f(-1)f(0)=0\Rightarrow f(-1)=0$ $P(-1,-1)\Rightarrow f(0)=1$ $P(x,0)\Rightarrow f(x+2)=f(x)+f(1)$ $P(x,-1)\Rightarrow f(-f(x))=-x\Rightarrow f$ is bijective Also, $f(-f(x)-f(1))=f(-f(x+2))=-x-2\Rightarrow f(-f(-f(x))-f(1))=f(x)-2\Rightarrow f(x+f(1))=f(x)+2\Rightarrow f(1)\ne0$. $P\left(f^{-1}\left(\frac{x-f(y)}y\right)+2,y\right)-P\left(f^{-1}\left(\frac{x-f(y)}y\right),y\right)\Rightarrow Q(x,y):f(x+yf(1))=f(x)+2y$ $Q\left(0,\frac x{f(1)}\right)\Rightarrow f$ is linear, using $f(-1)=0$ and $f(0)=1$ we have $\boxed{f(x)=x+1}$.

Observe that $P(-f(0),0)$ gives us $f(-f(0))=0.$ Let $\lambda=-f(0)$ for simplicity. Furthermore, $P(x,\lambda)$ yields $f(\lambda\cdot f(x))=\lambda\cdot x.$ Firstly, $x=\lambda$ gives $f(0)=\lambda^2.$ Now, $x=0$ implies $f(\lambda \cdot f(0))=0.$ Doing the same for $\lambda_0:=\lambda\cdot f(0)$ as we did for $\lambda,$ we get that\[f(\lambda_0\cdot f(x))=\lambda_0\cdot x\underset{x\to\lambda}{\implies}f(0)=\lambda_0\cdot\lambda\underset{\text{def}}{=}\lambda^2\cdot f(0).\]Hence, $\lambda^2=f(0)=\lambda^2\cdot f(0)$ so either $\lambda=0$ (so $f(0)=0$ in this case) or $f(0)=1.$ Before analysing the two cases, let us study the injectivity of $f.$ Assume that $f(a)=f(b).$ Then, $P(a,x)$ and $P(b,x)$ give us $f(a+2f(x))+ax=f(b+2f(x))+bx.$ However, note that \[a(a-b)=f(b+2f(a))-f(a+2f(a))=f(b+2f(b))-f(a+2f(b))=b(a-b)\]and therefore, $a=b$ which implies that $f$ is injective. Now, if $f(0)=0$ it follows by $P(x,0)$ that $f(f(x))=f(2f(x))$ for all $x$ so $f\equiv 0,$ a contradiction. When $f(0)=1,$ coming back to our work above, note that $\lambda=-1$ and thus $f(-f(x))=-x$ for all $x.$ Therefore, $f$ must be bijective. Simply observe that $P(x,0)$ implies $f(x+2)=f(x)+f(1).$ Then,\[2y=P(x+2,y)-P(x,y)=f(yf(x)+f(y)+yf(1))-f(yf(x)+f(y)).\]By the bijectivity of $f,$ when $f(y)\neq 0,$ as $x$ varies, $yf(x)+f(y)$ covers $\mathbb{R}.$ Furthermore, $f(y)\neq 0\iff y\neq 0.$ So for all $y\neq 0$ and all $x$ we have $f(x+yf(1))=f(x)+2y.$ By fixing $x=0$ for example, it follows that $f(y)$ is linear for $y\neq 0.$ A simple computation will yield that $f(x)=x+1$ is the only solution.

Let $P(x,y)$ denote the given assertion. We claim the only solution is $x\mapsto x+1$ which clearly works. Claim: $f(a)=f(b)\implies a=b.$ Proof. Combining $P(a,b)$ with $P(a,a)$ and combining $P(b,a)$ with $P(b,b)$ and then comparing the two gives injectivity. $\blacksquare$ Note that $f(0)\neq 0$ since then $f\equiv 0,$ absurd. So $P(-f(0),0)$ implies $f(0)=1.$ Moreover $P(x,-f(x))$ yields $f$ is an involution. Then $P(x,0)$ gives $f(x)+f(1)=f(x+2).$ In particular $f(-1)=0$ and thus $f(1)=2.$ Setting $f(y)=-f(x)/x$ for $x\neq 0$ and then comparing $P(y+2,x)$ with $P(y,x)$ yields $f(xf(1))=2x+1,$ as desired.

$$f(x)+f(yf(x)+f(y))=f(x+2f(y))+xy$$Only such function satisfying the equation is $f(x)=x+1$. Let $P(x,y)$ be the assertion. Claim: $f$ is injective. Proof: Suppose that $f(a)=c=f(b)$. Compare $P(a,y)$ and $P(b,y)$ to verify the following equation \[f(b+2f(y))+by=c+f(yc+f(y))=f(a+2f(y))+ay\]Choose $y=a$ which yields $f(b+2c)+ab=f(a+2c)=a^2$ and $y=b$ which implies $f(b+2c)+b^2=f(a+2c)+ab$. Thus, \[b^2-ab=f(a+2c)-f(b+2c)=ab-a^2\implies (a-b)^2=0\iff a=b\]Which shows injectivity.$\square$ Claim: $f(0)=1$. Proof: Let $f(0)=c$. $P(x,0)$ gives $f(x)+f(c)=f(x+2c)$. Pick $x=-c$ to get $f(-c)=0$. $y=-c$ in the original equation yields $f(x)+f(-cf(x))=f(x)-cx$ or $f(-cf(x))=-cx$. Choosing $x=0$ gives $f(-c^2)=0$ and $x=-c^2$ implies $c=f(0)=c^3$ hence $c\in \{0,1,-1\}$. If $f(0)=0$, then $P(0,y)$ implies $f(f(y))=f(2f(y))$ however by injectivity, we see that $f\equiv 0$ which is not a solution. Assume that $f(0)=-1$. Then, $f(-1)=0$ and plugging $y=0$ gives $f(x)+f(f(0))=f(x+2f(0))$ or $f(x)=f(x-2)$ which contradicts with injectivity again.$\square$ Claim: $f(1)=2$. Proof: First, see that $P(x,0)$ yields $f(x)+f(1)=f(x+2)$. This gives $f(-2)+f(1)=1$. Since $f(-1)=0$, subsituting $y=-1$ implies $f(x)+f(-f(x))=f(x)-x$ or $f(-f(x))=-x$. We get $f(-f(1))=-1$. \[P(0,-f(1)): \ \ 1+f(-f(1)-1)=f(-2)=1-f(1)\implies f(-f(1)-1)=-f(1)=f(-f(f(1)))\]By injectivity, $-1-f(1)=-f(f(1))$ or $f(2)=f(1)+1=f(f(1))$ or $f(1)=2$.$\square$ We have $f(x)+2=f(x+2)$. Compare $P(x,1)$ with $P(x+2,1)$ to observe \[f(x)+2+f(f(x)+4)=f(x+4)+2+x+2\]Or $f(x)+f(f(x))=f(x)+x+2$ or $f(f(x))=x+2$. Comparing $P(x,y)$ with $P(x+2,y)$ yields \[f(yf(x)+2y+f(y))-f(yf(x)+f(y))=2y\]Since $f$ is surjective, replace $yf(x)+f(y)\rightarrow x$ (for $y\neq 0 $ but it also holds for $y=0$) which implies \[f(x+2y)=f(x)+2y\]Thus, $f(x+y)=f(x)+y$. Plugging $x=0$ gives $f(y)=y+1$ as desired.$\blacksquare$