$P(x,y)$ here denotes the assertion $f(xf(x)+2y)=f(x)^2+x+2f(y)$. $P(0,0)\Rightarrow f(0)\in\{-1,0\}$ Case 1: $f(0)=0$ $P(0,x)\Rightarrow f(2x)=2f(x)$ $P(x,0)\Rightarrow f(xf(x))=x+f(x)^2$ $P(2x,0)\Rightarrow f(xf(x))=\frac x2+f(x)^2$ contradiction. Case 2: $f(0)=-1$ $P(0,x)\Rightarrow f(2x)=2f(x)+1$ $P(x,0)\Rightarrow f(xf(x))=f(x)^2+x-2$ $P(2x,0)\Rightarrow f\left(xf(x)+\frac x2\right)=f(x)^2+f(x)+\frac x2-2$ $P\left(x,\frac x4\right)\Rightarrow f\left(xf(x)+\frac x2\right)=f(x)^2+x+2f\left(\frac x4\right)$ So $f(x)=\frac x2+2+2f\left(\frac x4\right)$, spamming $f(2x)=2f(x)+1$ again gives: $$f(x)+\frac x2-2=x+\frac{f(x)-3}2\Leftrightarrow\boxed{f(x)=x-1}$$which is a solution.

@above, $f(x)=x-1$ is a possible solution

Sadly, I didn't have enough time to complete this problem in the test (still left with one step but made some typos and wrote some unnecessary stuffs) Answer: $f(x)=x-1 \ \forall x\in \mathbb{R}.$ It's easy to check that the above function is a solution to our FE, let $P(x,y)$ denote the given FE, we have \[P(0,0) \implies f(0)^2=-f(0) \implies f(0)=0 \ \text{or} -1.\]Case 1: $f(0)=0$ We have \[P(0,x)\implies f(2x)=2f(x)\]\[P(x,0)\implies f(xf(x))=f(x)^2+x.\]Now, \[P(2x,2yf(y)) \implies f(4xf(x)+4yf(y))=4f(x)^2+2x+2f(2yf(y))=4f(x)^2+2x+4f(yf(y))\]we then write \[4f(xf(x)+yf(y))=4(f(x)^2+\frac{x}{2}+f(yf(y)))\]\[f(xf(x)+yf(y))=f(x)^2+\frac{x}{2}+f(yf(y)).\]Choose $x\ne y$ and swap $x,y$ in the above equation, we get \[f(x)^2+\frac{x}{2}+f(yf(y))=f(y)^2+\frac{y}{2}+f(xf(x))\]\[f(x)^2+\frac{x}{2}+f(y)^2+y=f(y)^2+\frac{y}{2}+f(x)^2+x\]\[x=y\]which is a contradiction. $\square$ Case 2: $f(0)=-1.$ Let $f(x)=g(x)-1,$ we can rewrite $P(x,y)$ as \[g(xg(x)-x+2y)=g(x)^2-2g(x)+x+2g(y)\]and $g(0)=0$. We know that \[P(0,x)\implies g(2x)=2g(x)\]\[P\left(x,\frac{x}{2}\right) \implies g(xg(x))=g(x)^2+x-g(x)\]\[P(2x,x)\implies g(4xg(x)) = 4g(x)^2-2g(x)+2x.\]From $P(2x,x),$ \[g(xg(x))=g(x)^2-\frac{g(x)}{2}+\frac{x}{2}\]but comparing it with $P\left(x,\frac{x}{2}\right)$, we get $g(x)=x$ and so $f(x)=x-1$ for all real $x$.

Keith50 wrote:

good solution; good luck with your training

Keith50 wrote: @above, $f(x)=x-1$ is a possible solution I fixed my solution.

Let $P(x,y)$ be the assertion of $$f(xf(x)+2y) = f(x)^2+x+2f(y)$$Now, $$P(0,0) \implies f(0)=f(0)^2+2f(0)\implies f(0)\in\{0,-1\}$$ $\boxed{\textbf{Case 1: } f(0)=0}$ $P(x,0)\implies f(xf(x))= f(x)^2+x...........(1)$ $P(0,y) \implies f(2y)=2f(y)............(2)$ Here, $$P(\frac{x}{2},0)\implies f(\frac{x}{2}f(\frac{x}{2}))=f(\frac{x}{2})^2+\frac{x}{2} $$$$ \implies 4 f(\frac{x}{2}f(\frac{x}{2}))=4f(\frac{x}{2})^2+4\frac{x}{2} $$$$ \implies f(xf(x))=f(x)^2+2x\ \text{[By using (2)]}$$This contradicts with $(1)$. $\boxed{\textbf{Case 2: } f(0)=-1}$ $P(x,0)\implies f(xf(x))= f(x)^2+x -2 ...........(1)$ $P(0,y) \implies f(2y)=2f(y)+1...........(2)$ Here, $$P(2x,-x)\implies f(2xf(2x)-2x)=f(2x)^2+2x+2f(-x)$$$$ \implies f(2x(1+2f(x))-2x) = (1+2f(x))^2+2x+2f(-x)$$$$ \implies f(4xf(x)) = 1+4f(x)+4f(x)^2+2x+2f(-x)$$$$ \implies 1+2f(2xf(x))=3+4f(xf(x)) = 1+4f(x)+4f(x)^2+2x+2f(-x)$$$$ \implies 3+ 4f(x)^2+4x-8=1+4f(x)+4f(x)^2+2x+2f(-x) $$$$\implies 2x-6=4f(x)+2f(-x)$$$$\implies f(-x) = x-2f(x)-3........(3)$$ Now, $$P(x,-\frac{xf(x)}{2})\implies f(0) = f(x)^2+x+2f(-\frac{xf(x)}{2})$$$$\implies -1 = f(xf(x))+2+f(-xf(x))-1$$$$\implies -f(-xf(x)) = f(xf(x))+2$$$$\implies -(xf(x) -2f(xf(x)) -3) = f(xf(x))+2$$$$\implies -xf(x) +f(xf(x)) +1 = 0$$$$\implies f(x)^2 -xf(x) +x-1 = 0$$$$\implies f(x) = x-1 \text { or } f(x) =1$$ It is easy to check that $f(x) = x-1$ works where $f(x) = 1$ doesn’t work for $x\neq 2$.

$P(0,0) \implies f(0) \in {0,-1}$ 1) $f(0)=0$. $P(x,-xf(x)),P(x,0) \implies -f(xf(x))=f(-xf(x))$. Let $a=xf(x)$. $P(a,y) , P(-a,y) \implies a=0 \implies f(x)=0$ for all $x$ which does not fit. 2) $f(0)=-1$ $P(0,x) \implies f(2x)=2f(x)+1, P(x,0) \implies f(xf(x))=f(x)^2+x-2$ $P(2x,-x) \implies f(4xf(x))=(2f(x)+1)^2+2x+2f(-x) \implies f(2f(x))=2f(x)^2+2f(x)+x+f(-x) \implies 2(f(x)^2+x-2)+1=2f(x)^2+2f(x)+x+f(-x) \implies 2f(x)+f(-x)=x-3 \implies 2f(-x)+f(x)=-x-3 \implies f(x)=x-1$ for all $x$ which fits.

About the first case: $P(2x,4y)$ works as well.

We have another way to deduce $f(x)=x-1$ from $f(0)=-1$ as follows: $P(x,0)\Rightarrow f(xf(x))=f^2(x)+x-2$ $P(0,y)\Rightarrow f(2y)=2f(y)+1$ $\Rightarrow f(xf(x)+2y)=f^2(x)+x+f(2y)-1$ Substituting $y\rightarrow \frac{y}{2}$ $\Rightarrow f(xf(x)+y)=f^2(x)+x+f(y)-1\hspace{0.1cm}(1)$ Substituting $x\rightarrow 2x, y\rightarrow -2x$ to $(1)$ we get $$f(2xf(2x)-2x)=f(4xf(x)+2x-2x)=f(4xf(x))=(2f(x)+1)^2+2x+f(-2x)-1=4f^2(x)+4f(x)+2x+2f(-x)+1$$Also we have $$f(4xf(x))=2f(2xf(x))+1=4f(xf(x))+3=4f^2(x)+4x-5$$Thus we get $4f(x)+2f(-x)=2x-6\hspace{0.1cm}(2)$ Substituting $y\rightarrow -xf(x)$ to $(1)$ we get $f(-xf(x))=-f^2(x)-x$ So we get $f(xf(x))+f(-xf(x))=-2\hspace{0.1cm}(3)$ Now let $x\rightarrow xf(x)$ to $(2)$ using $(3)$ we get $2f(xf(x))=2xf(x)-2\Leftrightarrow f(xf(x))=xf(x)-1=f^2(x)+x-2$. Thus $f^2(x)-xf(x)+x-1=0$ So for each $x\in\mathbb{R}$ we must have $f(x)=x-1$ or $f(x)=1$ Let $x=2$, we have $f(2)=1$ and since it's easily seen that $f$ is an injective, we have $f(x)=x-1$ for all $x\in\mathbb{R}$

navi_09220114 wrote: Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x,y$, we have $$f(xf(x)+2y)=f(x)^2+x+2f(y)$$ $P(0,0)$ gives $f(0)=0,-1$ If $f(0)=0$ then : $P(0,y)$ gives $f(2y)=2f(y)$ $P(2x,4y)$ using the above gives: $4f(xf(x)+2y)=2f(2xf(x)+4y)=2f(xf(2x)+4y)=f(2xf(2x)+8y)=f(2x)^2+2x+2f(4y)=4f(x)^2+2x+8f(y)\Rightarrow f(xf(x)+2y)=f(x^2)+\frac{x}{2}+2f(y)$ contradiction at the first. If $f(0)=-1$ $P(0,y)$ gives $f(2y)=1+2f(y)$ (1) $P(x,\frac{x}{2}):f(xf(x)+x)=f(x)^2+x+2f(\frac{x}{2})=f(x)^2+x+f(x)-1$ $P(2x,x)$ with (1) gives: $2f(xf(2x)+x)+1=f(2xf(2x)+2x)=f(2x)^2+2x+2f(x)=4f(x)^2+4f(x)+1+2x+2f(x)\Rightarrow$ $f(xf(2x)+x)=2f(x)^2+2f(x)+x+f(x)\Rightarrow f(2xf(x)+x+x)=2f(x)^2+3f(x)+x+\Rightarrow f(xf(x)+x)=f(x)^2+\frac{3f(x)+x-1}{2}$ So$\frac{3f(x)+x-1}{2}=x+f(x)-1\Rightarrow f(x)=x-1$

Nice problem! I spend a lot of time to this problem(about 3 hours), and I finally solved (another solution, but longer), let's start. $P(0,0)$ gives $f(0)=0$ or $f(0)=-1$, I quickly solved the case $f(0)=0$: $P(0,x) \rightarrow f(2x)=2f(x)$ $(0)$. $P(x,0)$ and $P(x/2,0)$ gives $f(x)^2+2x=4f(x/2)^2+4(x/2)=4f((x/2)f(x/2))=f(xf(x))=f(x)^2+x$ (We used $(0)$), which is contradiction. Then $f(0)=-1$. Then we have $P(0,x)$ $\rightarrow$ $f(2x)=2f(x)-1$ $(1)$ $P(x,0) \rightarrow f(xf(x))=f(x)^2+x-2$ $(2)$ Using $(1)$ and $(2)$ we have $f(2xf(2x))=f(2x)^2+2x-2=(2f(x)+1)^2+2x-2=4f(x)^2+4f(x)+2x-1$ $(3)$ And again using $(1)$ and $(2)$ we have these: $P(x,x/2)=f(xf(x)+x)=f(x)^2+x+2f(x/2)=f(x)^2+x+f(x)-1=f(x)^2+f(x)+x-1$ Then using this we have: $P(x,xf(x)+x)=f(3xf(x)+2x)=f(x)^2+x+2f(xf(x)+x)=3f(x)^2+2f(x)+3x-2$. Then using this we also have: $P(x,(3xf(x)+2x)/2)=f(x)^2+x+2f((3xf(x)+2x)/2)= f(x)^2+x+f(3xf(x)+2x)=4f(x)^2+3f(x)+3x-2$ $(4)$. And we know that: $P(x,(3xf(x)+2x)/2)=f(4xf(x)+2x)=f(2x(2f(x)+1))=f(2xf(2x))$. Then we have $(3)=(4)$, then $4f(x)^2+4f(x)+2x-1=4f(x)^2+3f(x)+3x-2$, which finally gives that $f(x)=x-1$ and $f(x)=x-1$ satisfies the given condition. Then answer is $f(x)=x-1$ for all $x\in\mathbb{R}$.