This problem is proposed by me.

Put $y$ as $x-y$, $f(xf(y))+xf(x)=2x-y+f(x^2)+yf(x)$ Let this be $P(x,y)$ $P(0,1): f(0)=1$ $P(x,0): f(x)+xf(x)=2x+f(x^2)$ (1) $P(1,0): f(1)=2$ $P(1,y): f(f(y))=y+2$ and so $f(y+2)=f(y)+2$ Sub (1) into the original equation, $f(xf(y))+y=yf(x)+f(x)$ Replacing $y$ as $f(y-2)$ in above, $f(xy)=f(x)f(y)-f(x)-f(y)+2$ Put into (1), $f(x)^2-2f(x)+2+2x=xf(x)+f(x)$ Or $(f(x)-x-1)(f(x)-2)=0$. But if $f(c)=2$ for some $c$ then $3=f(2)=f(f(c))=c+1$ so $c=1$. This gives the only solution $f(x)=x+1$

gghx wrote: $P(0,1): f(0)=0$ This is a typo it should be $f(0)=1$. And also can you explain this following part how did you get I don't understand what did you do. gghx wrote: Sub (1) into the original equation, $f(xf(y))+y=yf(x)+f(x)$

@above i replaced the awkward $f(x^2)$ term by using $P(x,0)$. (I hope i didint make a mistake somewhere...) I think my use of the term original is confusing. I mean $P(x,y)$

Very nice problem Let $P(x,y)$ be the assertion $P(0,y)\implies f(0)=1$ $P(1,1)\implies f(1)=2$ $P(1,y)\implies f(f(1-y))=3-y$ implies bijection. Let $g(x)=f(x)-1$ so we have $Q(x,y): g(xg(x-y)+x)+yg(x)=x+g(x^2)$. $Q(x,x)\implies g(x)(x+1)=x+g(x^2)$ $Q\left(x,\frac{x}{g(x)}\right)\implies g(xg\left(x-\frac{x}{g(x)}\right)+x)=g(x^2)$ so bijection $xg\left(x-\frac{x}{g(x)}\right)+x=x^2$ since easy to check $g(0)=0$ say $g\left(x-\frac{x}{g(x)}\right)=x-1$. Now $f(f(1-y))=3-y\implies g(g(1-y)+1)=2-y$ so by $y=2-(x-1)$ we get $g((g(x)-1))=x-1$ thus bijection $$x-\frac{x}{g(x)}=g(x)-1$$$$x+1=g(x)+\frac{x}{g(x)}$$$$x+g(x^2)=g(x)(x+1)=g(x)^2+x$$$$g(x^2)=g(x)^2$$So we have $$g(x)x+g(x)=x+g(x)^2\iff g(x)^2-(x+1)g(x)+x$$Thus $$g(x)=\frac{x+1\pm \sqrt{x^2+2x+1-4x}}{2}=\frac{x+1\pm (x-1)}{2}$$Means $$g(x)=\begin{cases} x\\1\end{cases}$$Moreover $g(x)=1\iff x=1$ so we have $g(x)=x$ and we are done, $f(x)=x+1$.

pablock wrote: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(xf(x-y))+yf(x)=x+y+f(x^2),$$for all real numbers $x$ and $y.$ $P(0,1): f(0)=1$, $P(1,1): f(1)=2$, $P(1,1-x): f(f(x))=x+2..............(1)$ $(1)$ gives that the function is bijective and setting $x \rightarrow f(-x-2)$ gives, $f(-x)=f(-x-2)+2.................(2)$ Now setting $x \rightarrow -1$ we get $f(-1)=0$ from which we get, $P(-1,-1-f(x)): f(-x-2)=-f(x)$ Now plugging this in $P(x,x)-P(-x,-x)$ gives $f(x)=x+1$ as the only solution.

Maybe overcomplicated solution. Let $P(x,y)$ as usual denotes assertion of given functional equation. Step 1: Function $f$ is bijective Notice that $P(0,1)$ gives us $f(0) =1$ and consequently $P(1,1)$ reveals that $f(1) =2$. Observe that $P(1,y)$ yields to: $$ f(f(1-y)) = 3 -y \implies f(f(y))=y+2 $$Thus $f$ is bijective as desired. Step 2: $f(x+1) = f(x) +1 \quad (\star) $ Simply note that $P(x,x)$ gives us: $$ f(x) + xf(x) =2x + f(x)^2 \quad (1) $$Also observe that $P(x,x-1)$ gives us: $$ f(2x) + xf(x) - f(x) = 2x-1 +f( x^2) \quad (2) $$Subtracting $(1)$ from $(2)$ yields: $$ 2f(x) -1 =f(2x) \quad (3) $$In this relation replacing $x$ by $f(x)$ yields: $$ 2x + 3 =f(2f(x)) = f(f(2x+1)) $$Since $f$ is injective we conclude that $2f(x) =f(2x+1)$. Now $(3)$ becomes: $$ f(2x+1) = f(2x) + 1 \implies f(x+1) =f(x) + 1 $$as desired. Step 3: $f(x) + f(-x) =2$ From $(\star)$ it easily follows that $f(-1) = 0$ and that $f(x-1) = f(x) -1 \quad (4)$. Notice that $P(-1,y)$ gives us: $$ f(-f(-1-y)) =-1+y +2 = y+1 =f(f(y-1)) $$Since $f$ is injective we conclude that $-f(-1-y) = f(y-1)$. We can rewrite this as follows using $(4)$ and $(\star)$. $$ 1-f(-y) =f(y) -1 \implies 2 = f(y) + f(-y) $$as desired. Step 4: Finish The key step is to consider $P(-x, -2x)$: \begin{align*} f(-xf(x)) -2xf(-x) =-3x + f(x^2) \\ f(-xf(x)) -2x(2 - f(x)) =-3x + f(x^2) \\ f(-xf(x)) -4x + 2xf(x) = -3x + f(x^2) \\ f(-xf(x)) + 2xf(x) = x + f(x^2) \quad (5) \end{align*}On another hand $P(x,0)$ gives us: $$ f(xf(x)) = x + f (x^2) \quad (6) $$Adding $(5)$ and $(6)$ gives us: $$ 2 +2xf(x) = 2x + 2f(x^2) \implies 1+xf(x) = x +f(x^2 ) \quad (7) $$Finally we observe that $P(x,x)$ gives us: $$ f(x) + xf(x) =2x + f(x^2) \quad (8) $$Subtracting $(8)$ from $(7)$ yields: $$ f(x) = x+1 $$for all real numbers $x$, which clearly works.

After $f(f(x))=x+2$ you have bijectivity. Choose $y$ such that $f(x-y)=x$, then the original equation becomes $yf(x)=x+y$. Now $f(f(x-y))=x-y+2=f(x)$, then we have $y(x-y+2)=x+y$, which means that $y=1$ or $y=x$. The second case also implies $y=1$, so for every $x$ we have $f(x-1)=x$, which provides the solution $f(x)=x+1$

Iberoamerican 2020 Problem 5 wrote: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(xf(x-y))+yf(x)=x+y+f(x^2),$$for all real numbers $x$ and $y.$

Iberoamerican 2020 Problem 5 wrote: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(xf(x-y))+yf(x)=x+y+f(x^2),$$for all real numbers $x$ and $y.$ Nice problem.... My Solution Step 1.$f(0)=1$ Proof: $P(0,x)\implies f(0)=1$ Step 2.$f(1)=2$ Proof: $P(1,1)\implies f(1)=2$ Step 3.$f(f(x))=x+2$ Proof: $P(1,1-x)\implies f(f(x))=x+2,f-bijective$ Step 4.$f(-1)=0$ Proof: $step.3\implies f(f(x))=x+2\implies f(f(t-2))=t\implies f^{-1}(t)=f(t-2)$ $step.1\implies f(0)=1\implies 0=f^{-1}(1)=f(1-2)=f(-1)\implies f(-1)=0$ Srep 5.$f(x^2)=f(x)+xf(x)-2x$ Proof: $P(x,x)\implies f(x^2)=f(x)+xf(x)-2x$ Step 6.$f(-f(x))=-x$ Proof: $P(-1,-1-x)\implies f(-f(x))=-x$ Step 7.$f(x)+f(-x)=2$ Proof: $Step.3,Step.6\implies -x=f(-f(x))\implies f(-x)=f(f(-f(x)))=-f(x)+2\implies f(x)+f(-x)=2$ Step 8.$f(-x)(1-x)=f(x)+xf(x)-4x$ Proof: $Step.5\implies f(x)+xf(x)-2x=f(x^2)=f((-x)^2)=f(-x)+(-x)\cdot f(-x)+2x\implies f(-x)(1-x)=f(x)+xf(x)-4x$ Final: $Step.7,Step.8\implies (2-f(x))(1-x)=f(x)(x+1)-4x\implies f(x)=x+1$ Answer:$f(x)=x+1$

Iberoamerican Math Olympiad 2020 P5 wrote: Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(xf(x-y))+yf(x)=x+y+f(x^2),$$for all real numbers $x$ and $y.$ Let $P(x,y)$ be the given assertion. We prove some Claims. Claim 1: $f(0)=1$. Proof: Just consider $P(0,1)$. $\blacksquare$ Claim 2: $f(xf(y))=(y+1)f(x)-y$ for all $x,y$. Proof: $P(x,x)$ implies $f(x^2)=f(x)+xf(x)-2x \, (*)$, which after substituting to $P(x,y)$ implies $$f(xf(x-y))=(1+x-y)f(x)+y-x$$so by taking $y \rightarrow x-y$ we obtain $$f(xf(y))=(y+1)f(x)-y, \, (**)$$as desired $\blacksquare$ Claim 3: $f(f(x))=x+2$ for all $x$. Proof: $x \rightarrow 1$ at $(*)$ implies $f(1)=2$, hence $x \rightarrow 1$ at $(**)$ yields the desired $\blacksquare$. To the problem, take $x \rightarrow f(x)$ at $f(f(x))=x+2$ to obtain $f(x+2)=f(x)+2$. Now, take $y \rightarrow f(x-2)$ at $(**)$, to obtain $$f(xf(f(x-2)))=(f(x-2)+1)f(x)-f(x-2) \Rightarrow f(x^2)=(f(x)-1)f(x)-(f(x)-2) \Rightarrow f(x^2)=f^2(x)-2f(x)+2 \, (***)$$ Comparing $(*)$ with $(***)$ we easily conclude that $f(x) \in \{2, x+1 \}$ for all $x$. Suppose that there exists a $k$ such that $f(k)=2$. Then, $$k+2=f(f(k))=f(2)=f(0)+2=3,$$hence $k=1$. Consequently, $f(x) \neq 2$ for all $x\neq 1$, that is $f(x)=x+1$ for all $x \neq 1$. In fact, since $f(1)=1+1=2$, we have $f(x)=x+1$ for all $x$. To conclude, $f \equiv x+1$, which satisfies.

The only solution is $f(x)=x+1$. We set $y \rightarrow x-y$ to get that: $$f(xf(y))+xf(x)+y=2x+f(x^2)+yf(x)$$ Now we set $x=0$ to get that $y=y.f(0)$, this implies that $f(0)=1$ Now we set $x=y$ and plug it into the original equation to get that $f(x)+xf(x)=2x+f(x^2)$. Set $x=1$ to get that $2f(1)=2+f(1)$, this implies that $f(1)=2$ Now we plug in $x=1$ into the equation $f(xf(y))+xf(x)+y=2x+f(x^2)+yf(x)$, from which we get that: $$f(f(y))=2+y$$this implies that $f$ is a bijection. By induction we get that, if $x \in \mathbb{N}$ then we have that $f(x)=x+1$ From this fact we set $x$ to be a natural number and plug it into the equation $f(xf(y))+xf(x)+y=2x+f(x^2)+yf(x)$, to get that $f(xf(y))=x(y+1)+1$. Into that equation we plug in $y=-1$ to get that $f(xf(-1))=1=f(0)$. From injectivity we have that $f(-1)=0$. Now set $x=-1$ into the original equation to get that $f(-f(-1-y))=-1+y+2=y+1$, this implies that $f(-f(-t))=t$, for any real $t$. Since we have that $t=f(f(t-2))$, by injectivity we have that: $$-f(-t)=f(t-2)$$ If we set $t$ to be a natural number greater than $1$ to get that $f(-t)=1-t$. Thus if $x \in \mathbb{Z}$ we have that $f(x)=x+1$. Now let $x \in \mathbb{Z}$ and let's plug it into the first equation to get that: $$f(xf(y))=x(y+1)+1$$this implies that if $x=-t$, for some positive integer $t$ we get that: $$f(-tf(y))=1-t(y+1)$$ Thus this implies that $f(tf(y))+f(-tf(y))=2$, by surjectivity we set $x=tf(y)$ and we have that: $$f(x)+f(-x)=2$$for any real number $x$. Now to finish this off we plug in $y=f(x)-2$ into $f(xf(y))+xf(x)+y=2x+f(x^2)+yf(x)$, to get that: $$f(x)^2-(x+3)f(x)+2x+2=0$$by solving the quadratic we have that $f(x)\in\{ x+1,2\}$. Since we have that $f$ is a bijection the only point that can have $f(x)=2$ is only when $x=1$, if it were otherwise we wouldn't have that $f(f(y))=y+2$, thus we have that the only solution is $f(x)=x+1$. We check this by plugging $x+1$ into the equation.

The only working function which satisfies is $f(x)=x+1$ for all $x$. We now show that this is the only one. Let $P(x,y)$ denote the assertion. Note that $P(0,y)$ where $y\neq 0$ yields $f(0)=1,$ then $P(x,x)$ gives $$f(x)+xf(x)=2x+f(x^2)~(\clubsuit)$$Plugging $x=-1$ in last equation yields $f(1)=2$. Next we claim that $f(2x)=2f(x)-1$ holds. To see this, by $P(x,x-1)$ we have $$f(2x)+(x-1)f(x)=2x-1+f(x)^2\overset{(\clubsuit)}{=} f(x)+xf(x)-1$$which rearranges to yield $f(2x)=2f(x)-1,$ as desired. $\square$ Consequently, $$f(2x^2)=2f(x^2)-1\overset{(\clubsuit)}{=}2(f(x)+xf(x)-2x)-1=2f(x)+2xf(x)-4x-1$$ To finish, note by $P(2x,2x)$ and above reasoning we have \begin{align*} &f(2x)+2xf(2x)=4x+f(4x^2) \\ \implies &2f(x)-1+2x[2f(x)-1]=4x+2f(2x^2)-1\\ \implies &2f(x)-6x+\cancel{4xf(x)}=4f(x)+\cancel{4xf(x)}-8x-2\\ \implies &2x+2=2f(x) \implies f(x)=x+1 \\ \end{align*}as desired. $\blacksquare$

\[P(0,y) \implies f(0)+yf(0)=y+f(0) \implies f(0)=1\]\[P(x,,x) \implies f(x)+yf(x)=2x+f(x^2)\]\[P(1,1) \implies f(1)+f(1)=2+f(1) \implies f(1)=2\]\[P(1,y) \implies f(f(1-y))+2y=y+1+2 \implies f(f(1-y))=3-y \implies f(f(y))=y+2\]The last statement also gives bijectivity. Due to that ,for any $x \in R$ there exists some $z$ such that $f(x-z)=x$.Since we know the value for $0,1$ we are gonna assume $x \not=0,1$ \[P(x,z) \implies f(x^2)+zf(x)=x+z+f(x^2) \implies zf(x)=x+z\].we can assume $z \not =0$ since it gives $x=0$.Hence $f(x)=x/z+1$ also,\[x-z+2=f(f(x-z))=f(x)=x/z+1\]\[ \implies x(z-1)/z=z-1 \implies z=1 or x=z\].Later one gives $f(0)=1=x \implies x=1$contradicting our assumption. Hence $f(x-1)=x \forall x$(it is easy to do the case $x=0,1$).Hence only solution is $f(x)=x+1$

The only solution is $f(x)=x+1$. It is easy to see that it works. We now prove that it is the only function. As per usual, denote the assertion by $P(x,y)$. $P(0,y)$ gives that $yf(0)=y\implies f(0)=1$. $P(1,1)$ gives that $f(f(0))=2\implies f(1)=2$. From $P(2,1),$ we get that $f(2)=3$ and from $P(2,2)$ we get that $f(4)=5$. We now have enough to prove the first claim, which will help us prove the key claim that will finish the problem Claim: $f(2x)=2f(x)-1$ $P(x,x)$ gives that $f(x^2)=(x+1)f(x) -2x$ and $P(x,x-1)$ gives that $f(2x)+(x-1)f(x)=2x-1+f(x^2)$. Subsituting in the first equation for the value of $f(x^2)$ suffices. Claim: $f(f(x))=x+2$. $f(2,y)$ gives us that $f(2f(2-y))+yf(2)=y+2+f(4)\implies 2f(f(2-y))-1+3y=y+7\implies 2f(f(2-y))=-2y+8\implies f(f(x))=x+2$. We use the "DURR WE WANT THINGS TO CANCEL" method to set $xf(x-y)=x^2\implies f(x-y)=x$. Notice that there exists a value for which this is true for any $x$ because $f$ is surjective(this can either be seen from $f(f(x))=x+2$ or fixing $y$ in the original equation). This just leaves us with $yf(x)=x+y,$ and to find the value of $y$ from which this works, we have $f(f(x-y))=f(x)\implies x-y+2=f(x)\implies y=x+2-f(x).$ Substituting and solving the resulting quadratic gives the solutions of $f(x)=x+1, 2$. To get rid of the solution $2,$(including the possibility of point wise trap), we suppose that $f(k)=2\implies f(f(k))=f(2)\implies k+2=f(2)$. In the beginning, we proved that $f(2)=3\implies k=1$. Note that we cannot just note that 2 does not satisfy the original equation, because there could still be POINTWISE TRAP. We are now finally done yay.

gghx wrote: Put $y$ as $x-y$, $f(xf(y))+xf(x)=2x-y+f(x^2)+yf(x)$ Let this be $P(x,y)$ $P(0,1): f(0)=1$ $P(x,0): f(x)+xf(x)=2x+f(x^2)$ (1) $P(1,0): f(1)=2$ $P(1,y): f(f(y))=y+2$ and so $f(y+2)=f(y)+2$ Sub (1) into the original equation, $f(xf(y))+y=yf(x)+f(x)$ Replacing $y$ as $f(y-2)$ in above, $f(xy)=f(x)f(y)-f(x)-f(y)+2$ Put into (1), $f(x)^2-2f(x)+2+2x=xf(x)+f(x)$ Or $(f(x)-x-1)(f(x)-2)=0$. But if $f(c)=2$ for some $c$ then $3=f(2)=f(f(c))=c+1$ so $c=1$. This gives the only solution $f(x)=x+1$ I'm new to FE, and probably shouldn't be looking at High School Olympiad problems in the first place, but when you get that $f(y+2)=f(y)+2$, it basically says that for all real numbers, increasing the input by 2 increases the output by 2. Doesn't that mean that it has a constant rate of change, and is, therefore, a line? Once we know that it's a line, we can solve using $f(0)=1$ and $f(1)=2$ or by the fact that the slope is $2/2=1$ and using one point and get $f(x)=x+1$. We can say that it is the only solution since there is only one line that goes through any 2 points or goes through any point with a certain slope. Is there a problem with this solution or is it legit? oh, ok @below

No. For instance take $f\equiv\lfloor x\rfloor+1$.

Let $P(x,y)$ be the assertion $f(xf(x-y))+yf(x)=x+y+f(x^2)$. $P(0,1)\Rightarrow f(0)=1$ $P(-1,-1)\Rightarrow f(1)=2$ $P(1,-x+1)\Rightarrow f(f(x))=x+2\Rightarrow f$ is bijective and $f(x)+2=f(f(f(x)))=f(x+2)$ $P\left(x,x-f^{-1}\left(\frac yx\right)-2\right)-P\left(x,x-f^{-1}\left(\frac yx\right)\right)\Rightarrow Q(x,y):f(y+2x)=f(y)+2f(x)-2\forall x\ne0$, but it clearly holds for $x=0$ also. $Q(x,0)\Rightarrow f(2x)+1=2f(x)$ Then we have $f(y+2x)=f(y)+f(2x)-1$, or $f(x+y)=f(x)+f(y)-1$. Let $g(x)=f(x)-1$, then $g$ is additive and $g(0)=0$. $P(x,y)$ gives $g(xg(x-y)+x)+yg(x)=x+g(x^2)$, and upon setting $x=y$, we see that $g(x)+xg(x)=x+g(x^2)$. Substituting $x\mapsto x+y$ and using additivity, yields, after some simplification, $R(x,y):g(2xy)=xg(y)+yg(x)$. Now $R\left(x,\frac12\right)\Rightarrow g$ is linear, so $f$ also is, and using the fact that $f(x)=x+1\forall x\in\{0,1\}$, we conclude that $\boxed{f(x)=x+1}$ for all $x$.

Actually, I am wondering, in my solution could I just disregard the solution of $2$ as extraneous because it clearly does not work in the original equation? Or do I still have to prove it does not work like I did?

Yes, quadratic formula gives that for each $x$, either $f(x)=2$ or $f(x)=x+1$, so it is a pointwise trap.

Wow this is pretty nice! Let $P(x,y)$ be the given assertion. $P(0,1)$ gives that $f(0) = 1$ $P(1,1)$ gives that $f(1) = 2$ $P(1,1-x)$ gives that $f(f(x)) + 2(1-x) = 2-x + 2 \implies f(f(x)) = x+2$ and so $f$ is bijective. Now, make the substitution $g(x) = f(x) - 1$ to get the new assertion $P'(x,y)$ as $g(x(g(x-y)+1)) + yg(x) = x + g(x^2)$ and we have $g$ is bijective as well, and that $g(0) = 0, g(1) = 1$ and $f(f(x)) = x+2$ becomes $g(g(x)+1) = x+1$ Now, we try to make $x(g(x-y)+1)$ and $x^2$ equal, this happens only when $g(x-y)+1 = x \implies g(g(x-y)+1) = g(x) \implies g(x) = x-y+1 \implies y = x+1-g(x)$ So, $P'(x,x+1-g(x))$ gives that $x = (x+1-g(x))g(x)$, which factorizes as $(g(x)-x)(g(x)-1) = 0$. Since $g(1) = 1$ and $g$ is injective, $g(x) =1$ has only one solution $x = 1$. So, we get that $g(x) = x$ for all $x$ So, substituting back, we get that the only solution is $f(x) = x+1$ for all $x \in \mathbb{R}$

Let $P(x,y)$ be the assertion $P(0,1): f(0)=1$ $P(1,1): f(1)=2 ...(1)$ $P(1,1-x): f(f(x))=x+2$ which implies bijectivity and $f(f(x-y))=x-y+2 ...(2)$ Assume $f(x-y)=x$ for some $y$. This implies $f(f(x-y))=f(x) ...(3)$ and $yf(x)=x+y ...(4)$ (by putting $f(x-y)=x$ in the original equation. Combining $(2)$ and $(3)$ we get, $f(x)=x-y+2 ...(5)$ Combining $(4)$ and $(5)$ we get, $y(1-y)=x(1-y)$ Either $x=y$ or $y=1$ For the sake of contradiction suppose $x=y$. Putting $x=y$ in $(5)$ we get $f(x)=2$. Comparing $f(x)=2$ with $(1)$ and using injectivity we get $x=1$. $x$ is a constant. Contradiction. Thus, $y=1$ is the only possible case. Putting $y=1$ in $(5)$ we get, $f(x)=x+1$

$P(x, y) : f(xf(x-y))+yf(x) = x+y+f(x^2)$ $P(0, 1) : f(0) = 1$ $P(x, x) : f(x^2) = f(x)(x+1) - 2x \implies f(1)=2$ $P(x, 0) : f(xf(x)) = x + f(x^2)$ $P(x, x-y) : f(xf(y)) = f(x)(y+1) - y \implies f(f(y)) = y+2$ $x \leftarrow f(y) \text{ in above} : f(f(y)^2) = f(f(y))(y+1)-y \implies f(f(y))(f(y)+1) - 2f(y) = f(f(y))(y+1)-y \implies \\ (y+2)(f(y)+1) - 2f(y)=(y+2)(y+1)-y \implies yf(y)=y(y+1) \implies f(y)=y+1$

$P(0,y) : f(0) + yf(0) = y + f(0) \implies f(0) = 1$ $P(1,1) : f(1) + f(1) = 2 + f(1) \implies f(1) = 2$ $P(1,1-x) : f(f(x)) + 2 - 2x = 2 - x + 2 \implies f(f(x)) = x + 2$ so we can assume for every $x$ there exists $k$ such that $f(k) = x$. $P(x,x-k) : f(x^2) + (x-k)f(x) = 2x - k + f(x^2) \implies (x-k)f(x) = 2x - k$ $f(x) = f(f(k)) = k + 2$ so $(x-k)(k+2) = 2x-k \implies k(x-k) = k$ so either $k = 0$ or $x - k = 1$. but if $k = 0$ then $x = 1$ so contradiction. so we have $x - k = 1$ or $x = k+1$ so $f(x) = x + 1$.

Let $P(x,y)$ be the given assertion. Trivially we get $f(0)=1$ and $f(1)=2.$ Combining $P(x,x)$ and $P(x,x-1)$ yields $f(2x)=2f(x)-1.$ $P(2x,x)\implies f(2x)+2xf(2x)=4x+f(4x^2)$ $\implies 2f(x)-1+2x(2f(x)-1)=4x+4f(x^2)-3. ~~~~~(*)$ Combining $(*)$ and $P(x,x)$ readily yields, $\boxed{f(x)=x+1}, ~~\forall x\in R,$ which obviously works.

I claim the answer is $f(x)=x+1$, which works upon checking. Now we prove it is the only solution. Let $P(x,y)$ be the given assertion. $P(0,1)$ gives that $f(0)+f(0)=1+f(0),$ so $f(0)=1.$ $P(1,1)$ gives that $f(1)+f(1)=2+f(1),$ so $f(1)=2.$ $P(x,0)$ gives that $f(xf(x))=x+f(x^2).$ $P(x,x)$ gives that $f(x)+xf(x)=2x+f(x^2).$ Claim 1: $f(2x)=2f(x)-1.$ Proof: $P(x,x-1)$ gives that $$f(2x)+(x-1)f(x)=2x-1+f(x^2)\iff$$$$\iff f(2x)+xf(x)-f(x)=f(x)+xf(x)-1 \iff$$$$\iff f(2x)=2f(x)-1.$$ Now from the claim it follows that $f(4x^2)=2f(2x^2)-1=2(2f(x^2)-1)-1=4f(x^2)-3.$ Also, we know that $f(2xf(x))=2f(xf(x))-1=2(x+f(x^2))-1.$ Now consider $P(2x,x):$ $f(2xf(x))+xf(2x)=3x+f(4x^2).$ We have $$2x+2f(x^2)-1+x(2f(x)-1)=3x+4f(x^2)-3\iff$$$$\iff xf(x)+1=x+f(x^2)\iff 2x+f(x^2)-f(x)+1=x+f(x^2)\iff f(x)=x+1$$and we are done.

The only solution is $f(x) = x + 1$, which works. Now we prove it is the only solution. Let $P(x,y)$ denote the given assertion. $P(0,x): f(0) + xf(0) = x + f(0)$, so $xf(0) = x \implies f(0) = 1$. Claim: $f$ is injective. Proof: Suppose $f(a) = f(b)$. $P(x, x - a): f(x f(a)) + (x-a) f(x) = x + (x - a) + f(x^2)$, so $f(xf(a)) + (x-a)(f(x) - 1) = x + f(x^2)$. Similarly, $P(x,x-b)$ gives that $f(xf(b)) + (x-b)(f(x) - 1) = x + f(x^2)$, so\[f(xf(a)) + (x-b)(f(x) - 1) = x + f(x^2) = f(xf(a)) + (x-a) (f(x) - 1) \] Since $f$ is clearly isn't constant at $1$, picking $x$ with $f(x) \ne 1$ gives that $(x-a) (f(x) - 1)) = (x-b) (f(x) - 1)$, so $a = b$. $\square$ (this wasn't even needed oops) $P(1,1): f(f(0)) = 2$, so $f(1) = 2$. $P(1, y): f(f(1 - y)) + 2y = y + 3$, so $f(f(1-y)) = 3 - y$, which implies that $f(f(x)) = x + 2$ for all reals $x$. Thus, $f$ is bijective. Thus, $f(x) + 2 = f^2(f(x)) f^3(x) = f(x+2)$. $P(x, y + 2): f(xf(x-y) - 2x) + yf(x) + 2f(x) = x + y + f(x^2) + 2$. Comparing this with $P(x,y)$ gives that $f(xf(x-y)) - f(xf(x-y) - 2x) - 2f(x) = -2$, so $f(xf(x-y)) - f(xf(x-y) - 2x) = 2f(x) - 2$. Note that for $x\ne 0$, $xf(x-y)$ is surjective over reals as $y$ varies (as $f$ is surjective), so $f(y) - f(y - 2x) = 2f(x) - 2$ for all reals $x,y$ (note that this is trivially true when $x = 0$). Setting $y = 2x$ here gives that $f(2x) - 1 = 2f(x) - 2$, so $f(2x) = 2f(x) - 1$. Hence $f(y) - f(y - 2x) = f(2x) - 1$, so $f(y) - f(y-x) = f(x) - 1 $ for all reals $x,y$. Setting $y \to x + y$ gives that\[f(x+y) - f(y) = f(x) - 1 \implies f(x+y) = f(x) + f(y) - 1\]Now letting $g(x) = f(x) - 1$, we have $g(x+y) + 1 = g(x) + 1 + g(y) + 1 - 1$, so $g(x+y) = g(x) + g(y)$ and thus $g$ is additive. Note that since $f(1) = 2$, $g(1) = 1$. $P(x,x): f(x) + xf(x) = 2x + f(x^2)$. Thus, $g(x) + 1 + x(g(x) + 1) = 2x + g(x^2) + 1$, so $g(x^2) = (x+1) g(x) - x$. Now, $g((x+1)^2) = (x+2) g(x+1) - (x+1) = (x+2) (g(x) + 1) - x - 1$. The LHS becomes\[g(x^2) + g(2x) + g(1) = (x+1) g(x) - x + 2g(x) + 1 = (x+3) g(x) - x + 1\]Thus, $(x+3) g(x) - x + 1 = (x+2) (g(x) + 1) - x - 1$, implying $(x+2) g(x) + g(x) - x + 1 = (x+2) g(x) + 1$, so $g(x) = x$, and therefore $f(x) = x + 1$.