Hey could you post all other problems from turkey tst 2018 BTW: $x=f(y) - 2y$ gives a lot.

$f (x) $ is surjective , so $\forall y \in \mathbb {R} $ , $\exists x \in \mathbb{R} $ s.t. $ f (x)=y $ Let $z=y-2x $ so $z=f (x)-2x $ From $P (z,x) \Rightarrow zf (z) =0 $ Since $f (x) =0 $ is not surjective$ \Rightarrow z=f (x)-2x=0 \Leftrightarrow y=f (x)=2x $ wich satisfy our problem Is it correct , or I missed something?

After getting $\left(f\left(x\right)-2x\right)\cdot f\left(f\left(x\right)-2x\right)=0$, you are not done, as this implies that either $f\left(x\right)=2x$ and/or $f\left(f\left(x\right)-2x\right)=0$. It does not imply only one or only the other for all $x$.

Valikk202 wrote: Since $f (x) =0 $ is not surjective In addition to the problem specified in the above post, it was also assumed that $z$ is surjective which isn't always the case.

This FE is so hard Here is a proof that $f(0)=0$: Suppose that $f(0)\ne0$. As $f$ is surjective, there exists an $a$ such that $f(a)=0$. Set $x=a$ and $y=b$ so that $f(a)=f(b)=0$. The equation reduces to $f(b^2)=f((a+b)^2)$. By symmetry, this means that $f(a^2)=f(b^2)$ for any $f(a)=f(b)=0$. Let $c=f(a^2)$ which is constant over all $a$ with $f(a)=0$. Now, set $y=a$, so the equation becomes $f(a^2)=f((x+a)^2)-xf(x)$. If $a\ne b$, then setting $x=b-a$ gives $f(b-a)=0$ as well. We also have $f((b-a)-b)=f(-a)=0$. This means that for any $f(a)=f(b)=0$ with $a\ne b$, $f(b-a)=0$ and $f(-a)=0$. Also, this tells us that $f((x+a)^2)=f((x+b)^2)$ for all $f(a)=f(b)=0$, so $f((x+a)^2)=f((x+a-b)^2)$. Setting $x=z-a$ gives $f(z^2)=f((z-b)^2)=0$. Now, as $f(b^2)=f((x-b)^2)-xf(x)=f(x^2)-xf(x)$, $f(x^2)-xf(x)=c$ for all $x$. Setting $y=0$ gives $f(xf(0))=f(x^2)-xf(x)=c$ for all $x$, so if $f(0)\ne0$, $f$ is constant, contradicting surjectivity. Now if $f$ were also injective, then setting $x=f(y)-2y$ gives $(f(y)-2y)f(f(y)-2y)=0$, so $f(y)-2y=0$ or $f(f(y)-2y)=0\implies f(y)-2y=0$. This gives $f(x)=2x$ as the only solution. Maybe someone can help finish...

Just an idea, rather than proving $f$ to be injective at for all $x$, proving it injective at $0$ should be easier and that suffices too. Edit: I just realized that your proof is complete. Assuming $a\neq b$ gave $f(x)$ to be constant, thus, $f$ is injective at $0$. Since now, $P(f(x)-2x, x)$ gives \[\begin{cases} f(x)-2x = 0 &\forall x\in\mathbb{A}\\ f(f(x)-2x)=0 &\forall x\in\mathbb{R}\backslash\mathbb{A}\end{cases}\]where $\mathbb{A}$ is an arbitrary subset of $\mathbb{R}$. By injectivity at $0$, $f(f(x)-2x)=0\implies f(x)-2x=0$. Thus, \[f(x)=2x\ \forall x\in\mathbb{R}\]

ABCDE wrote: This FE is so hard Here is a proof that $f(0)=0$: Suppose that $f(0)\ne0$. As $f$ is surjective, there exists an $a$ such that $f(a)=0$. Set $x=a$ and $y=b$ so that $f(a)=f(b)=0$. The equation reduces to $f(b^2)=f((a+b)^2)$. By symmetry, this means that $f(a^2)=f(b^2)$ for any $f(a)=f(b)=0$. Let $c=f(a^2)$ which is constant over all $a$ with $f(a)=0$. Now, set $y=a$, so the equation becomes $f(a^2)=f((x+a)^2)-xf(x)$. If $a\ne b$, then setting $x=b-a$ gives $f(b-a)=0$ as well. We also have $f((b-a)-b)=f(-a)=0$. This means that for any $f(a)=f(b)=0$ with $a\ne b$, $f(b-a)=0$ and $f(-a)=0$. Also, this tells us that $f((x+a)^2)=f((x+b)^2)$ for all $f(a)=f(b)=0$, so $f((x+a)^2)=f((x+a-b)^2)$. Setting $x=z-a$ gives $f(z^2)=f((z-b)^2)=0$. Now, as $f(b^2)=f((x-b)^2)-xf(x)=f(x^2)-xf(x)$, $f(x^2)-xf(x)=c$ for all $x$. Setting $y=0$ gives $f(xf(0))=f(x^2)-xf(x)=c$ for all $x$, so if $f(0)\ne0$, $f$ is constant, contradicting surjectivity. Now if $f$ were also injective, then setting $x=f(y)-2y$ gives $(f(y)-2y)f(f(y)-2y)=0$, so $f(y)-2y=0$ or $f(f(y)-2y)=0\implies f(y)-2y=0$. This gives $f(x)=2x$ as the only solution. Maybe someone can help finish... And İ proved that $f (1)$ is either $0$ or $2$.(if f (0)=0) And it is easy to get $f (x)=2x $ in second case but İ dont know what to do in first case

Answer. The function $x\mapsto 2x$ is the unique solution. Solution. Denote the given equation by $P(x,y)$. $P(f(y)-2y,y)$ gives, for all $y\in\mathbb R$, $$(f(y)-2y)f(f(y)-2y)) = 0$$hence for all $y\in\mathbb R$, either $f(y)=2y$ or $f(f(y)-2y)=0$. Now we consider the set of roots of $f$, denoted by $A$. These are the relevant properties of $A$: 1. If $a\in A$ then Since $f(a) = 0 \neq 2a$, $f(f(a)-2a)=0$ so $-2a \in A$. 2. If $a,b\in A$, from $P(a,b), P(b,a)$ we have $f(a^2) = f((a+b)^2) = f(b^2)$. 3. If $a,b\in A$, comparing $P(x-a,a)$ and $P(x-a,b)$ gives $f(x^2) = f((x+a-b)^2)$ for all $x$. 4. If $a,b\in A$, $P(b-a,a)$ gives $b-a\in A$. Suppose that $A$ contain a nonzero element $u$. Setting $(a,b)=(u,-2u)$ in 3. gives us $f(x^2)=f((x+3u)^2)$ for all $x$. Also, from $4$. we have $u-(-2u)=3u\in A$. Compare $P(x,0)$ and $P(x,3u)$ gives $$f(xf(0)) = f(9u^2).$$If $f(0)\neq 0$ then it follows that $f$ is constant, which is a contradiction, so $f(0)=0$. Now $P(x,0)$ gives $xf(x)=x^2$ for all $x\in\mathbb R$, hence we can rewrite the original equation as $$f(xf(y)+y^2) = f((x+y)^2) - f(x^2).$$ Now setting $x$ as $x+3u$ in the above equation, then comparing, gives $$f((x+3u)f(y) + y^2) = f(xf(y) + y^2)$$for all $x,y\in\mathbb R$. Finally, setting $x=-\frac{y^2}{f(y)}$ gives $f(3uf(y)) = f(0)$ for all $y\in\mathbb R$ such that $f(y)\neq 0$. However, $f$ is surjective, so $3uf(y)$ is surjective, so $f$ must be constant, which is a contradiction. Therefore $A$ cannot contain a nonzero element. Now if $f$ is not the function $x\mapsto 2x$ then, for some $x_0$, $f(x_0)\neq 2x_0$, which implies $f(x_0)-2x_0\in A$, which is impossible. Therefore, the function $x\mapsto 2x$ is the only solution. $\blacksquare$

What an excellent FE, here's a different approach. First note that $P(f(y)-2y,y)$ implies that $(f(y)-2y)f(f(y)-2y))=0$ for all real $y$. Suppose $f(a) = 0$ with $a \neq 0$. The above gives $f(-2a) = 0$ and similarly we get $f(4a) =0$. Next we have $P(a,a)$ implies $f(a^2)=f(4a^2)$. Combining $P(-3a,a)$ with the above we have $f(-3a)=0$. Then we have $P(3a,-2a)$ gives $f(3a)=0$. Let $b=3a$ and note that $P(x,b)$ is $f(b^2)=f((x+b)^2)-xf(x)$. While $P(-x,-b)$ gives $f(b^2) = f((x+b)^2) +xf(-x)$ which means $f(x) = -f(-x)$ for all $x \neq 0$. Thus $f(-a) = f(a) = 0$. We have $P(x+a,a)$ gives $f(a^2) = f((x+2a)^2) -(x+a)f(x+a)$ while $P(x,2a) = f(4a^2)=f(a^2) = f((x+2a)^2) -xf(x)$ so we have $(x+a)f(x+a) = xf(x)$. Meanwhile $P(x+a,-a)$ gives $f(a^2)=f(x^2)-(x+a)f(x+a) = f(x^2)-xf(x)$. We can finally find $f(0)$ by $P(x,0)$ which gives $f(xf(0))=f(x^2)-xf(x) = f(a^2)$ which contradicts surjectivity for $f(0) \neq 0$. Therefore $f(0) = 0$ and we have $f(x^2)=xf(x)$ for all $x$. Furthermore $f(a^2) = af(a) = 0$ so $P(x,a)$ gives $f(a^2) = 0 = f((x+a)^2) - xf(x)$ so $f((x+a)^2) = f(x^2)$. Since $f$ is surjective we can find $c$ such that $f(c)=1$. Note that $c \neq 0$ by the above and $f(c^2)=cf(c)=c \neq 0$. $P(a,x)$ gives $f(af(x)+x^2)=f((a+x)^2)= f(x^2)$. And substituting $x=c$ we get $f(a+c^2) = f(c^2)$. Meanwhile $(a+c^2)f(a+c^2) = c^2f(c^2)$ and since $f(c^2) \neq 0$ we have $a+c^2 = c^2$ which means $a=0$, a contradiction. Therefore $f(0) =0$ is the only zero of the function. Thus $P(f(x)-2x,x)$ implies that $(f(x)-2x)f(f(x)-2x))=0$ for all real $x$ which means $f(x) \neq 2x \rightarrow f(f(x)-2x)=0 \rightarrow f(x)-2x = 0$ a contradiction. Therefore $f(x) = 2x$ for all real $x$ which indeed works.

Any solution without using surjectivity?

I have a solution without surjectivity i will post it soon

Let $P(x,y)$ be the assertion we will proceed without using surjectivity Let $A$ be a set of reals like $x$ such that $f(x)=0$ $CLAIM(1)$ : for every real $x$ we have $A$ contains $f(x)-2x$ $Proof(1)$ : see $P(f(x)-2x,x)$ $CLAIM(2)$ : if $x^2>0$ and $f(x)=0$ then we have $f$ is constant $Proof$ : $SUBCLAIM(1)$ : if $A$ contains $x$ it contains $-2x$ as well $Sub Proof(1)$ : from $P(f(x)-2x,x)$ we have that $(f(x)-2x)f(f(x)-2x))=0$ Let us call that assertion $Q(x)$ if $A$ contains $x$ from $Q(x)$ we have that it contains $-2x$ aswell $SUBCLAIM(2)$ : if $A$ contains $x,y$ Then we have $f(x^2)=f(y^2)$ $Sub Proof(2)$ : let us have that $x,y$ are elements of $A$ then from $P(x,y)$ we have $f(x^2)=f((x+y)^2)$ and by switching $x$ and $y$ we have $f(x^2)=f(y^2)$ $SUBCLAIM(3)$ : if $x$ and $y$ are distinct elements of $A$ then so is $x-y$ $Sub Proof(3)$ : check $P(x-y,y)$ Take note that now if $a$ is an element of $A$ and $z$ is an integer then we have that $az$ is an element of $A$ We shall call a function $mamdali$ if for a constant $a,b$ we have $f(x)=f(ax+b)$ $SUBCLAIM(4)$ : $f$ is $mamdali$ $Sub Proof(4)$ : by comparing $P(x,y) , P(x,z)$ such that $y,z$ are distinct elements of $A$ we have $f((x+y)^2)=f((x+z)^2)$ so we have $f(x^2)=f((x+az)^2)$ for integer $z$ and $a$ from $A$ By comparing $P(x,y) , P(x,y+a)$ such that $a$ is an element of $A$ we have : $f(xf(y)+y^2)=f(xf(y+a)+(y+a)^2)$ so the function is $mamdali$ if the function is not $0$ constant but the zero constant function is also $mamdali$ so we have $f$ is $mamdali$ $SUBCLAIM(5)$ : for a nonzero $k$ we have $f(x)=f(x+k)$ $Sub Proof$ : by checking $P(x+na,a)$ such that $n$ is integer and $a$ is an element of $A$ we have $f((x+(n+1)a)^2) - (x+na)f(x+na)=f(x^2)-xf(x)$ But we also have $f((x+(n+1)a)^2)=f(x^2)$ So it follows $xf(x)=(x+na)f(x+na)$ call this $S(x,n)$ Now by comparing $P(x,y)$ and $P(x+a,y)$ we have $f(xf(y)+y^2)=f(xf(y)+y^2+af(y))$ now we have $f$ is $0$ constant or there exists $k$ such that $f(x)=f(x+k)$ Now by checking $S(x+k,n) -S(x,n)$ we have $kf(x)=kf(x+na)$ so $f(x)=f(x+na)$ so $(na)f(x+na)=0$ so $f$ is constant function Proof complete Now we have $f(x)=0$ if and only if $x=0$ $CLAIM(3)$ : $f(x)=2x$ $Proof(3)$ : By $Q(x)$ we have that either $f(x)=2x$ or $f(f(x)-2x)=0$ both of them are true only when $f(x)=2x$ and we are done

Can someone please show me how to prove (only) injectivity at this problem

Can pco help... or any other aopser

Maxito12345 wrote: Can pco help... or any other aopser You wait somone to answer you after 2 years??

Let $f(0) = c$ $P(1,0) : f(f(0)) = 0 \implies f(c) = 0$ Now Assume there exists $k$ such that $f(k) = 0$ and $k \neq c$. $P(k,c) , P(c,k) : f(c^2) = f((c+k)^2) = f(k^2)$ $P(k-c,c) : f(c^2) = f(k^2) - (k-c)f(k-c) \implies (k-c)f(k-c) = 0 \implies f(k-c) = 0$ so we have $f((k-c)-k) = 0$ so $f(-c) = 0$. $P(x,c) , P(x,-c) : f(c^2) = f((x+c)^2) - xf(x) = f((x-c)^2) - xf(x) \implies f((x+c)^2) = f((x-c)^2)$ $P(x,c) , P(x,k) , P(x,k-c) : f((x+c)^2) = f((x+k)^2) = f((x+k-c)^2)$ so $P(x-k,k) , P(x-k,k-c) : f(x^2) = f((x-c)^2) = f((x+c)^2)$ $P(x,0) : f(xc) = f(x^2) - xf(x) = f((x-c)^2) - xf(x) = f(c^2) \implies f(x) = f(c^2)$ so $f$ is constant but $f$ is surjective so contradiction so there exists no such $k$ so if $f(t) = 0 \implies t = c$. $P(f(y)-2y,y) : -(f(y)-2y)f(f(y)-2y) = 0$ so either $f(y) = 2y$ or $f(f(y)-2y)) = 0$. $f(y) = 2y$ is clearly an answer for $f$ so let's assume $f(f(y)-2y) = 0$. since we had if $f(t) = 0 \implies t = c$, Now we have that $f(y)-2y = c$ so $f(y) = 2y+c$ but now $P(y,y)$ gives contradiction unless $c = 0$ which in this case we have $f(y) = 2y$ so for all $y$, $f(y) = 2y$ and it's the only answer.

Let $P(x,y)$ denote $f(xf(y)+y^2)=f((x+y)^2)-xf(x)$. Lemma: There exists $m$ such that $f(t)=0$. Moreover if $f(s)=f(t)=0$ then $s=t$. Proof. Existence is obvious. Let $f(s)=f(t)=0$, then $P(s,t)$ and $P(t,s)$ yields $f(s^2)=f(t^2)$. Let $p=s-t$, and compare $P(x-t,s)$ and $P(x-t,t)$, then $f((x+p)^2)=f(x^2)$. Now $P(x,s)$ and $P(x+p,t)$ yields $xf(x)=(x+p)f(x+p)$. Finally compare $P(x,y)$ and $P(x+p,y)$ and use surjectivity, $f(x)=f(x+py)$, forcing $p=0$. $\blacksquare$ Using Lemma and $P(f(x)-2x,x)$ yields $f(x)\in \{2x,2x+t\}$. Casework in $P(x,y)$, we can easily conclude $f(x)=2x$ for all $x$. This clearly works.

Mahdi_Mashayekhi wrote: $P(x,0) : f(xc) = f(x^2) - xf(x) = f((x-c)^2) - xf(x) = f(c^2) \implies f(x) = f(c^2)$ so $f$ is constant but $f$ is surjective so contradiction so there exists no such $k$ so if $f(t) = 0 \implies t = c$. This proves $c=f(0)=0$ but not "so there exists no such $k$", no? For the "$f(t) = 0 \implies t = c$", you'll have to show a little more, as done in above solutions.

The only solution is $\boxed{f(x) = 2x}$, which works. Let $P(x,y)$ denote the given assertion. $P(f(x) - 2x,x): (f(x) - 2x) f(f(x) - 2x) = 0$. Let $f(0) = a$. $P(1,0): f(a) = 0$. Claim: $f$ is injective at $0$. Proof: Suppose there existed $b\ne a$ such that $f(b) = 0$. $P(a,x): f(af(x) + x^2) = f((a+x)^2)$. $P(b,x): f(bf(x) + x^2) = f((b+x)^2)$. Comparing $P(a,b)$ and $P(b,a)$, we get $f(a^2) = f(b^2)$. $P(x,a): f(a^2) = f((a+x)^2) - xf(x)$. $P(x,b): f(b^2) = f( (b+x)^2) - xf(x)$. So $f((a+x)^2) = f((b+x)^2) $, which implies\[f(x^2) = f((x+c)^2),\]where $ c = b-a\ne 0 $. By $P(x,a)$, \[f(a^2) = f((a+x)^2) - xf(x)\]Setting $x\to x+c$ here gives $xf(x) = (x+c)f(x+c)$. Comparing $P(x,y)$ with $P(x+c,y)$ gives\[f(xf(y)+ y^2) = f((x+c)f(y) + y^2)\]Let $f(k) = 1$ by surjectivity. Then we have $f(x + k^2) = f(x+k^2 + c)$. Since $x + k^2$ can take any real number, we have $f(x) = f(x+c)$. So\[xf(x)= (x+c) f(x+c) = (x+c) f(x).\]Since $c\ne 0$, $f\equiv 0$, which is not surjective, contradiction. $\square$ Now using $P(f(x) - 2x,x)$ we get that $f(x) \in \{2x, 2x+a\}$ for each $x$. So $0 = f(a) \in \{2a, 3a\}$, which implies $a=0$. Thus $f(x) = 2x$ for all $x$.

We'll prove the only solution is $f(x)=2x$ for all real $x$, which apparently works. Let $P(x,y)$ be the given equation. $P(f(0),0)\colon\text{ } f(0)f(f(0))=0\implies f(f(0))=0.$ $P(f(x)-2x,x)\colon\text{ } (f(x)-2x)f(f(x)-2x)=0\implies f(x)=2x \text{ or }f(x)-2x\text{ is a zero of }f$ $(*).$ Our goal is to prove that the only zero of $f$ is $f(0)$. Assume the contrary, so there are reals $p\neq q$, such that $f(p)=f(q)=0$. $P(p,q)\colon\text{ }f(q^2)=f((p+q)^2)$ $P(q,p)\colon\text{ }f(p^2)=f((p+q)^2)$ The last provides us that $f(p^2)=f(q^2)$. Now from $P(x,p)\text{ and }P(x,q)$ we get that $f((x+p)^2)=f((x+q)^2)$, equivalently, $f(x^2)=f((x+r)^2)\text{, }\forall x\in\mathbb{R}$, where $r=|p-q|\neq 0.$ Comparing $P(x,p)$ and $P(x+r,p)$ we get that $xf(x)=(x+r)f(x+r)\text{, }\forall x\in\mathbb{R}$. Now we compare $P(x,y)$ and $P(x+r,y)$. It gives us that \[f(xf(y)+y^2)=f(xf(y)+y^2+rf(y)).\]By surjectivity we can take reals $\alpha, \beta\colon f(\alpha)=1, f(\beta)=2$. Also, lets take $y_0$ in way that $f(y_0)=\frac{\beta-\alpha}{r}$ and $x_0=\frac{(\alpha-y_0^2)r}{\beta-\alpha}=\frac{\alpha-y_0^2}{f(y_0)}$. Now plug $x_0\text{ and }y_0$ in the last equation to get \[1=f(\alpha)=f(\beta)=2\text{ -- contradiction!}\]That proves the desired uniqueness. Now, assume that $f(0)\neq 0$. Because of $f(f(0))=0\neq2f(0)$, then $f(0)=f(f(0))-2f(0)=-2f(0)\implies f(0)=0$ -- contradiction. So, $f(0)=0$ and from $(*)$ the desired conclusion easily follows.

Denote the assertion as $P(x, y)$. First, plugging in $P(1, 0)$ shows that $f(f(0)) = 0$. Claim: $f(0)$ is the only real number satisfying $f(u) = 0$, and $f(0) = 0$. Proof: Assume by contradiction that $f(u) = 0$, $u \neq f(0)$. Note that $P(u, f(0))$ and $P(f(0), u)$ implies that \[ f(u^2) = f((u+f(0))^2) = f(f(0)^2) \]However, note that $P(x, u)$ implies that \[ f(u^2) = f((x+u)^2) - xf(x) \]and $P(x, f(0))$ implies that \[ f(f(0)^2) = f((x+f(0))^2) - xf(x) \]Thus, we must have that $f(x^2) = f((x+c)^2)$ where $c = u - f(0)$. Note that plugging in $x = c$ into above implies that \[ f(f(0)^2) = f(u^2) - cf(c) \implies f(c) = 0 \]But then, using $P(x, 0)$ and $P(x, c)$, we have that \[ f(xf(0)) + xf(x) = f(x^2) = f((x+c)^2) = f(c^2) + xf(x) \]Thus $f(xf(0)) = f(c^2)$ is constant, which is only possible if $f(0) = 0$. But then, this implies that $f(x^2) = xf(x)$. Thus, since $f(x^2) = f((x+c)^2)$, we must have that $xf(x) = (x+c)f(x+c)$. Then, taking $P(x, y)$ and $P(x+c, y)$ implies that \[ xf(x) + f(xf(y)+y^2) = f((x+y)^2) = f((x+y+c)^2) = (x+c)f(x+c) + f((x+c)f(y) + y^2) \]This implies that $f(xf(y) + y^2) = f((x+c)f(y) + y^2)$. But taking the $w$ such that $f(w) = 1$ (as $f$ is surjective), we must have that \[ f(x+w^2) = f(xf(w) + w^2) = f((x+c)f(w) + w^2) = f(x+c+w^2) \]This implies that $f(x) = f(x+c)$ for all reals $x$. But since $xf(x) = (x+c)f(x+c)$, then $x = x+c$, and thus $c = 0$, a contradiction. $\square$ Then if we use $P(f(y) - 2y, y)$, terms cancel and we have that \[ (f(y)-2y)f(f(y)-2y) = 0 \]But this implies that $f(y) = 2y$, which works. $\blacksquare$

Hi @megarnie !! Denote the assertion by $P(x,y).$ First suppose there exist $a,b$ distinct with $f(a)=f(b)=0.$ First $P(a,b)$ and $P(b,a)$ gives $f(a^2)=f(b^2).$ Next, $P(x,a)$ and $P(x,b)$ give $f((x+a)^2)=f((x+b)^2)$ so letting $c=b-a$ we get $f(x^2)=f((x+c)^2).$ Now $P(x,a)$ and $P(x+c,a)$ give $xf(x)=(x+c)f(x+c).$ Now take $P(x+c,y)$ to get $f(xf(y)+y^2)=f((x+c)f(y)+y^2).$ Since $f$ is surjective, we get $f(kx+h^2)=f(kx+kc+h^2)$ for any real $k$ and constant $h,$ so $f$ is periodic with any period, so $f$ is constant at $f(x)=0$ but this is not surjective, contradiction. Thus, there is exactly one $a$ with $f(a)=0.$ Take $P(f(y)-2y,y)$ to get that for all $y$ either $f(y)=2y$ or $f(y)=2y+a.$ Setting $y=a$ gives $f(a)=2a$ or $3a,$ so $a=0$ and we get $f(y)=2y$ for all $y,$ which works. Thus, we have $f(x)=2x$ is the only solution.

I will find all functions satisfying the above without assuming surjectivity. Let $P(x,y)$ be the assertion that $f(xf(y)+y^2) = f((x+y)^2) - xf(x)$ for all $x,y\in\mathbb R$. The two solutions are $\boxed{f(x) = 0}$ and $\boxed{f(x) = 2x}$ which clearly work. Let us now assume that $f$ is not constant ($f(x) =0$ is the only constant solution). $P(f(0),0)\implies f(f(0)) = 0$. Let $a = f(0)$ Let $Z = \{z:f(z)=0\}$ Let's assume that $Z$ has more than one distinct element. Throughout we will denote $u\ne v\in Z$ Claim 1: $f(u^2) = f(v^2)$ for all $u,v\in Z$ Let $c = f(f(0)^2)$. Claim 1 says that $f(u^2) = c$ for all $u\in Z$

Claim 2: $f(u-v) = 0$ for all $u\ne v\in Z$

Claim 3: $f(x^2) = f((x+v-u)^2)$ for all $x\in\mathbb R$, and $u,v\in Z$

Claim 4: $0\in Z$

Claim 5: $xf(x) = (x+u)f(x+u)$ for all $x\in\mathbb R$, $u\in Z$ Call this expression $Q(x, u)$.

Now, let $u\in Z$ and $f(s) = t\ne 0$ $P\left(\frac{x-s^2}{t} + u, s\right) - P\left(\frac{x-s^2}{t}, s\right)\implies f(x+tu) = f(x)\, \forall x\in\mathbb R,\, t\in f(\mathbb R),\,u\in Z$ Now, $Q(x,u)-Q(x + tu, u)\implies f(x) = f(x+u)\implies uf(x) =0$ which is a contradiction. So $Z$ has only one element, namely $a$. $P(f(x)-2x, x)\implies (f(x)-2x)f(f(x)-2x) = 0$ Thus $f(x) \in \{2x, 2x+a\}$ for all $x$. But $0 = f(a) \in \{2a,3a\}$ which can only work if $a=0$. Thus $\boxed{f(x) = 2x}$ $\square$

Denote the assertion as $A(x,y)$. Substituting $A(f(y)-2y,y)$ tells us either $f(y)=2y$ or $f(f(y)-2y) = 0$ for each $y \in \mathbb{R}$. Now suppose the set of zeros to this function is $S$: If $a, b \in S$, then \begin{align*} A(a,b), A(b,a) &\implies f(a^2) = f(b^2), \\ A(x-a,a), A(x-a,b) &\implies f(x^2) = f((x-a+b)^2), \\ A(b-a,a), a \neq b &\implies b-a \in S. \end{align*} If $0 \neq a \in S$, then $-2a, 3a \in S$, $f(x^2) = f((x+3a)^2)$, and \begin{align*} A(x,3a), A(x,0) \implies f(xf(&0)) = f(9a^2) \implies f(0) = 0 \implies f(x^2)=xf(x), \\ A(x,3a), A(x+3a,y), A(x,y) &\implies f(xf(y)+y^2+3a \cdot f(y)) = f(xf(y)+y^2). \end{align*} Thus we can cancel by substituting $x=-\frac{y^2}{f(y)}$, where $f(y) \neq 0$, to find $f(3a \cdot f(y)) = 0$, or $f(x)=0$ for all $x \neq 0$ by surjectivity, contradiction. Hence the only zero of $f(x)$ is $x=0$. From here, our pointwise trap has a simple finish; suppose some $t$ exists such that $f(t) \neq 2t$. Then we must have \[f(f(t)-2t) = 0 \implies f(t) - 2t = 0,\] contradiction. Thus our only solution is $\boxed{f(x)=2x}$, which works. $\blacksquare$

The answer is $f(x)=2x$ only, which works. We first try to make things cancel by setting $$xf(y)+y^2=(x+y)^2\rightarrow x=f(y)-2y$$to get that for each $y$, either $f(y)=2y$ or $f(f(y)-2y)=0$. From $(x,y)=(1,0)$ we have $f(f(0))=0$. Note that injectivity at $0$ would finish here, as then we would have $f(y)-2y=f(0)$ so $f(y)=2y+c$ but we can see only $c=0$ works. Now, the rest of the problem is showing injectivity at $0$. Suppose for the sake of contradiction that $f(a)=f(b)=0$ for $a\neq b$. Of course, now the most natural substitutions to make are $x=a$, $x=b$, $y=a$, and $y=b$. Here they are: $$f(af(y)+y^2)=f((a+y)^2)$$$$f(bf(y)+y^2)=f((b+y)^2)$$$$f(a^2)=f((x+a)^2)-xf(x)$$$$f(b^2)=f((x+b)^2)-xf(x)$$and label these equations $(1),(2),(3),(4)$ in that order. By plugging $x=b$ into (3) we have $$f(a^2)=f((a+b)^2).$$However, plugging $x=a$ into (4) yields $$f(b^2)=f((a+b)^2),$$and thus $f(a^2)=f(b^2).$ Thus, $$f((x+a)^2)=f((x+b)^2).$$In other words, $g(x)=f(x^2)$ is periodic with period $b-a$. Let $P=b-a$. Now, consider equation (3), which is $$f(a^2)=f((x+a)^2)-xf(x).$$If $x$ is increased by $P$, then we know $f((x+a)^2)$ does not change, therefore $xf(x)$ does not change either. Thus, $h(x)=xf(x)$ is periodic with period $P$ as well. Finally, go back to our original equation, $$f(xf(y)+y^2)=f((x+y)^2)-xf(x).$$Since $f((x+y)^2)$ and $xf(x)$ are both periodic with period $P$, this means that $$k(x)=f(xf(y)+y^2)$$also is for any fixed $y$. Thus, $$f(xf(y)+y^2)=f(xf(y)+y^2+Pf(y)).$$For any fixed $y$ such that $f(y)\neq 0$, $xf(y)+y^2$ can take on any real value, meaning that $f$ is periodic with period $Pf(y)$. Finally, since $f$ is surjective, $f(y)$ can be any nonzero value, so $f$ is periodic with period any nonzero real number, meaning it is constant, contradiction. remarks: need hints on this solution, but now that I think about it, it is very motivated. Injectivity at 0 should come to mind because what we are trying to get rid of are "bad" cases where $f(f(y)-2y)=0$, aka $f$ has funky roots, and the work from there is relatively standard with the periodicity tricks. Another observation I had it felt like the goal is to "periodicize" the original FE. First, we show that one component of the FE, $f(x^2)$ was periodic. This kind of led to a chain reaction where we then used this along with out equations from plugging roots to get that $xf(x)$ also was periodic with the same period, then followed by $f(xf(y)+y^2)$ for any fixed $y$. This was what, at least for me, truly felt like was the main idea of the problem, although I am not sure about whether this is a good viewpoint. i am going to sleep now but this was very :blobheart:

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