Solution avoiding pointwise trap: The solutions are $f \equiv 0, f \equiv x^2, f \equiv -x^2$. We can verify these all work, so now we prove they are the only ones. Let $P(x, y)$ denote the original assertion. From $P(0, 0)$ we have $f(f(0)) = 0$. Then, from $P(x, \tfrac{x^2}{2})$ we get $x^2f(x) = f(f(x))$. Claim 1: $f$ is even Proof: Substituting $x^2f(x) = f(f(x))$ into the original equation, we have $$f(x^2 - y) + 2yf(x) = x^2f(x) + f(y).$$Then, substituting $x \mapsto -x$ yields $$f(x^2 - y) + 2yf(-x) = x^2f(-x) + f(y).$$Subtracting the two equations, we have $$x^2f(x) - x^2f(-x) - 2yf(x) + 2yf(-x) = 0 \implies (f(x) - f(-x))(x^2 - 2y) = 0,$$so taking $y$ such that $x^2 \neq 2y$ gives $f(x) = f(-x)$. $\Box$ Now, taking $P(0, y)$ for any nonzero $y$ yields $2y f(0) = 0$, so $f(0) = 0$. Then, from $P(x, 0)$, we have $f(x^2) = f(f(x))$. Claim 2: There exists a constant $k$ such that $f(x) = kx^2$ for all $x$ Proof: Since $f$ is even and $f(0) = k \cdot 0^2$ for any constant $k$, it suffices to prove the claim for positive $x$. Notice that $xf(\sqrt{x}) = f(f(\sqrt{x})) = f(x)$, so from $P(\sqrt{x}, y)$ for positive $y$ we have $$f(x - y) + \frac{2yf(x)}{x} = f(x) + f(y).$$Similarly, $P(\sqrt{y}, x)$ yields $$f(y - x) + \frac{2xf(y)}{y} = f(y) + f(x).$$Since $f$ is even, we have $f(x - y) = f(y - x)$, so $\tfrac{2yf(x)}{x} = \tfrac{2xf(y)}{y} \implies \tfrac{f(x)}{x^2} = \tfrac{f(y)}{y^2}$, and from varying $y$ it follows that $f(x) = kx^2$ for some constant $k$. $\Box$ Now, notice that $$kx^4 = f(x^2) = f(f(x)) = k^3x^4,$$so by taking nonzero $x$ we have $k^3 = k \implies k \in \{-1, 0, 1\}$, which implies the solution set. $\blacksquare$

$f(x^2-y)+2yf(x)=f(f(x))+f(y)$ We claim the solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2.$ Denote the assertion as $P(x,y).$ Take $P(x,0)$ to get $f(x^2)=f(f(x))+f(0).$ Take $P(x,x^2)$ to get $f(0)+2x^2f(x)=f(f(x))+f(x)^2=2f(f(x))+f(0).$ Thus $x^2f(x)=f(f(x)).$ In particular, $f(f(0))=0.$ From $P(x,0)$ and $P(-x,0)$ we see $f(f(x))=f(f(-x)).$ Thus $x^2f(x)=x^2f(-x),$ so for $x\ne 0$ we have $f(x)=f(-x)$ and thus $f$ is even. Take $P(0,y)$ to get $f(-y)+2yf(0)=f(f(0))+f(y).$ This simplifies to $2yf(0)=0,$ so $f(0)=0.$ Now suppose $f(k)=0.$ From $P(k,0)$ we get $f(k^2)=0.$ Taking $P(k,x^2)$ gives $f(x^2)=f(k^2-x^2)=f(x^2-k^2).$ Then $P(x,k^2)$ gives $f(x^2-k^2)+2kf(x)=f(f(x))=f(x^2),$ so $2kf(x)=0,$ so either $f$ is identically zero or $k=0.$ Now suppose $f(a)=f(b)\ne 0.$ From $x^2f(x)=f(f(x))$ we have $a^2f(a)=f(f(a))=f(f(b))=b^2f(b)$ so $a^2=b^2$ and $a=\pm b.$ From $P(x,0)$ we get $f(x)=\pm x^2$ for all $x.$ Taking $P(x,f(y))$ gives $f(x^2-f(y))=f(f(x))+f(f(y))-2f(x)f(y).$ This is symmetric, so $f(x^2-f(y))=f(y^2-f(x))$ so $x^2-f(y)=\pm(y^2-f(x)).$ Now suppose $f(a)=a^2,f(b)=-b^2.$ Taking $x=a,y=b$ in the above, we get $a^2+b^2=\pm(b^2-a^2),$ but clearly this only holds when either $a=0$ or $b=0.$ Thus the only possible solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2$ and we may easily check that these work.

how many points if i got $x^6f(x)=x^2f(x)^3$ and said this means $f(x)=0,x^2$ and did pointwise trap on these ( i forgot about $-x^2$)

Let $P(x,y)$ the assertion of the following F.E., i claim that $f(x)=0,x^2,-x^2$ work. Indeed it is trivial by replacing, now we prove those are the only ones that work. $P(0,0)$ gives $f(f(0))=0$, $P(x,f(y))$ with symetry gives $f(x^2-f(y))=f(y^2-f(x))$ call this $Q(x,y)$. Now $P(0,x)$ gives $f(-x)-f(x)=-2xf(0)$, $P(x,0)$ gives $f(x^2)=f(f(x))+f(0)$ which gives $f (f(x))=f(f(-x))$. Also $P(-x,y)$ for $y \ne 0$ gives $2yf(x)=2yf(-x)$ therefore $f$ is even and thus $f (0)=0$. Now we have $f(x^2)=f(f(x))$ as well, if $f(a)=f(b)$ y $P(a,x)-P(b,x)$ we get $f(a^2-x)=f (b^2-x)$. Therefore if $f$ is periodic (say its period is $T$), by $P(x,y)-P(x,y+T)$ we get $2yf(x)=2 (y+T)f(x)$. From here we can conclude that $f(x)=0$ or $a^2=b^2$, suppose the latter then $f(x)=x^2$ or $f(x)=-x^2$. Suppose $f(a)=a^2$ and $f(b)=-b^2$ for $a,b \ne 0$, then by $Q(a,b)$ we get $f(a^2+b^ 2)=f(a^2-b^2)$ If $f(x)=x^2$ for both then $b=0$, if $f(x)=-x^2$ then $b=0$ so suppose $(a^2+b^2)^2=-(a ^2-b^2)^2$, then $a=b$ which cannot happen, therefore $f(x)=x^2$ or $f(x)=-x^2$ as desired thus we are done .

Here is my cursed solution which somehow got a 7/7. The only solutions are $\boxed{f(x) = 0}, \boxed{f(x) = x^2},$ and $\boxed{f(x) = -x^2}.$ These can all be sought out to satisfy our functional equation. Now we show that these are the only solutions. Let $P(x,y)$ denote the given equation. $P\left(x,\frac{x^2}{2}\right)$ gives \[ f(f(x)) = x^2 f(x). \]$P(x,0)$ gives \[ f(x^2) = f(f(x)) + f(0). \]On the other hand, $P(-x,0)$ gives \[ f(x^2) = f(f(-x)) + f(0). \]Combining these two gives \[ f(f(-x)) = f(f(x)). \]Using $f(f(x)) = x^2 f(x)$ gives \[ x^2 f(x) = x^2 f(-x) \rightarrow \boxed{f(x) = f(-x) \text{ for all } x}. \]Then $P(0,x)$ gives \[ f(-x) + 2x f(0) = f(f(0)) + f(x). \]Since $f(f(0)) = f(0) \cdot 0^2 = 0$ and $f(-x) = f(x),$ we get $2x f(0) = 0,$ so $f(0) = 0.$ Plugging into $f(x^2) = f(f(x)) + f(0)$ thus yields \[ \boxed{f(f(x)) = f(x^2) = x^2 f(x)}. \]We now find the value of $f(f(f(x)))$ in two different ways. The first is to use the above identity on $f(x)$ to get \[ f(f(f(x)) = f(x)^2 f(f(x)) = f(x)^2 \cdot x^2 \cdot f(x) = f(x)^3 \cdot x^2. \]The second way is to find $f(f(x))$ and stick it into $f.$ We get \[ f(f(f(x)) = f(f(x^2)) = (x^2)^2 f(x^2) = x^4 \cdot x^2 \cdot f(x) = x^6 \cdot f(x). \]Therefore, \[ x^6 \cdot f(x) = x^2 \cdot f(x)^3. \]This implies that $f(x) = 0, x^2, -x^2$ for all individual $x.$ We now deal with the pointwise trap. There are $3$ cases that we consider: Case 1: $f(a) = a^2$ and $f(b) = 0$ for nonzero $a$ and $b.$ Then $P(b,a)$ yields \[ f(b^2 - a) + 2a f(b) = f(f(b)) + f(a) \rightarrow f(b^2 - a) = f(0) + a^2 = a^2, \]while $P(b,-a)$ yields \[ f(b^2+a) - 2a f(b) = f(f(b)) + f(-a) \rightarrow f(b^2 + a) = 0 + f(a) = a^2. \]Since $a^2 > 0,$ we require that $f(b^2 - a) = (b^2 - a)^2 = a^2$ and $f(b^2+a) = (b^2 + a)^2 = a^2.$ Thus $(b^2 - a)^2 = (b^2 + a)^2,$ which simplifies down to $4a^2 b = 0.$ This is impossible since $a,b \ne 0.$ Case 2: $f(a) = -a^2$ and $f(b) = 0$ for nonzero $a$ and $b.$ Then $P(b,a)$ yields \[ f(b^2 - a) + 2a f(b) = f(f(b)) + f(a) \rightarrow f(b^2 - a) = f(0) - a^2 = -a^2, \]while $P(b,-a)$ yields \[ f(b^2+a) - 2a f(b) = f(f(b)) + f(-a) \rightarrow f(b^2 + a) = 0 + f(a) = -a^2. \]Since $-a^2 < 0,$ we require that $f(b^2 - a) = -(b^2 - a)^2 = -a^2$ and $f(b^2+a) = -(b^2 + a)^2 =- a^2.$ Thus $(b^2 - a)^2 = (b^2 + a)^2,$ which simplifies down to $4a^2 b = 0.$ This is impossible since $a,b \ne 0.$ Case 3: $f(a) = a^2$ and $f(b) = -b^2$ for nonzero $a$ and $b.$ Then $P(a,b)$ yields \[ f(a^2 - b) + 2b f(a) = f(f(a)) + f(b) \rightarrow f(a^2-b) = f(a^2) - 2ba^2 - b^2 = a^2 f(a) - 2ba^2 - b^2 = a^4 - 2ba^2 - b^2 \]while $P(a,-b)$ yields \[ f(a^2 + b) - 2b f(a) = f(f(a)) + f(-b) \rightarrow f(a^2+b) = f(a^2) + 2ba^2 - b^2 = a^4 + 2ba^2 - b^2. \]If both of these are equal to $0,$ this clearly implies that $4ba^2 = 0,$ which is impossible. Thus there are $4$ subcases from here: Subcase 3a: $f(a^2 - b) = (a^2-b)^2 = a^4 - 2ba^2 - b^2.$ This simplifies to $2b^2 = 0,$ impossible. Subcase 3b: $f(a^2 + b) = (a^2 + b)^2 = a^4 + 2ba^2 - b^2.$ This also yields $2b^2 = 0,$ impossible. Subcase 3c: $f(a^2 - b) = -(a^2 - b)^2 = a^4 - 2ba^2 - b^2.$ This simplifies to $2a^4 - 4ba^2 = 0,$ so $a^2 = 2b.$ Plugging into $P(a,-b)$ gives \[ f(3b) = (2b)^2 + 2b(2b) - b^2 = 7b^2 \ne 0, 9b^2, -9b^2, \]which is a contradiction. Subcase 3d: $f(a^2 + b) = -(a^2 + b)^2 = a^4 + 2ba^2 - b^2.$ This yields $2a^4 + 4ba^2 = 0,$ so $a^2 = -2b.$ Plugging into $P(a,b)$ gives \[ f(-3b) = (-2b)^2 - 2b(-2b) - b^2 = 7b^2 \ne 0, 9b^2, -9b^2, \]which is impossible. We have exhausted through all cases, so Case $3$ is impossible. Therefore, the pointwise trap is impossible, and so the only solutions to the given functional equation are those given at the beginning of the solution, as desired.

Here is my solution: no pointwise

Attachments:

How many points for getting the right answer and most of the proof??

I was nearly trapped by pointwise and fixed it in last 15 min; answer is still $$f(x)=x^2,-x^2,0$$

how many points for doing everything like post #5 but only providing like 70 percent detail in the proof of no pointwise trap

oopsy daisies

Solution Sketch: P(0, 0) -> f(f(0)) = 0 Also get, f(x^2) = x^2 f(x) = f(f(x)) So f(0) = 0 $f$ is even from taking $x = 0$. If f(n) = 0 for some $n \neq 0$, we plug in $x = n$ to get, $f(n^2 - y) = f(y)$ Then, plug in $y = x^2 - n^2$ to eventually get $-2n^2 f(x) = 0 \iff f(x) = 0$ for all $x$. Prove injectivity over positives by FTSOC f(m) = f(n) but $m \neq n > 0$... Now look at $f(f(x)) = f(x^2)$ to get $f(x) = \pm x^2$ depending on its sign (use even). $f(x) \in {x^2, -x^2}$ pretty straight-forward pointwise trap.

how many points for showing $x^6*f(x)=x^2f(x)^3$

I showed that $f(f(x))=f(x^2)$, and I wrote that $f(x)=x^2$, $0$, or $-x^2$. I also made a reference to the pointwise trap and wrote a little bit of garbage about avoiding it; I think it's obvious enough that if you have $f(a)=a^2$ and $f(b)=0$ or $f(a)=a^2$ and $f(b)=-b^2$ or $f(a)=-a^2$ and $f(b)=0$, then it doesn't satisfy the original equation, but I don't think I made that clear enough since I had literally 1 minute left while writing that. I also showed that $f(0)=0$. I talked about how if $f(x)$ was injective, then $f(x)=x^2$, but clearly $f$ is not injective since we have $f(x)=f(-x)$. I couldn't figure out how to rigorously prove that $f$ could only be $x^2$, $0$, and $-x^2$ though hoping for a 5 or somehow 6 on this one

i missed -x^2 help but i proved f(0)=0, f(1)=0 or 1, f(x^2-x)=(x-1)^2f(x), f(x)=f(-x), and f(x^2)=x^2f(x)=f(x^2). how many points would i get?

FE! I LOVE FE! no convolution sketch: 1. Prove $f(0)=0$ 2. Prove $f$ is even 3. Prove $f(x^2)=f(f(x))=x^2f(x)$ 4. Prove $f$ is bijective over $x\in\mathbb{R}^+$ and $f(x)\neq0$ 5. Plug in $x=x_0$ for some $f(x_0)=0$ and use (4) to prove $f\equiv0$ or $f(x)\neq0$ for all $x\neq0$ (pointwise trap pt. 1) 6. Deal with $\pm x^2$ (pointwise trap pt. 2)

requested by sixoneeight. apparently using calc on this fe was kinda funny but idk :shrug: also this avoids pointwise trap very nicely anyways im kinda tired rn but heres an outline of my sol in contest 1. Show that f(0) = 0 2. Show that f(x) = f(-x) 3. Show that f(x^2) = f(f(x)) 4. This is the main idea. Subtracting P(x, y) from P(x, -y), we obtain $f(x^2 + y) - f(x^2 - y) = 4yf(x)$, and taking $y$ arbitrarily small means that we can easily show $f(x)$ is continuous over positive reals and hence by part $2$ it is also continuous over all reals. From this, it is straightforward to show that $f(x^2)$ is differentiable with derivative $2f(x)$. 5. Finish (the problem is kinda trivial from here but iirc my finish wasnt elegant at all so im not gonna bother reconstructing it)

Mathandski wrote: Solution Sketch: P(0, 0) -> f(f(0)) = 0 Also get, f(x^2) = x^2 f(x) = f(f(x)) So f(0) = 0 $f$ is even from taking $x = 0$. howdid you prove injectivity If f(n) = 0 for some $n \neq 0$, we plug in $x = n$ to get, $f(n^2 - y) = f(y)$ Then, plug in $y = x^2 - n^2$ to eventually get $-2n^2 f(x) = 0 \iff f(x) = 0$ for all $x$. Prove injectivity over positives by FTSOC f(m) = f(n) but $m \neq n > 0$... Now look at $f(f(x)) = f(x^2)$ to get $f(x) = \pm x^2$ depending on its sign (use even). $f(x) \in {x^2, -x^2}$ pretty straight-forward pointwise trap.

When the word "plug-in" is used without context, I am referring to the original equation. We claim that $f(x) \equiv 0$, $f(x) \equiv -x^2$, $f(x) \equiv x^2$ are the three unique solutions. Plugging these in, they work. It suffices to prove that no other functions satisfying the original equation exist. Plug in $\bigl(0, \tfrac{x^2}{2}\bigr)$ to get \[f\left(x^2 - \frac{x^2}{2}\right) + x^2 f(x) = f(f(x)) + f\left(\frac{x^2}{2}\right) \implies x^2f(x) = f(f(x)).\]Plug in $x = 1$ into the above equation to get $f(1) = f(f(1))$. Plug in $(x, 0)$ to get \[f(x^2) = f(f(x)) + f(0).\]Plug in $x = 1$ into the above equation to get $f(1) = f(f(1)) + f(0)$, so $f(0) = 0$. Plug in $(0, x)$ to get \[f(0^2 - x) + 0 = f(f(0)) + f(x) \implies f(x) = f(-x).\]So, $f$ is an even function. Combining our two expressions for $f(f(x))$, we get that \[f(f(x)) = f(x^2) = x^2f(x).\]By tripling the involution (more specifically, plugging in $f(x)$ in $f(f(x)) = x^2f(x)$ and taking $f$ of both sides in $f(x^2) = x^2f(x)$), we get that \begin{align*} f(f(f(x))) &= f(x)^2f(f(x)) = x^2f(x)^3 \\ f(f(f(x))) &= f(f(x^2)) = x^4f(x^2) = x^6 f(x). \end{align*}By equating, we get that \[x^2 f(x)^3 = x^6 f(x) \implies f(x) \in \{0, -x^2, x^2\} \ \forall \ x \in \mathbb{R}.\]We will not be fooled by the pointwise trap--we still need to show that $f(x) \equiv 0$, $f(x) \equiv -x^2$, and $f(x) \equiv x^2$ are the only three solutions out of the infinitely many remaining. For the first case, suppose that there existed some nonzero real number $a$ such that $f(a) = 0$. Plug in $(a, -x)$ and use the fact that $f(x) = f(-x)$ to get \[f(a^2 + x) = f(-x) = f(x).\]Suppose there existed some nonzero real number $b$ such that $f(b) \ne 0$. Then, $f(b) = \pm b^2$. But, this is impossible, because we can repeatedly apply the equation $f(a^2 + x) = f(x)$ to obtain \[\pm b^2 = f(b) = f(a^2 + b) = f(2a^2 + b) = \cdots.\]Since $\pm b^2$ cannot equal $\pm (\ell a^2 + b)^2$ for all integers $\ell$, this is a contradiction. Thus, in this case, we get that $f(x) \equiv 0$. For the second case, we can now assume that $f(x) \in \{-x^2, x^2\}$ for all $x \in \mathbb{R}$. Suppose, for the sake of contradiction, that there existed some nonzero real numbers $a$ and $b$ such that $f(a) = a^2$ and $f(b) = -b^2$. Then, plugging in $(a, -b)$, we get that \[f(a^2 + b) - 2bf(a) = f(f(a)) + f(b).\]We know that $f(f(a)) = a^2f(a) = a^4$. Plugging these in, we get that \[f(a^2 + b) = a^4 - 2a^2b - b^2.\]We have two possibilities for $f(a^2 + b)$, namely $(a^2 + b)^2$ and $-(a^2 + b)^2$. Testing the two, we get that $b = 0$, $b = -2a^2$, or $a = 0$, each of which are impossible. Thus, this is a contradiction, and the two solutions for this case are $f(x) \equiv -x^2$ and $f(x) \equiv x^2$. We have proven that the only three possible solutions are $f(x) \equiv 0$, $f(x) \equiv -x^2$, and $f(x) \equiv x^2$. We are done.

Missed f(x) = -x^2. The big sad :O(

I think I'm understanding how to do these type of problems... 35 minutes of solve-time. Solution: The answers are $f \equiv 0$, $f \equiv +x^2$ and $f \equiv -x^2$. Its very easy to verify that all of these work! We'll now prove that they are the only ones. Let $P(x,y)$ denote the assertion to the functional equation. The only constant solution is the trivial solution which is not hard to prove. Henceforth, assume $f$ is non-constant. Start by $P(0,0)$ to find out that $\exists \alpha \in \mathbb{R}$ such that $f(\alpha) = 0$. From $P\left(\alpha,\frac{\alpha^2}{2}\right)$ we get $f(0) = 0$. Its also noteworthy that $P(0,y)$ reveals that $f$ is even. This finishes the basic substitutions. Claim: $f$ is partially injective, i.e, $f(a) = f(b) \implies a = \pm b$. Proof: Assume there are reals $|a| > |b|$ such that $f(a) = f(b)$. From $P(a,x)$ and $P(b,x)$ we would get \[ f(x + a^2) = f(x+b^2)\]which means $f$ is periodic with period $T \coloneqq a^2 - b^2$. Comparing $P(x,y)$ and $P(x,y+T)$, we would get that $T = 0$(since $f$ is non-constant) which is a contradiction to the assumption $|a| > |b|$. $\square$ We will now use the "bump in symmetry". Consider $P(x,f(y))$ and then swap $x,y$. This on comparison would give \[f\left(x^2 - f(y)\right) = f\left(y^2-f(x)\right) \tag{1}\]Setting $y = 0$ in above equation would yield $f(x) \in \{+x^2, -x^2\}$ via the partial injectivity for any $x \in \mathbb{R}$. To resolve the point-wise trap, consider $f(a) = a^2$ and $f(b) = -b^2$ for non-zero $a,b \in \mathbb{R}$. Plug $x = a$ and $y = b$ in $(1)$. This results in \[f(a^2 + b^2) = f(b^2 - a^2).\]Again invoking partial injectivity in the above equation, we find that one of $a$ or $b$ must be $0$ which is impossible. Therefore, either $f \equiv +x^2$ or $f \equiv -x^2$ as desired. $\blacksquare$

Let's let $P(x, y)$ denote the assertion. Our proof proceeds in a couple of steps. Claim: $f(x^2) = x^2f(x)$ Proof. Note that $P(0, 0)$ gives that $f(f(0)) = 0$. Then from $P(x, 0)$ we find, \begin{align*} f(x^2) = f(f(x)) + f(0) \end{align*}Additionally, from $P(x, x^2)$ we find, \begin{align*} f(0) + 2x^2f(x) &= f(f(x)) + f(x^2) \end{align*}Then setting this equal, \begin{align*} x^2f(x) = f(f(x)) \end{align*}as desired. $\square$ Claim: $f(0) = 0$ Proof. Consider taking $P(x, 0)$ and using our prior identity to find, \begin{align*} f(x^2) &= f(f(x)) + f(0)\\ \iff f(x^2) &= f(x^2) + f(0)\\ \iff f(0) &= 0 \end{align*}as desired. $\square$ Claim: $f$ is even. Proof. Note that from $P(0, y)$ we find, \begin{align*} f(-y) = f(y) \end{align*}as desired. $\square$ To finish consider taking $P(x, y^2)$ to find, \begin{align*} f(x^2 - y^2)= - 2y^2f(x) + f(x^2) + f(y^2) \end{align*}However swapping $x$ and $y$ gives, \begin{align*} f(x^2 - y^2) = -2x^2f(y) + f(x^2) + f(y^2) \end{align*}and hence $\frac{x^2}{f(x)} = \frac{y^2}{f(y)} = k$ for some constant $k$. Upon verifying solutions we find $f \equiv x^2$, $f \equiv - x^2$ and $f \equiv 0$.

If we put $y=0$ then we get $f(x^2)=f(f(x))+f(0)$. Now put $y=x^2$ then $x^2f(x)=f(x^2)-f(0)=f(f(x))$. In particular if we put here $x=f(0)$ we get $f(0)=0$, since $f(f(0))=0$. Now since $x^2f(x)=f(x^2)$ it follows that $f$ is even. Suppose $f(a)=f(b)\neq 0$ then $a^2f(a)=f(f(a))=f(f(b))=b^2f(b)$, so $a=\pm b$. In particular since $f(f(1))=1$, we get that $f(1)\in\{0,1,-1\}$. If $f(1)=0$ then $(x,y)=(\sqrt x,1)\Rightarrow f(x-1)+2f(x)/x=f(x)$ and $x=1\Rightarrow f(1-y)=f(y)$. Thus $2f(x)/x=0$ for all $x$, so $f\equiv 0$, which is indeed a solution. If $f(1)=1$ then the same two substitutions yield $f(x-1)+2f(x)/x=f(x)+1$ and $f(1-y)+2y=1+f(y)$, so $f(x)=x^2$ for all $x$, which is also a solution. If $f(1)=-1$ the similarly $f(x)=-x^2$ is a solution.

they gave 0 for just including the functions

We claim the only such functions are $\boxed{f(x)\equiv -x^2}$, $\boxed{f(x)\equiv 0}$, and $\boxed{f(x)\equiv x^2}$. It is easy to check that these work. Let $P(x,y)$ be the given assertion. $P(0,0)$ gives us $f(f(0))=0$. We have \begin{align*} P(1,0)&\implies f(1)=f(f(1))+f(0)\\ P(1,1)&\implies f(0)+2f(1)=f(f(1))+f(1) \end{align*}so $f(0)=0$. Now \begin{align*} P(x,0)&\implies f(x^2)=f(f(x))\\ P(x,x^2)&\implies 2x^2f(x)=f(f(x))+f(x^2) \end{align*}so $f(f(x))=f(x^2)=x^2f(x)$. It follows that $f(x)=f(-x)$. Claim 1. If $f(a)=f(b)\neq 0$ then $a=\pm b$. Proof. Let $a,b\in\mathbb{R}$ satisfy $f(a)=f(b)\neq 0$. Then \[ P(a,b^2)\implies f(a^2-b^2)+2b^2f(a)=a^2f(a)+b^2f(b)\implies f(a^2-b^2)=f(a)(a^2-b^2) \]so \[ f(a)(a^2-b^2)=f(a^2-b^2)=f(b^2-a^2)=f(a)(b^2-a^2). \]It follows that $a^2=b^2$, as desired. Thus $f(x)\in\{-x^2,0,x^2\}$. $\square$ Claim 2. There exists no $a,b\in\mathbb{R}\setminus\{0\}$ such that $f(a)=a^2$ and $f(b)=-b^2$. Proof. Assume the contrary. Then \begin{align*} P(a,b^2)&\implies f(a^2-b^2)=a^4-b^4-2a^2b^2\\ P(b,a^2)&\implies f(b^2-a^2)=a^4-b^4+2a^2b^2 \end{align*}so $ab=0$, a contradiction. $\square$ Claim 3. There exists no $a,b\in\mathbb{R}\setminus\{0\}$ such that $f(a)=a^2$ and $f(b)=0$. Proof. Assume the contrary. Then \begin{align*} P(a,b^2)&\implies f(a^2-b^2)=a^4-2a^2b^2\\ P(b,a^2)&\implies f(a^2-b^2)=a^4 \end{align*}so $ab=0$, a contradiction. $\square$ Claim 4. There exists no $a,b\in\mathbb{R}\setminus\{0\}$ such that $f(a)=-a^2$ and $f(b)=0$. Proof. Assume the contrary. Then \begin{align*} P(a,b^2)&\implies f(a^2-b^2)=-a^4+2a^2b^2\\ P(b,a^2)&\implies f(a^2-b^2)=-a^4 \end{align*}so $ab=0$, a contradiction. $\square$ The conclusion follows. $\square$

Awesomeness_in_a_bun wrote: they gave 0 for just including the functions Yea I wrote the correct answer along with full proof, only error being assuming polynomial and got a 0 rips

It is obvious that the functions $f(x) = x^2 \thinspace \forall x \in \mathbb{R}$ and $f(x)\equiv 0$ satisfy the given equation. Now, we will prove that these are the only solutions. Let $P(x, y)$ denote the given assertion. \begin{align*} P(0, 0)&: f(f(0)) = 0 \\ P(f(0), f(0)^2/2)&: f(f(0)^2/2) = f(0) + f(f(0)^2/2) \implies f(0) = 0 \\ P(x, 0)&: f(x^2) = f(f(x)) \\ P(0, x)&: f(-x) = f(x) \\ P(x, x^2/2)&: f(f(x)) = x^2 f(x). \end{align*}Then, we can rewrite $P(x, y)$ as \[f(x^2 - y) = (x^2 - 2y) f(x) + f(y).\]Replacing $y$ by $-y^2$ gives \[f(x^2 + y^2) = x^2 f(x) + y^2 f(y) + 2y^2 f(x).\]By symmetry, we get $y^2 f(x) = x^2 f(y)$. Setting $y \to 1$, we have $f(x) = cx^2$ for some constant $c$. Since $f(f(x)) = x^2 f(x)$, it follows that $c = 0$ or $1$.

The solutions are $f(x) = x^2$, $f(x) = -x^2$ and $f(x) = 0$. Claim: We have $f(0) = 0$. Proof: The claim follows if we subtract $P(1, \tfrac{1}{2} )$ from $P(1,1)$. Now, taking $P(x,0)$ gives us $f(x^2) = f(f(x))$, so our FE is equivalent to \[f(x^2 - y) = (x^2 - 2y) f(x) + f(y).\]Let $Q(x,y)$ denote the above assertion. Claim: If there exists a nonzero root of $f$, then $f(x) = 0$ for all $x$. Proof: Let this nonzero root be $r$. Subtracting $Q(0,x)$ from $Q(r,x)$ gives us \[f(r^2 - x) = f(-x),\]so $f$ is periodic with period $r^2$. Now, subtracting $Q(x, r^2)$ from $Q(x,0)$ gives us $r^2 f(x) = 0$ for all $x$, so $f$ is identically $0$. From hereon, assume $0$ is the only root of $f$. Taking $P(x, \tfrac{x^2}{2} )$ where $x \neq 0$ gives us $x^2 f(x) = f(f(x))$, implying "quasi-injectivity": that if $f(a) = f(b)$, then $|a| = |b|$. Now, taking $P(x,0)$ gives us $f(x^2) = f(f(x))$, so $f(x) = \pm x^2$ for each $x$. Claim: Either $f(x) = x^2$ for all $x$ or $f(x) = -x^2$ for all $x$. Proof: Suppose otherwise. Notice that if $f$ is a solution to our rewritten FE, then $-f$ is also a solution. So, assume WLOG that $f(1) = 1$. Now, let $x$ and $y$ be nonzero reals with such that $x^2 - y = 1$: If, FTSOC, $f(x) = x^2$ and $f(y) = -y^2$, taking $Q(x,y)$ gives us $y = 0$, which is a contradiction. If, FTSOC, $f(x) = -x^2$ and $f(y) = y^2$, taking $Q(x,y)$ gives us \[x^2(x^2 - 2y) = 0 \implies y = x^2 - y = 1 \implies x = \sqrt{2}.\]But now, taking $Q(1, \sqrt{2})$ gives us $\left| (1 - \sqrt{2})^2 \right| = \left| -1 - 2 \sqrt{2} \right|$, contradiction. If, FTSOC, $f(x) = -x^2$ and $f(y) = -y^2$, taking $Q(x,y)$ gives us $2(x^2 - y)^2 = 0$, contradiction. It follows that $f(x) = x^2$ or $f(x) = -x^2$.

The solutions are $f(x)=0$, $f(x) \equiv x$, and $f(x) \equiv -x^2$, which work. Let $P(x,y)$ denote the assertion in the problem statement. $P(0,0)$ gives $f(f(0))=0$. $P(f(0),\tfrac{f(0)^2}{2})$ gives $f(0)=0$. $P(0,x)$ gives $f(x)=f(-x)$. $P(x,0)$ gives $f(f(x))=f(x^2)$. $P(x,y^2)$ gives \[f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)=f(x^2)+f(y^2).\]Swapping $x$ and $y$ gives \[f(x^2-y^2)+2y^2f(x)=f(y^2-x^2)+2x^2f(y)=f(x^2-y^2)+2x^2f(y),\]so $y^2f(x)=x^2f(y)$. Plugging in $y=1$ gives $f(x)=f(1)x^2$. Since $f(f(x))=f(x^2)$, we have \[f(1)(f(1)x^2)^2=f(1)x^4 \implies f(1)^3=f(1),\]so $f(1) \in \{0,1,-1\}$, as desired. $\blacksquare$

We claim the only answers are $\boxed{f(x) = 0}$, $\boxed{f(x) = x^2}$, and $\boxed{f(x) = - x^2}$. It is easy to verify that these work. It remains to show that they the only solutions. Claim 1: $x^2 f(x) = f(f(x))$ Proof. $\underline{P}(x, 0) \implies f(x^2) = f(f(x)) + f(0)$ $\underline{P}(x, x^2) \implies f(0) + 2x^2f(x) = f(f(x)) + f(x^2)$ Adding, we get $2x^2 f(x) = 2f(f(x)) \implies \boxed{x^2 f(x) = f(f(x))}$ Claim 2: $f(0) = 0$ Proof. By Claim 1, we have $f(f(1)) = f(1)$. Therefore: $\underline{P}(1, 1) \implies f(0) + \cancel{2f(1)} = \cancel{2f(1)}$ $\implies \boxed{f(0) = 0}$. Claim 3: $f$ is even. Proof. Using Claim 2: $\underline{P}(0, y) \implies \boxed{f(-y) = f(y)}$ Claim 4: If $f \not\equiv 0$, then $f(x) = f(y) \implies x = \pm y$. Proof. By Claim 1, $f(x) = f(y) \implies f(f(x)) = f(f(y)) \implies \frac{f(f(x))}{f(x)} = \frac{f(f(y))}{f(y)} \implies x^2 = y^2 \implies x = \pm y$. It is easy to see that Claim 4 then finishes the problem: If $f \not\equiv 0$, then we have $f(x^2) = f(f(x)) \implies \boxed{f(x) = \pm x^2}$. It remains to take care of the pointwise trap, which is not hard to do. However, for the sake of compactness, I have excluded the details from my post.

I claim the only solutions are $\boxed{f(x) \equiv 0, x^2, -x^2}$ which can easily be verified to work. Let $P(x, y)$ denote the given assertion. $P(x, \frac{x^2}{2})$ gives $f(f(x)) = x^2f(x)$. $P(-x, y)$ and $P(x, y)$ and equating gives $x^2f(x)-2yf(x)=x^2f(-x)-2yf(-x)$. $y=0$ so $x^2f(x)=x^2f(-x)$ or $f(x)=f(-x)$ for nonzero $x$ and thus for all $x$. Therefore $f$ is even. Call this statement $1$. Let $x=0$ immediately gives us $2yf(0)=0$ so $f(0)=0$, coupled with the fact that $f(f(0))=0$. Now let $y=0$, so $f(x^2)=f(f(x))$, and we have the chain: $$f(x^2)=f(f(x))=x^2f(x)$$which is of tremendous use. Let $x=f(x)$ in the latter inequality to get $$f(f(f(x)))=f(x)^2f(f(x))$$$$f(x)^2 f(x^2) = x^4 f(x^2)$$so $f(x) = 0, x^2, -x^2$ as mentioned before for an individual $x$. Now we bash out the pointwise trap which I don't feel like typing out.

The only solutions that satisfy the given are $f(x) \equiv x^2, -x^2, 0$. Claim 1: $f(0) = 0$ Proof: Let $P(x, y)$ denote the given equation. From $P(x, 0)$, we obtain \[f(x^2) = f(f(x)) + f(0)\] Let $c = f(0)$. From $P(0, 0)$, we obtain \[c = f(c) + c \implies f(c) = 0 \implies f(f(c)) = c\] From $P(c, 0)$, we obtain \[f(c^2) = f(f(c)) + c \implies f(c^2) = 2c\] However, from $P(c, c^2)$, we know that \[f(0) + 2c^2f(c) = f(f(c)) + f(c^2)\]\[c = c + f(c^2) \implies f(c^2) = 0\]\[f(c^2) = 2c = 0 \implies c = 0\] as desired. Claim 2: $f(f(x)) = f(x^2) = x^2f(x)$. Proof: From $P(x, 0)$, we have \[f(x^2) = f(f(x)) + f(0) \implies f(x^2) = f(f(x))\] From $P(x, x^2)$, we have \[f(x^2 - x^2) + 2x^2f(x) = f(f(x)) + f(x^2)\]\[2x^2f(x) = 2f(x^2) \implies x^2f(x) = f(x^2)\] as desired. Claim 3: $f(x)$ is even. Proof: From $P(0, y)$, we have \[P(0 - y) + 2yf(0) = f(f(0)) + f(y)\]\[f(-y) = f(y)\] as desired. Using claims $2$ and $3$, we will prove that $f(x) \equiv x^2, -x^2, 0$. After we finish doing so, we will show these are the only solutions by avoiding the pointwise trap. Note that \[f(f(f(x))) = f(f(x^2)) = x^4f(x^2) = x^6f(x)\] and that \[f(f(f(x))) = (f(x))^2f(f(x)) = x^2(f(x))^3\] We can then equate the two to obtain \[x^6f(x) = x^2\left(f(x)\right)^3\]\[x^2f(x)\left(f(x)^2 - x^4\right) = 0\]\[x^2f(x)\left(f(x) + x^2\right)\left(f(x) - x^2\right) = 0\] The only way for this equation to be satisfied for all $x$ is if $f(x) \equiv -x^2, x^2, 0$. Avoiding the pointwise trap There are three cases to tackle: $f(x) \equiv 0, x^2$, $f(x) \equiv 0, -x^2$, and $f(x) \equiv -x^2, x^2$. Because these take a while to type out, I will edit this solution later to include them. The basic idea is to have non-zero variables $a$ and $b$ such that $f(a) = 0$ and $f(b) = b^2$ (for case 1; same logic applies for cases 2 and 3) and show that this leads to a contradiction with the given. Claim 3 ($f(x)$ is even) will prove useful for this.

OronSH wrote: $f(x^2-y)+2yf(x)=f(f(x))+f(y)$ We claim the solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2.$ Denote the assertion as $P(x,y).$ Take $P(x,0)$ to get $f(x^2)=f(f(x))+f(0).$ Take $P(x,x^2)$ to get $f(0)+2x^2f(x)=f(f(x))+f(x)^2=2f(f(x))+f(0).$ Thus $x^2f(x)=f(f(x)).$ In particular, $f(f(0))=0.$ From $P(x,0)$ and $P(-x,0)$ we see $f(f(x))=f(f(-x)).$ Thus $x^2f(x)=x^2f(-x),$ so for $x\ne 0$ we have $f(x)=f(-x)$ and thus $f$ is even. Take $P(0,y)$ to get $f(-y)+2yf(0)=f(f(0))+f(y).$ This simplifies to $2yf(0)=0,$ so $f(0)=0.$ Now suppose $f(k)=0.$ From $P(k,0)$ we get $f(k^2)=0.$ Taking $P(k,x^2)$ gives $f(x^2)=f(k^2-x^2)=f(x^2-k^2).$ Then $P(x,k^2)$ gives $f(x^2-k^2)+2kf(x)=f(f(x))=f(x^2),$ so $2kf(x)=0,$ so either $f$ is identically zero or $k=0.$ Now suppose $f(a)=f(b)\ne 0.$ From $x^2f(x)=f(f(x))$ we have $a^2f(a)=f(f(a))=f(f(b))=b^2f(b)$ so $a^2=b^2$ and $a=\pm b.$ From $P(x,0)$ we get $f(x)=\pm x^2$ for all $x.$ Taking $P(x,f(y))$ gives $f(x^2-f(y))=f(f(x))+f(f(y))-2f(x)f(y).$ This is symmetric, so $f(x^2-f(y))=f(y^2-f(x))$ so $x^2-f(y)=\pm(y^2-f(x)).$ Now suppose $f(a)=a^2,f(b)=-b^2.$ Taking $x=a,y=b$ in the above, we get $a^2+b^2=\pm(b^2-a^2),$ but clearly this only holds when either $a=0$ or $b=0.$ Thus the only possible solutions are $f(x)=0,f(x)=x^2,f(x)=-x^2$ and we may easily check that these work. Oron orz

Let $P(x, y)$ denote the given assertion. We claim that the only solutions are $f\equiv0$, $f(x)=x^2$ and, $f(x)=-x^2$ which clearly work. Claim 1.4 We have $f(x^2)=f(f(x))$ Proof $P(x,0)$ yields $f(x^2)=f(f(x))+f(0)$, so it suffices to prove that $f(0)=0$. Observe that by $P(1,1)$, we get $$f(0)+2f(1)=f(f(1))+f(1)$$but after cancelling out terms and substituting $f(f(1))$ for $f(1)-f(0)$, we get $f(0)+f(1)=f(1)-f(0)$ which directly implies the result. Notice how by using $f(0)=0$ we can now easily prove that $f$ is even, as $P(0,y)$ gives $f(-y)=f(y)$. The key claim is as follows. Claim 1.5 $f(x)=f(y)$ implies $\vert x\vert=\vert y\vert$ or $f(x)=f(y)=0$ Proof $P(x, y^2)$ states $$f(x^2-y^2)+2y^2f(x)=f(f(x))+f(y^2)$$and $P(y, x^2)$ states $$f(y^2-x^2)+2x^2f(y)=f(f(x))+f(x^2)$$but using both that $f(x^2)=f(f(x))$ and that $f$ is even, this is equivalent to $y^2f(x)=x^2f(y)=x^2f(x)$ so either $f(x)=0$ or $x^2=y^2\implies \vert x\vert=\vert y\vert y$ Claim 1.6 If $f(c)=0$ for some positive real $c$, than $f\equiv0$ Proof $P(c,x)$ gives $$f(c^2-y)=f(f(c))+f(y)$$but note that $f(f(c))=f(0)=0$ so we in fact have $f(c^2-y)=f(y)$. However, by Claim 1.5 either $f(c^2-y)=f(y)=0$ or $\vert c^2-y\vert=\vert y\vert$, the latter being only true when $y=\frac{c^2}{2}$. Thus, we have $f(x)=0$ for all $x\neq \frac{c^2}{2}$. But one can now easily find $\frac{c^2}{2}$ taking any $x, y$ satisfying $x^2-y=\frac{c^2}{2}$, as then $P(x,y)$ gives $f(\frac{c^2}{2})=0$. From now on, suppose that $f(c)\neq0$ for any $c$ greater than $0$. Recall that $f(x^2)=f(f(x))$ so we can apply Claim 1.5 to get $\vert f(x)^2\vert=\vert x^4\vert$, which yields $f(x)=\pm x^2$. We now aim to prove that $f$ can't change of sign, thus leaving us with only the initially stated solutions. Assume for the sake of contradiction that $f$ changes in sign. By Pigeonhole principle, one of these must repeat an infinite number of times, WLOG say it is $f(x)=x^2$, as the other is completely symmetric. Now choose a sufficiently large $a$ such that $f(a)=a^2$ and a small $b\neq 0$ in comparison which satisfies $f(b)=-b^2$. Now notice that $P(a,b)$ gives $$f(a^2-b)+2a^2b=a^4-b^2$$where we used $f(f(x))=x^2f(x)$, which is a consequence of $P(x^2, y^2)$. We can rewrite this as $f(a^2-b)=a^4-b^2-2a^2b$ but we also know $f(a^2-b)=\pm (a^2-b)^2$. So we can now distinguish two cases. Case 1. $f(a^2-b)=(a^2-b)^2$ Combining this with $f(a^2-b)=a^4-b^2-2a^2b$ we get $a^4-b^2-2a^2b=a^4-2a^2b+b^2$ so $-b^2=b^2\implies b=0$. But we supposed $b\neq 0$, so this case leads to a contradiction. Case 2. $f(a^2-b)=-(a^2-b)^2$ Notice how $a^4-b^2-2a^2b$ is positive for sufficiently large $a$, however $-(a^2-b)^2$ is always negative, so we can immediately discard this case.

Plug in $y = 0$ to get $f(f(x)) = f(x^2) - f(0)$, substitute in the original equation to get $f(x^2 - y) + 2yf(x) = f(x^2) + f(y) - f(0)$*. Plug in $x = y = 1$ to get $f(0) + 2f(1) = f(1) + f(1) - f(0) \implies f(0) = 0$. Substitute into * to get $f(x^2 - y) + 2yf(x) = f(x^2) + f(y)$**. Plug in $x = 0$ to get $f(y) = f(-y)$. Substitute $y \mapsto -y$ to get $f(x^2 + y) - 2yf(x) = f(x^2) + f(-y) = f(x^2) + f(y)$. Subtract ** from this to get $f(x^2 + y) - f(x^2 - y) = 4yf(x)$. Substitute $y \mapsto y^2$ to get $f(x^2 + y^2) - f(x^2 - y^2) = 4y^2f(x)$. Note that $f(x^2 - y^2) = f(y^2 - x^2)$ (because $f$ is even), so $4y^2f(x) = f(x^2 + y^2) - f(x^2 - y^2) = f(y^2 + x^2) - f(y^2 - x^2) = 4x^2f(y)$. Simplifying, we get $y^2f(x) = x^2f(y)$. Plug in $y = 1$ to get $f(x) = x^2f(1)$. Therefore, $f(x) = cx^2$ for some constant $c$. Going back to the original equation, we have $f(x^2 - y) + 2yf(x) = cx^4 - 2cx^2y + cy^2 + 2x^2y = cx^4 + cy^2$, and $f(f(x)) + f(y) = c^3x^4 + cy^2$. Equating these, we have $(c^3 - c)x^4 = 0$ for all real $x$; solving $c^3 - c = 0$, we get $c \in \{-1, 0, 1\}$. The desired functions are $f(x) = -x^2$, $f(x) = 0$, and $f(x) = x^2$. $\square$