We show that all solutions to the functional equation are $f(x) = 0$ for all $x \in \mathbb{R}$ and $f(x) = x$ for all $x \in \mathbb{R}$. It is clear they satisfy the equation, we now prove that they are the only ones. As usual, let $P(x, y)$ denote the given assertion. Note that from $P(x, 0)$ we deduce $f(x^2) = f(x)^2$. A very crucial bound we get from this is that $f(x) \geq 0$ whenever $x \geq 0$. We will use this freely in what follows. Denote the set of zeroes of $f$ by $S$, that is, $S$ is the set of all $x \in \mathbb{R}$ with $f(x) = 0$. Note that we are given $0 \in S$. We will first show that either $S \equiv \{0\}$ or $S \equiv \mathbb{R}$. Suppose that $S$ has some element other than $0$. We will show that $S$ must be $\mathbb{R}$. Claim 1: Properties of $S$: $S$ is closed under addition. If $a \in S$ then $-a$ in $S$. If $a \in S$ then $a^2 \in S$. If $a \in S$ with $a > 0$ then $\pm \sqrt{a} \in S$. Proof: If $f(t) = 0$, then we have $f(t^2) = f(t)^2 = 0$ and thus the third condition is proven. Since $f(-t)^2 = f(t^2)$ the second condition follows as well. Finally, if $a, b \in S$ then from $P(a, b)$ we obtain $f(a+b)^2= 0$, proving the first condition as well. From the first three, we can easily get the fourth one as well. This proves the claim. Claim 2: For all $x \in \mathbb{R}$ we have $x^2-xf(x) \in S$ Proof: Note that from Claim 1, we know that $S$ has additive structure, and since we can negate as well, we can actually inductively get any linear combination, as long as the coefficients are integers. Suppose $x \ge 0$ for now, we handle the other sign later. Pick any $r \in S$ such that $r \geq x$. This is clearly possible, otherwise replace $r$ with $N|r|$ for large enough $N$. Set $y = r-x$. Note that $x, y \geq 0$ and thus \[x^2+yf(x), y^2+xf(y) \geq 0\]Hence, from $P(x, y)$ we obtain that \[f(x^2+yf(x)) = f(y^2+xf(y)) = 0\]In particular, \[f(x^2-xf(x) + rf(x)) = 0\]Repeating the above argument with $r = r+r'$ instead where $r' \geq 0$ is also in $S$, we obtain that $x^2-xf(x)+(r+r')f(x) \in S$. Subtracting, we obtain that $r'f(x) \in S$ for all $r' \geq 0$ in $S$. Since we can negate, we actually get that $tf(x) \in S$ for all $t \in S$. This will be crucial as well. Setting $t = r$ and simply subtracting, we obtain that $x^2-xf(x) \in S$. Now from $P(x, -x)$ we obtain that \[f(x^2+xf(-x)) = 0\]implying the result for $x \leq 0$ as well. Claim 3: $S$ is closed under multiplication. Proof: We first show that if $a \in S$ then $a/2 \in S$. Obviously, we may assume $a \geq 0$. Write $a = b^2$ (note that $b \in S$) and use Claim 2 for $x = b/2$. We obtain \[\frac{a}{4}-\frac{b}{2}f(b/2) \in S\]Multiplying by $2$ and adding $bf(b/2)$ (which is in $S$ by the way, from before) we get that $a/2 \in S$ as required. Now if $a, b \in S$ then \[(a+b)^2-a^2-b^2 = 2ab \in S\]and dividing by $2$ gets the desired result. Claim 4: $1 \in S$. In particular, $f(x) \in S$ for all $x \in \mathbb{R}$. Proof: Let $m \in S$ with $m \ne 0$. Now using Claim 2 for $1/m$ we obtain \[\frac{1}{m^2}-\frac{f(1/m)}{m} \in S\]Multiplying with $m^2$, and adding $mf(1/m)$ we obtain that $1 \in S$. It now follows that $f(x) \in S$ for all $x \in \mathbb{R}$. Claim 5: $S$ is closed under division. (no dividing by zero, obviously) Proof: Very similar to the proof above. Let $a, b \in S$ with $b \ne 0$. Using Claim 2 for $a/b$ we obtain \[\frac{a^2}{b^2}-\frac{af(a/b)}{b}\]Multiplying with $b$ and adding $af(a/b)$ we obtain that $a^2/b \in S$. Replacing $a$ with $\sqrt{|a|}$ if necessary, we can easily see now that $S$ is closed under division. Claim 6: Suppose $x \not \in S$ and $r \in S$. Then $f(x+r) = f(x)+2r$. Proof: We use Claim 2 on $x$ and $x+r$ to obtain: \[x^2-xf(x), (x+r)^2-(x+r)f(x+r) \in S\]Since $r^2, rf(x+r) \in S$ we may obtain that \[x^2+2xr-xf(x+r) \in S\]Subtracting, we now obtain \[x[f(x+r)-f(x)-2r] \in S\]Since $f(x), f(x+r), -2r \in S$ it follows that $f(x+r)-f(x)-2r \in S$. Therefore, if the previous quantity is nonzero, from Claim 5 we obtain that $x \in S$ by dividing. This would be a contradiction, thus \[f(x+r) = f(x)+2r\]for all $x \not \in S$ and $r \in S$. Claim 7: There is no $x \not \in S$. Proof: Let $x \not \in S$. Take any positive integer $n$. From $P(x, n)$ we have: \[f(x^2+nf(x))+f(n^2+xf(n)) = f(x+n)^2\]Note that $f(n) = 0$, $f(x+n) = f(x)+2n$ and that $f(x^2+nf(x)) = f(x)^2+2nf(x)$ (recall that $x^2 \not \in S$ and that $nf(x) \in S$) This gives us \[f(x)^2+2nf(x) = f(x)^2+4nf(x)+4n^2\]We then obtain $f(x) = -2n$, which is impossible since $n$ varies. With Claim 7, we see that $S \equiv \mathbb{R}$ and this implies $f(x) = 0$ for all $x \in \mathbb{R}$, retrieving the first solution mentioned at the start. Now (after so long ), we may assume that $f(x) = 0$ if and only if $x = 0$. Fortunately, the rest of the problem is not that complicated at all. Claim 8: $f(x) \leq x$ for all $x \in \mathbb{R}_{\geq 0}$. Proof: Suppose otherwise, and let $t \geq 0$ be such that $f(t) > t$. Replacing $t$ with $t^2$ repeatedly, eventually we will have \[f(t) > 2t\]Let $v = f(t)-2t$. Note that \[t^2+vf(t) = t^2+f(t)^2-2f(t)t = (f(t)-t)^2 = (v+t)^2\]and hence from $P(t, v)$ we obtain \[f(v^2+tf(v)) = 0\]which forces $v^2+tf(v) = 0$. However, we have $t, v > 0$, impossible. For $x \geq 0$ we have \[f(-x)^2 = f(x^2) \leq x^2\]implying that $f(-x) \in [-x, x]$. In particular, $f(-x) \geq -x$. From $P(x, -x)$ we obtain \[f(x^2-xf(x)) + f(x^2+xf(-x)) = 0\]for all $x \geq 0$. Note that $x^2-xf(x) = x \cdot (x-f(x)) \geq 0$ and $x^2+xf(-x) = x(x+f(-x)) \geq 0$. It follows that both terms on the left hand side above are nonnegative. Since they sum to zero, we obtain that \[f(x^2-xf(x)) = f(x^2+xf(-x)) = 0\]For $x > 0$ this tells us that $f(x) = x$ and $f(-x) = -x$. Adding in $x = 0$, we therefore obtain that $f(x) = x$ for all $x \in \mathbb{R}$; retrieving the second solution mentioned at the start. To reiterate, $f \equiv 0$ and $f \equiv \text{id}$ are the only solutions. $\square$

wow, that was quite a journey!

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