Let $P(x,y)$ be the assertation $f(xy+f(x^2))=xf(x+y)$ $P(-x,-y) \Rightarrow xf(x+y)=-xf(-x-y) \Rightarrow -f(x)=f(-x) \Rightarrow f(0)=0$ $P(x,y+z) \Rightarrow f(xy+xz+f(x^2))=xf(x+y+z)$ $P(y,x+z) \Rightarrow f(xy+yz+f(y^2))=yf(x+y+z)$ By comparing these $2$ assertations we get $R(x,y,z); x,y\neq0$: $$yf(xy+f(x^2)+xz)=xf(xy+f(y^2)+yz)$$Let there exist $t\neq0$ such that $f(t)=0 \Rightarrow f(-t)=0 \Rightarrow WLOG(t>0)$ $R(t^{1/2},y,z)\Rightarrow yf(t^{1/2}(y+z))=t^{1/2}f(t^{1/2}y+f(y^2)+yz)$ $R(-t^{1/2},y,z)\Rightarrow yf(-t^{1/2}(y+z))=-t^{1/2}f(-t^{1/2}y+f(y^2)+yz)$ Comparing these $2$ we get: $f(t^{1/2}y+f(y^2)+yz)=f(-t^{1/2}y+f(y^2)+yz)$ By taking $z=-\frac{f(y^2)}{y}$ we get: $f(t^{1/2}y)=f(-t^{1/2}y)=-f(t^{1/2}y) \Rightarrow f(t^{1/2}y)=0$ Beacuse we also have $f(0)=0$ we get $\boxed{f(x)=0,\forall x}$ which indeed works If there doesn't exist such a $t$ then $f$ is injective at $0 \Rightarrow$ $P(x,-x)\Rightarrow f(f(x^2)-x^2)=0 \Rightarrow f(x^2)=x^2 \Rightarrow $ Using $-f(x)=f(-x) \Rightarrow \boxed{f(x)=x,\forall x}$ which indeed works

This problem was proposed by me (Dorlir Ahmeti, Albania)

The only solutions are $\boxed{f\equiv 0}$ and $\boxed{f(x) = x}$, which clearly work. Now we prove they are the only solutions. Let $P(x,y)$ be the given assertion. Assume $f$ is nonconstant. $P(x,0): f(f(x^2)) = xf(x)$. For any $x\ne 0$, comparing $P(x,0)$ and $P(-x,0)$ gives that $f(-x) = -f(x)$. Claim: $f(0) = 0$. Proof: Suppose otherwise. Let $c = |f(0)|$. $P(0,0): f(f(0)) = 0$, so $f(c) = \pm f(f(0)) = 0$ and $c > 0$. $P(\sqrt{c}, -\sqrt{c}): f(-c) = \sqrt{c} f(0)$. However, $f(-c) = -f(c) = 0$, so $f(0) =0$. $\square$ This also implies that $f$ is odd. Claim: There does not exist any $x\ne 0$ with $f(x) = 0 $. Proof: Suppose otherwise. Since $f(-x) =- f(x)$, there exists some $k > 0$ with $f(k^2) = 0$. $P(k,x): f(kx) = k f(x + k)$. Now, for any real number $x$, $P(k,x)$ compared with $P(k,-x)$ gives that $f(x + k) = -f(-x + k) = f(x - k)$, so $f(x) = f(x + d)$ for all $x$, where $d = 2k \ne 0$. For all $x\ne 0$, $P\left( x, \frac{d}{x} - x\right): f(f(x^2) - x^2) = x f\left( \frac dx \right)$. $P(x,-x): f(f(x^2) - x^2) = 0$. Hence for any $x\ne 0$, $f \left( \frac dx \right) = 0$, so $f$ is constant, contradiction. $\square$ $P(x,-x): f(f(x^2) - x^2) = 0\implies f(x^2) = x^2$. Since $f$ is odd, $f(x) = x$ must hold for all reals $x$.

Not actually difficult, mostly since this so standard. The answers are $f(x)=0$ for all $x \in \mathbb{R}$ and $f(x)=x$ for all $x\in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(xy+f(x^2))=xf(x+y)$. Then, $P(0,0)$ gives us that $f(f(0))=0$. It is easy to see that if $f$ is constant, then it must be constant 0. In what follows, we assume that $f$ is non-constant. We start off by proving the following important property of $f$. Claim : The function $f$ is odd. Proof : Comparing $P(x,y)$ and $P(-x,-y)$ we have, \[xf(x+y)=f(xy+f(x^2))=f((-x)(-y)+f((-x)^2))=-xf(-x-y)\]Thus, $f(-(x+y))=-f(x+y)$ for all $x\neq 0$, and thus, setting $x=1$ and $y=t-1$, we have that $f(-t)=-f(t)$ for all $t\in \mathbb{R}$, which proves the claim. Claim : $f(x)=0$ if and only if $x=0$. Proof : We know that there exists $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha)=0$ and thus, $f(-\alpha)=0$ since $f$ is odd. Then, note that $P(\sqrt{\alpha},-\sqrt{\alpha})$ implies, \[0=f(-\alpha)=f(-\alpha + f(\alpha))=\sqrt{\alpha }f(\sqrt{\alpha} - \sqrt{\alpha})=\sqrt{\alpha}f(0)\]which implies that $\alpha=0$ or $f(0)=0$, either way we have that $f(0)=0$. Now, say there exists $x_0\neq 0 \in \mathbb{R}$ such that $f(x_0)=0$. Again since $f$ is odd, $f(-x_0)=0$ and we WLOG, assume that $x_0>0$. Note that $P\left(\sqrt{x_0},\frac{-f(x_0)}{\sqrt{x_0}}\right)$ gives us \[0=f(0)=f\left(\sqrt{x_0}\left(\frac{-f(x_0)}{\sqrt{x_0}}\right)+f(x_0)\right)=f\left(\frac{\sqrt{x_0}-f(x_0)}{\sqrt{x_0}}\right)=f\left(\frac{\sqrt{x_0}}{\sqrt{x_0}}\right)=f(1)\]Thus, $f(1)=0$. Now, looking at $P(1,x)$ gives us that $f(x)=f(x+1)$ for all $x\in \mathbb{R}$. Thus, $f$ is periodic with period $T=1$. Now, $P(x,y+1)$ gives us that \[f(xy+f(x^2)+x)=xf(x+y+1)=xf(x+y)=f(xy+f(x^2)+x)\]Then, considering $x\neq 0$ and $y=\frac{t-f(x^2)}{x}$, we have that $f(t+x)=f(t)$ for all $t \in \mathbb{R}$ and $x\neq 0$, thus, we must have that $f$ is in fact constant, which is a clear contradiction. Thus, such $x_0$ cannot exist and our claim is proved. Now, note that $P(x,-x)$ gives \[f(-x^2+f(x^2))=xf(x-x)=xf(0)=0\]which by our previous claim implies that $-x^2=f(x^2)$, so $f(t)=t$ for all $t \in \mathbb{R}_{>0}$. Combining this with the fact that $f$ is odd, we can conclude that $f(x)=x$ for all $x\in \mathbb{R}$ which shows that all solutions to the given equation are indeed of the claim types.

Let $P(x, y)$ denote the assertion $f(xy+f(x^2))=xf(x+y)$. $P(0,0) : f(f(0))=0$ $P(x,0) : f(f(x^2))=xf(x)$ $P(-x,-y) : f(xy+f(x^2))=-xf(-x-y)=xf(x+y)$ so $f(x)=f(-x)$ for all $x$, and this implies that $f$ is odd. Let $a=|f(0)|$ so how $f(|f(0)|)=0$ we have that $f(a)=f(-a)=0$. $P(\sqrt{a}, -\sqrt{a}): f(-a) = \sqrt{a} f(0)=0$ so $f(0)=0$. $P(x,-x) : f(-x^2+f(x^2))=xf(0)=0$ so $ f(f(x^2)-x^2)=0$. Claim: If exist $c\ne 0$ such that f(c)=0 we have that $f \equiv 0$ Proof: Let $d=|c|$.$P(\sqrt{d},x) : f(x\sqrt{d})=\sqrt{d}f(x+\sqrt{d})$ for all $x$. $P(\sqrt{d},-x) : -f(x\sqrt{d})=-\sqrt{d}f(x-\sqrt{d})$ for all $x$ so $f(x+\sqrt{d})=f(x-\sqrt{d})$ and this implies that $f(x)=f(x+2\sqrt{d})$ for all $x$. Let $2\sqrt{d}=t\ne 0$ and from $P(x,\frac{t}{x}-x)$ we have : $f(t-x^2+f(x^2))=f(f(x^2)-x^2)=x f(\frac{t}{x})=0$ for all $x$ so $f(\frac{t}{x})=0$. If we put $x=\frac {t}{x}$ we have that $f(x)=0$ for all $x$.So the claim is proved. If $f(x)$ is not $0$ for all $x$ we have that $f(x^2)=x^2$ for all $x$ and how $f$ is odd we obtained that $f(x)=x$ for all $x$. So we've shown that either $f(x) = x$ for all $x\in \mathbb{R}$, or $f\equiv 0$, and that both functions work, so we're done.

The only solution are $f(x)=0 \ \forall \ x \in \mathbb{R}$ and $f(x)=x \ \forall \ x \in \mathbb{R}$. We will prove that they are the only solutions, let $P(x,y)$ be the assertion of $f(xy+f(x^2))=xf(x+y)$. From $P(0,0)$, we have $$f(f(0))=0$$If $f$ is constant then $f(x)=0 \ \forall \ x \in \mathbb{R}$. Which satisfies. Assume otherwise, from now on assume there exist $a \in \mathbb{R}$ such that $f(a) \neq 0$. Claim 1. For every $x \neq 0$, we have $f(x)=-f(-x)$. Proof. From $P(x,0)$, we have $$f(f(x^2))=xf(x)$$From $P(-x,0)$, we have $$f(x^2)=-xf(-x)$$Comparing the two equations above for $x \neq 0$, we must have $f(x)=-f(-x)$. As desired. Claim 2. $f(x)=0$ if and only if $x=0$ Proof. First, we will prove that $f(0)=0$. FTSOC, assume $f(0) \neq 0$. Then, we have $f(-f(0))=-f(f(0))=0$. From $P(\sqrt{|f(0)|},-\sqrt{|f(0)|})$, we have $$f(\pm f(0) + f(\mp f(0))) = \sqrt{|f(0)|} \cdot f(0) \rightarrow 0 = \sqrt{|f(0)|} \cdot f(0)$$Since $f(0) \neq 0$, this is a contradiction. Thus $f(0)=0$. Now we can say that $f$ is $odd$. Now we will prove that there doesn't exist $t \in \mathbb{R} / \{ 0\}$ such that $f(t)=0$. FTSOC, assume there exist such $t$. We have $f(-t)=-f(t)=0$, WLOG assume $t>0$. From $P(\sqrt{t},y)$, we have $$f(\sqrt{t} \cdot y + f(t))= \sqrt{t} \cdot (y + \sqrt{t}) \Rightarrow f(\sqrt{t} \cdot y)= \sqrt{t} \cdot (y + \sqrt{t})$$From $P(-\sqrt{t},y)$, we have $$f(-\sqrt{t} \cdot y + f(t))= -\sqrt{t} \cdot (y - \sqrt{t}) \Rightarrow f(-\sqrt{t} \cdot y)= -\sqrt{t} \cdot (y - \sqrt{t})$$Since $f$ is $odd$ and $t>0$, by comparing the two equations above we have $f(y + \sqrt{t})=f(y - \sqrt{t})$. Hence $f$ is $periodic$ with period $2\sqrt{t}$. There exist a sufficiently large $n_0 \in \mathbb{Z}$ such that for any $x,y \in \mathbb{R}$ with fixed $x+y=a$, we have $$a^2-4(f((a-y)^2-n_0 \cdot 2\sqrt{t}) \ge 0$$So, the equation $y^2-a\cdot y +f((a-y)^2)-n_0 \cdot 2\sqrt{t}=0$ or equivalently $xy+f(x^2)=x+y+n_0 \cdot 2\sqrt{t}$ has at least one real solution , say $(x_0,y_0)$. From $P(x_0,y_0)$ and using the fact that $f$ is $periodic$ with period $2\sqrt{t}$, we have $$f(x_0y_0 + f(x_{0}^2))=x_0f(x_0+y_0) \Rightarrow f(x_0+y_0+n_0 \cdot 2\sqrt{t})=x_0f(x_0 + y_0) \Rightarrow f(a)=x_0 \cdot f(a)$$Since $f(a) \neq 0$, we must have $x_0=1$. But substituting this to $x_0y_0+f(x_{0}^2)=x_0+y_0+n_0 \cdot 2\sqrt{t}$ gives $f(1)=n_0 \cdot 2\sqrt{t}+1$. But we can choose another value $n_1 \in \mathbb{Z}$ greater than $n_0$ as $n_0$, and similarly get $f(1)=n_1 \cdot 2\sqrt{t}+1$ since $n_1$ satisfy $$a^2-4(f((a-y)^2-n_1 \cdot 2\sqrt{t}) \ge a^2-4(f((a-y)^2-n_0 \cdot 2\sqrt{t}) \ge 0$$for fixed $x+y=a$. Thus, this is a contradiction. As desired. For the finishing blow, from $P(x,-x)$ we have $$f(f(x^2)-x^2))=xf(0) \Rightarrow f(f(x^2)-x^2))=0$$Therefore $f(x^2)-x^2=0$ or $f(x^2)=x^2$ for every $x \in \mathbb{R}$. This implies $f(x) = x \ \forall \ x \in \mathbb{R}_{\ge 0}$, since $f$ is $odd$ we must have $f(x) = x \ \forall \ x \in \mathbb{R}$. Hence, the only solution are $f(x)=0 \ \forall \ x \in \mathbb{R}$ and $f(x)=x \ \forall \ x \in \mathbb{R}$.

Let $P(x,y)$ denote the assertion of $(x,y)$ into the original FE. Comparing $P(x,0)$ and $P(-x,0)$ shows $f$ is odd. Also\[P\left(z, \frac{y-f(z^2)}{z}\right)\Longrightarrow f(y) = zf\left(\frac{y-f(z^2)}{z}+z\right).\]Using the fact that $f$ is odd implies that: \[zf\left(\frac{y-f(z^2)}{z}+z\right) = f(y) = -f(-y) = -zf\left(\frac{-y-f(z^2)}{z}+z\right)=zf\left(\frac{y+f(z^2)}{z}-z\right).\]Changing $y\to yz$ implies that if for some $z\neq 0$ we have $\frac{f(z^2)-z^2}{z}=\frac{d}{2}\neq 0$, then $f$ would be periodic with period $d$. We now deal with the two cases separately: If no such $z$ exists, then $f(z^2)=z^2$ for all $z\neq 0$, but $f$ is odd, so we get $f(-z^2)=-z^2$ as well. Plugging in $P(x,-x)$ also shows that $f(0)=0$, so this case leads to $f\equiv \text{id}$, which is indeed a solution. If such a $z$ exists, then $f$ is periodic with period $d>0$. Now compare $P(x_1, x_2)$ and $P(x_2, x_1)$ where $x_2^2-x_1^2=d$. That is, for every $x_1\geq 0$, we pick $x_2=\sqrt{x_1^2+d}$. These substitutions give: \[x_1f(x_1+x_2)=f(x_1x_2+f(x_1^2))=f(x_2x_1+f(x_2^2))=x_2f(x_1+x_2)\Longrightarrow f(x_1+x_2)=0.\]However, this shows that $f(x+\sqrt{x^2+d})=0$ for all $x\geq 0$, but as $x+\sqrt{x^2+d}$ is continuous and strictly increasing, we have $f(x)=0$ for all $x\geq \sqrt{d}$. As $f$ is periodic, this trivially translates to $f\equiv 0$ over the entire $\mathbb{R}$, which is also a solution.

The only solutions are $f \equiv 0$ or $f \equiv x$. Those clearly work, so we will prove that these are the only solutions. As usual, let $P(a,b)$ denote the assertion of the given statement. Assume $f$ is non constant from henceforth. Claim: $f$ is an odd function Proof: $P(x,0)$ gives that $f(f(x^2))=xf(x)$. Here plug $-x$ to prove the claim. Claim: There exists a non negative integer (denoted henceforth by $k$) such that $f(k)=0$ Proof: By the above claim, it is sufficient to show that $0$ is in the image of $f$. $P(0,0)$ gives that $f(f(0))=0$. Claim: $k=0$ and $f(x)=0$ iff $x=0$ Proof: Suppose $k>0$. Then $P\left(\sqrt{k}, \frac{\sqrt{k}}{\sqrt{k}-1}\right)$ gives that $k=1$ or $k=0$. If $k=1$, then $P(1,y)$ gives that $f(y+1)=f(y)$. Thus, we can conclude that $f(2)=0$. This is a contradiction because $k$ can only be $0$ or $1$. Thus, $k=0$ Claim: $f\left( \frac{f(x^2)-x^2}{1-x} \right)=0$ for $x \ne 1$ Proof: $P\left(x,\left( \frac{f(x^2)-x^2}{1-x} \right)\right)$ immediately proves the claim. From the above claim, we get that $f(x^2)=x^2$, and by the first claim, we get that $f(x)=x$ for all $x \ne \pm 1$. Now $P(3,-2)$ gives that $f(\pm 1) = \pm 1$

nyaaa The solutions are the identity and $0$. These trivially work. Checking constant functions, $f$ being $0$ works. Else, let $f$ not be constant. Let $P(x,y)$ be the assertion given by the problem! Firstly, taking $P(x,-x)$ gives $f(-x^2+f(x))=xf(0)$ but taking $P(-x,x)$ gives $f(-x^2+f(x^2))=-xf(0)$ hence $xf(0)=-xf(0)$. Pick $x\neq 0$, so $f(0)=0$. We'll show $f$ is odd. Take $P(x,y)$ and $P(-x,-y)$, the LHS on each is the same, but the RHS is $-xf(-x-y)=xf(x+y)$. Pick $x=x_0\neq 0$, and $f(-y-x_0)=f(y+x_0)$. But if $y+x_0=z$, $f(-z)=f(z)$, for all $z$ therefore $f$ is odd ($y+x_0$ spans all reals.) We claim $f$ is not periodic. Assume it is, that it has period $k$. Then, $f(x+k)=f(x)$ for all $x$. Consider $P\left(x, y\right)$ and $P\left(x,y-\frac{k}{x}\right)$. The LHS is $f(\text{stuff})$ versus $f(\text{stuff}-k)$, hence equal. But the RHS, for nonzero $x$, would imply that \[f(x+y)=f(x+y-\frac{k}{x})\]Let $x+y=z$, making an equation in $z,x$, which is possible as for each choice of $z,x$, we get $y$ from $z-x$. Then, \[f(z)=f(z-\frac{k}{x})\]as $k\neq 0$, $\frac{k}{x}$ spans the whole of $\mathbb{R}$, excluding $0$ [but $f(z)=f(z+0)$ anyways], hence fix $z$, and $f$ is constant, contradiction. Now, we claim $f$ is injective at $0$. Assume otherwise, that $f(c)=0$, for $c\neq 0$. As $f$ is odd, $f(-c)=0$ too, so WLOG $c>0$. Then $\sqrt{c}$ exists, so consider $P(\sqrt{c}, y)$ and $P(\sqrt c, -y)$. Then \[\sqrt{c}f(y+\sqrt c)=f(\sqrt{c}y)=-\sqrt{c}f(\sqrt{c}-y)=\sqrt{c}f(y-\sqrt{c})\]by oddness, and as we can cancel by $\sqrt{c}\neq 0$. Thus, $f(y-\sqrt{c})=f(y+\sqrt{c})$, and shifting $y$ gives $f$ is periodic, noting $\sqrt{c}\neq 0$, contradiction. Now, from $P(x,-x)$ before, $f(f(x^2)-x^2)=0=f(0)$, hence by injectiveness at $0$, $f(x^2)=x^2$. Thus, $f(xy+x^2)=xf(x+y)$. Factor $xy+x^2$ as $x(x+y)$. Now, let $z=x+y$ again, noting $y=z-x$, so $f(xz)=xf(z)$, for all $x,z$ real. Then let $z=1$, so $f(x)=xf(1)$. Now, $P(1,y)$ gives $f(y+f(1))=f(y+1)$. As $f$ isn't periodic, $f(1)=1$, so $f(x)=x$, yay.

Okay, this is very nice! We claim the only solutions are $f\equiv 0$ and $f \equiv id$. Both of these functions obviously work; now we prove these are the only such ones. Let $P(x, y)$ denote the assertion. Note that if $f$ is constant, then $f \equiv 0$. Therefore, assume that $f$ is nonconstant. Claim 1: $f$ is odd. Proof: Note that $P(x, 0)$ implies that $f(f(x^2)) = xf(x) = -xf(-x) \implies f(-x) = -f(x) \forall x \ne 0$. Now we prove that $f(0) = 0$ to complete the proof. Assume that $f(0) \ne 0$. Note that $f(f(0)) = 0$. Now, $P\left(\sqrt{|f(0)|}, -\sqrt{|f(0)|}\right) \implies 0 = f(f(0)) = \sqrt{|f(0)|}f(0) \implies f(0) = 0$ anyways, so $f(0) = 0$. $\square$ Claim 2: $f$ is injective at zero. Proof: Assume $\exists k > 0 : f(k) = 0$. Then $P(x, \sqrt k) \implies f(\sqrt ky) = \sqrt k f(y + \sqrt k) \implies f(\sqrt k -y) = -f(\sqrt k + y) \implies f(-sqrtk - y),$ i. e. $f$ is periodic. Therefore let $p$ be the period. $P(x, y+p) \implies f(xy+f(x^2) + xp) = xf(x+y) = f(xy+f(x^2))$. For $x \ne 0$, $\{xy+f(x^2): y \in \mathbb{R}\} = \mathbb{R}$. Thus $f(m+xp) = f(m) \forall m, x$. But this means $f$ is constant, as we can force $m+xp = a$ for each $a$. Contradiction. $\square$ Now that we have the above two claims, $P(x, -x) \implies f(f(x^2)-x^2) = 0 \implies f(x^2) = x^2 \implies f(x) = x \forall x \ge 0 \implies f(x) = x \forall x$, as desired. $\blacksquare$