The only solutions are $f(x) = 0$ and $f(x) = \frac{1}{4}x^2$, which indeed work. Let $P(x, y)$ be the assertion $f(x^2+f(y))=f(f(y)-x^2)+f(xy)$. $P(0, 0) : f(0) = 0$ $\Longrightarrow P(x, 0) : f(x^2) = f(-x^2) \Longrightarrow f(x) = f(-x)$ If there exists $t \neq 0$ such that $f(t) = 0$, then $P(x, t) : f(xt) = 0 \Longrightarrow \boxed{f(x) = 0}$ From now we assume $f(x) = 0$ if and only if $x = 0$. Suppose $f(a) = f(b)$ for some $0 < a < b$, then comparing $P(x, a)$ with $P(x, b)$ gives $f(ax) = f(bx)$. Let $t \neq 0$ be an arbitrary real number, then there exists $x$ such that $ax + bx = 2f(t)$. Then, $bx > \frac{ax + bx}{2} = f(t)$ and $P(\sqrt{bx - f(t)}, t) : f(t\sqrt{bx - f(t)}) = f(bx) - f(ax) = 0,$ contradiction since $t\sqrt{bx - f(t)} \neq 0$. Therefore, if $f(a) = f(b)$, then $|a| = |b|$ (combined with the fact that $f$ is even). By the Intermediate Value Theorem, if there exists $a, b > 0$ such that $f(a) > 0 > f(b)$, then there exists $c > 0$ such that $f(c) = 0$, contradiction, so either $f(x) \geq 0$ for all $x$, or $f(x) \leq 0$. If $f(x) \geq 0$, $P(\sqrt{f(x)}, x) : f(2f(x)) = f(x\sqrt{f(x)}) \Longrightarrow 4f(x)^2 = x^2f(x) \Longrightarrow \boxed{f(x) = \frac{1}{4}x^2}$ And if $f(x) \leq 0$, $P(\sqrt{-f(x)}, x) : f(2f(x)) + f(x\sqrt{-f(x)}) = 0$ contradiction when $x \neq 0$. Hence, we are done.

navi_09220114 wrote: Find all continuous functions $ f : \mathbb{R} \rightarrow \mathbb{R} $ such that for all real numbers $ x, y $ $$ f(x^2+f(y))=f(f(y)-x^2)+f(xy) $$ Here is a solution which does not need continuity : Let $P(x,y)$ be the assertion $f(x^2+f(y))=f(f(y)-x^2)+f(xy)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x^2)=f(-x^2)$ and $f(x)$ is an even function If $f(u)=0$ for some real $u\ne 0$, then $P(\frac xu,u)$ $\implies$ $\boxed{\text{S1 :}f(x)=0\quad\forall x}$, which indeed fits So let us from now consider that $f(x)=0$ $\iff$ $x=0$ If $f(a)=f(ca)$ for some $a\ne 0,c\ne \pm 1$, comparing $P(\frac xa,a)$ with $P(\frac xa,ca)$ implies $f(x)=f(cx)$ $\forall x$ And so $f(c^nx)=f(x)$ $\forall x$, $\forall n\in\mathbb Z$ Let then $y\ne 0$ and any (of infinitely many) $n\in\mathbb Z$ such that $c^{2n}\ge \frac{4f(y)}{y^2}$ Quadratic $x^2-c^nxy+f(y)=0$ has nonzero roots $x_1,x_2$ and $x_i^2+f(y)=c^nx_iy$ So $P(x_i,y)$ implies $f(f(y)-x_i^2)=0$ and so $f(y)=x_i^2$ And so (plugging this in quadratic $x_i^2-c^nx_iy+f(y)=0$) : $2x_i^2=c^nx_iy$ And so $x_i=c^n\frac y2$ and $f(y)=x_i^2=c^{2n}\frac {y^2}4$ But this can not be true for infinitely many $n$ So no such $a,c$ And so $f(a)=f(b)$ implies $a=\pm b$ If $f(y)\le \frac {y^2}4$ for some $y\ne 0$, then quadratic $x^2+f(y)=xy$ has at least one real root $x_1\ne 0$ $P(x_1,y)$ is $f(f(y)-x_1^2)=0$ and so $f(y)=x_1^2$ Plugging this in quadratic $x_1^2+f(y)=x_1y$, we get $x_1=\frac y2$ and so $f(y)=x_1^2=\frac {y^2}4$ So $f(x)\ge \frac{x^2}4$ $\forall x\ne 0$, still true when $x=0$ As a consequence, $f(x)>0$ $\forall x\ne 0$ Let then $x\ne 0$ : $P(\sqrt{f(x)},x)$ $\implies$ $f(2f(x))=f(x\sqrt{f(x)})$ And so $2f(x)=\pm x\sqrt{f(x)}$ and so $4f(x)^2=x^2f(x)$ And so $\boxed{\text{S2 : }f(x)=\frac{x^2}2\quad\forall x}$ which indeed fits (note that we got this only for $x\ne 0$ but this remains trivially true when $x=0$)

pco wrote: navi_09220114 wrote: Find all continuous functions $ f : \mathbb{R} \rightarrow \mathbb{R} $ such that for all real numbers $ x, y $ $$ f(x^2+f(y))=f(f(y)-x^2)+f(xy) $$ Here is a solution which does not need continuity : Let $P(x,y)$ be the assertion $f(x^2+f(y))=f(f(y)-x^2)+f(xy)$ $P(0,0)$ $\implies$ $f(0)=0$ $P(x,0)$ $\implies$ $f(x^2)=f(-x^2)$ and $f(x)$ is an even function If $f(u)=0$ for some real $u\ne 0$, then $P(\frac xu,u)$ $\implies$ $\boxed{\text{S1 :}f(x)=0\quad\forall x}$, which indeed fits So let us from now consider that $f(x)=0$ $\iff$ $x=0$ If $f(a)=f(ca)$ for some $a\ne 0,c\ne \pm 1$, comparing $P(\frac xa,a)$ with $P(\frac xa,ca)$ implies $f(x)=f(cx)$ $\forall x$ And so $f(c^nx)=f(x)$ $\forall x$, $\forall n\in\mathbb Z$ Let then $y\ne 0$ and any (of infinitely many) $n\in\mathbb Z$ such that $c^{2n}\ge \frac{4f(y)}{y^2}$ Quadratic $x^2-c^nxy+f(y)=0$ has nonzero roots $x_1,x_2$ and $x_i^2+f(y)=c^nx_iy$ So $P(x_i,y)$ implies $f(f(y)-x_i^2)=0$ and so $f(y)=x_i^2$ And so (plugging this in quadratic $x_i^2-c^nx_iy+f(y)=0$) : $2x_i^2=c^nx_iy$ And so $x_i=c^n\frac y2$ and $f(y)=x_i^2=c^{2n}\frac {y^2}4$ But this can not be true for infinitely many $n$ So no such $a,c$ And so $f(a)=f(b)$ implies $a=\pm b$ If $f(y)\le \frac {y^2}4$ for some $y\ne 0$, then quadratic $x^2+f(y)=xy$ has at least one real root $x_1\ne 0$ $P(x_1,y)$ is $f(f(y)-x_1^2)=0$ and so $f(y)=x_1^2$ Plugging this in quadratic $x_1^2+f(y)=x_1y$, we get $x_1=\frac y2$ and so $f(y)=x_1^2=\frac {y^2}4$ So $f(x)\ge \frac{x^2}4$ $\forall x\ne 0$, still true when $x=0$ As a consequence, $f(x)>0$ $\forall x\ne 0$ Let then $x\ne 0$ : $P(\sqrt{f(x)},x)$ $\implies$ $f(2f(x))=f(x\sqrt{f(x)})$ And so $2f(x)=\pm x\sqrt{f(x)}$ and so $4f(x)^2=x^2f(x)$ And so $\boxed{\text{S2 : }f(x)=\frac{x^2}2\quad\forall x}$ which indeed fits (note that we got this only for $x\ne 0$ but this remains trivially true when $x=0$) Very nice