amparvardi wrote: Find all functions $f : \mathbb R \to \mathbb R$ such that for all $x, y,$ \[f(f(x) + y) = f(x^2 - y) + 4f(x)y.\] Let $P(x,y)$ be the assertion $f(f(x)+y)=f(x^2-y)+4f(x)y$ $P(x,\frac{x^2-f(x)}2)$ $\implies$ $f(x)(f(x)-x^2)=0$ and so : $\forall x$, either $f(x)=0$, either $f(x)=x^2$ Suppose now that $\exists a,b\ne 0$ such that $f(a)=0$ and $f(b)=b^2$ $P(a,b)$ $\implies$ $b^2=f(a^2-b)$ and so either $b^2=0$, either $b^2=(a^2-b)^2$ $\iff$ $a^2(a^2-2b)=0$ $\implies$ $b=\frac{a^2}2$ Choose then $c\in\mathbb R\setminus\{0,a,-a,\frac{a^2}2\}$ If $f(c)=0$, then $P(c,b)$ $\implies$ $b=\frac {c^2}2$ and so $c^2=a^2$, impossible If $f(c)=c^2$, then $P(a,c)$ $\implies$ $c=\frac {a^2}2$, impossible So either $f(x)=0$ $\forall x\ne 0$, either $f(x)=x^2$ $\forall x\ne 0$ So either $f(x)=0$ $\forall x$, either $f(x)=x^2$ $\forall x$ and it's easy to check that both are solutions. Hence the answer : $f(x)=0$ $\forall x$ $f(x)=x^2$ $\forall x$

I have explained this problem on my channel: Link: Video Solution

As usual, let $P(x,y)$ be the assertion of $f(f(x) + y) = f(x^2 - y) + 4f(x)y$. $P(0,0)\implies f(f(0))=f(0)$ $P(0,-f(0))\implies f(0)=f(0)-4f(0)^2\implies f(0)=0$. $P(x,x^2)\implies f(f(x)+x^2)=4f(x)x^2$ $P(x,-f(x))\implies f(f(x)+x^2)=4f(x)^2x$, hence $f(x)=0$ or $f(x)=x^2$. Let $\mathcal{S}=\{s\mid f(s)=0\}$ and $\mathcal{T}=\{t\mid f(t)=t^2\}$. Obviously, either $\vert \mathcal{S}\vert$ or $\vert \mathcal{T}\vert $ is infinite and assume that both sets are nonempty, also delete $0$ from the both sets since $0$ is bad. Now take $s$ and $t$, $P(s,t)\implies f(f(s) + t) = f(s^2 - t) + 4f(s)t\implies f(s^2 - t)=t^2\implies s^2(s^2-2t)=0\implies s^2=2t$ $P(t,t^2-s)\implies f(f(t) + t^2-s) = f(s) + 4f(t)(t^2-s)\implies f(2t^2-s)=4t^2(t^2-s)\implies t^2=s$, therefore $t^4=2t\implies t^3=2$. Thus, $\vert \mathcal{T}\vert \leq 3$. And by $s=t^2$, we also have that $\vert\mathcal{S}\vert \leq 3$. This contradicts the fact that one of them is infinite set. Therefore, our assumption was wrong, hence either $\mathcal{S}$ or $\mathcal{T}$ is empty ($0$ excluded). That means we have possibilities $f(x)=0\forall x\in\mathbb R\setminus 0$ and $f(x)=x^2\forall x\in\mathbb R\setminus 0$, note that since $f(0)=0$, it is easy to verify that following solutions work: $\boxed{f(x)=0\forall x\in\mathbb R,}$ $\boxed{f(x)=x^2 \forall x\in\mathbb R.}$

Let $P(x,y)$ denote the assertion. \[P(x,0)) : f(f(x)) = f(x^2)\]\[P(x,f(x) - x^2) : f(f(x)) = f(x^2) + 4f(x)(f(x) - x^2) \Longrightarrow f(x) = x^2 \text{ or } 0\]Assume that $\exists a,b \in \mathbb{R}$ such that $a,b \neq 0$ and $f(a) = 0$ and $f(b) = b^2$. $P(a,b) : f(a^2-b)=f(b)=b^2$. If $f(a^2-b) = 0$, then $b=0$, contradiction. Hence $(a^2-b)^2=b^2 \Longrightarrow a^2(a^2-2b)=0 \Longrightarrow a^2=2b$. Since $f(x^2) = f(f(x))$ for all reals $x$, $f(a^2) = f(f(a)) = f(2b) = 0$. $P(2b,b) : f(b) = f(4b-b^2)$. Again, $f(4b-b^2) \neq 0$ since $b\neq 0$. Hence \[(4b-b^2)^2=b^2\Longrightarrow 8b^3(2b-1)=0\Longrightarrow b=\tfrac 12 \text{ and } a=1.\]\[P\left(\tfrac 12,-\tfrac 34\right) : f\left(-\tfrac 12\right) = f(1)-\tfrac 34 = -\tfrac 34.\]Which is a contradiction since $f(x) \in \{x^2,0\}$. Hence the only solutions are $f(x) = x^2$ $\forall x$ $\in \mathbb{R}$ and $f(x)=0$ $\forall$ $x$ $\in$ $\mathbb{R}$.

I actually solved it, but it's just the same as pco's solution, so I won't take the pain of $\LaTeX$ compiling it @below okay, sorry.

ZETA_in_olympiad wrote: I actually solved it, but it's just the same as pco's solution, so I won't take the pain of $\LaTeX$ compiling it so what's the point in the bump?

megarnie wrote: ZETA_in_olympiad wrote: I actually solved it, but it's just the same as pco's solution, so I won't take the pain of $\LaTeX$ compiling it so what's the point in the bump? Pun: pointwise trap.