djb86 wrote: Find all functions $f$ such that \[f(f(x) + y) = f(x^2-y) + 4yf(x)\] for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $f(f(x)+y)=f(x^2-y)+4yf(x)$ $P(x,\frac{x^2-f(x)}2)$ $\implies$ $f(x)(f(x)-x^2)=0$ $\forall x$ 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 since $f(a^2-b)$ is either $0$, either $(a^2-b)^2$, we get $b^2=(a^2-b)^2$ and so $a^2=2b$ let then $c\notin\{0,a,b,\frac{a^2}2,\pm\sqrt{|2b|}\}$ : If $f(c)=0$, previous path using $c$ instead of $a$ implies $c^2=2b$, impossible If $f(c)=c^2$, previous path using $c$ instead of $b$ implies $a^2=2c$, impossible So no such $a,b$ exist and we got two solutions : $f(x)=0$ $\forall x$, which indeed is a solution $f(x)=x^2$ $\forall x$, which indeed is a solution

Is the similarity with http://www.artofproblemsolving.com/Forum/viewtopic.php?p=827178&sid=87879fa80ebb0714ee55f93e2120b749#p827178 accidental?

Let's make $f(f(x) + y) = f(x^2-y)$ $P(x,\frac{x^2-f(x)}{2}) : 4f(x)\frac{x^2-f(x)}{2} = 0 \implies f(x) = 0$ or $f(x) = x^2$. Let $a,b \neq 0$ such that $f(a) = 0$ and $f(b) = b^2$. $P(a,b) : b^2 = (a^2-b)^2 \implies b = \frac{a^2}{2}$. $P(a,b) , P(b,a)$ gives contradiction so no such $a,b$ exists. Answers : $f(x) = 0$ and $f(x) = x^2$.

Let $P(x,y)$ denote the assertion. $P(x,x^2)-P(x,-f(x))\implies 4f(x)^2=4x^4f(x).$ So $f(x)=0$ or $f(x)=x^2.$ Suppose $\exists a\in \mathbb{R} : a\neq 0, f(a)\neq 0.$ Then $f(a)=a^2\implies f(a^2+y)=f(a^2-y)+4ya^2.$ If $f(a^2-y)=0$ for some $y\neq 0,$ it follows that $f(a^2+y)=4ya^2=(a^2+y)^2\neq 0 \implies a^2-y=0.$ So $f(a)\neq 0$ for some $a \implies f(x)=0,$ when $x=0.$ So $f(x)=0 ~~\forall x\in \mathbb{R} \text{ or } f(x)=x^2~~\forall x\in \mathbb{R},$ obviously these fit.