djb86 wrote: Find all functions $f : \mathbb{R} \to \mathbb{R}$ that satisfy \[f(x^2 + y) + f(f(x) - y) = 2f(f(x)) + 2y^2\quad\text{ for all }x, y \in \mathbb{R}.\] Let $P(x,y)$ be the assertion $f(x^2+y)+f(f(x)-y)=2f(f(x))+2y^2$ $P(x,0)$ $\implies$ $f(x^2)=f(f(x))$ $P(x,f(x)-x^2)$ $\implies$ $f(x^2)=f(f(x))+2(f(x)-x^2)^2$ Subtracting, we get $\boxed{f(x)=x^2}$ $\forall x$ which indeed is a solution.

Let $P(x,y)$ be the assertion \[ f(x^2+y)+f(f(x)-y)=2f(f(x))+2y^2 .\] $P(x,-x^2)$ $ \Rightarrow $ $f(0)+f(f(x)+x^2)=2f(f(x))+x^4$ and $P(x,f(x))$ $ \Rightarrow $ $ f(x^2+f(x))+f(0)=2f(f(x))+f(x)^2 $. So, we get that $ f(x)^2=x^4 $, $ \forall x .$ Answer: $f(x)=x^2$. :

mathuz wrote: Let $P(x,y)$ be the assertion \[ f(x^2+y)+f(f(x)-y)=2f(f(x))+2y^2 .\] $P(x,-x^2)$ $ \Rightarrow $ $f(0)+f(f(x)+x^2)=2f(f(x))+x^4$ and $P(x,f(x))$ $ \Rightarrow $ $ f(x^2+f(x))+f(0)=2f(f(x))+f(x)^2 $. So, we get that $ f(x)^2=x^4 $, $ \forall x .$ Answer: $f(x)=x^2$. : 1) two typos errors : ============== $f(0)+f(f(x)+x^2)=2f(f(x))+2x^4$ and not $f(0)+f(f(x)+x^2)=2f(f(x))+x^4$ $ f(x^2+f(x))+f(0)=2f(f(x))+2f(x)^2 $ and not $ f(x^2+f(x))+f(0)=2f(f(x))+f(x)^2 $ 2) a missing step : ============ You cant immediately conclude from $f(x)^2=x^4$ that $f(x)=x^2$. A lot of other functions are such that $f(x)^2=x^4$ : $f(x)=-x^2$ $f(x)=x^2$ when $x\in\mathbb Q$ and $f(x)=-x^2$ when $x\notin\mathbb Q$ $f(x)=\lim_{t\to x^+}\frac{|t^3-t|}{t-1}$ ...