Problem 5

Determine all functions

f:\mathbb{R}\to\mathbb{R}

such that

f(x^2+f(y))=f(f(x))+f(y^2)+2f(xy)

for all real numbers

x

and

y

Step 1 of 9: Basic values and a symmetric identity

Previous step Next step

f(0)=0;\qquad f(x^2)=f(f(x));\qquad f(x^2+f(y))=f(x^2)+f(y^2)+2f(xy)=f(y^2+f(x))

Detailed analysis

Setting $x=y=0$ gives $f(f(0))=f(f(0))+3f(0)$ , so $f(0)=0$ . Setting $y=0$ then gives $f(x^2)=f(f(x))$ for all $x$ . Using this, the right side of the original equation equals $f(x^2)+f(y^2)+2f(xy)$ , which is symmetric in $x,y$ , so $f(x^2+f(y))=f(y^2+f(x))$ for all $x,y$ ; call this identity $(*)$ .