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 9 of 9: Extend to all reals and conclude

Previous step Next step

x<0:\ f(x)=f(-x)=(-x)^2=x^2;\qquad f(0)=0=0^2\ \Longrightarrow\ f(x)=x^2\ \text{for all}\ x

Detailed analysis

For $x<0$ , $-x>0$ so $f(-x)=(-x)^2=x^2$ by step 8, and $f(x)=f(-x)=x^2$ by evenness (step 2). Together with $f(0)=0=0^2$ (step 1), $f(x)=x^2$ holds for every real $x$ . This function satisfies the original equation, since $f(x^2+f(y))=(x^2+y^2)^2=x^4+2x^2y^2+y^4$ equals $f(f(x))+f(y^2)+2f(xy)=x^4+y^4+2x^2y^2$ . Together with the trivial solution $f\equiv0$ from step 4, the solutions are exactly $f(x)=0$ for all $x$ , and $f(x)=x^2$ for all $x$ .