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 2 of 9: f is even

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2f(x)=f(x^2+f(1))-f(x^2)-f(1)\ \Longrightarrow\ f(-x)=f(x)

Detailed analysis

Setting $y=1$ in the original equation and using $f(f(x))=f(x^2)$ gives $f(x^2+f(1))=f(x^2)+f(1)+2f(x)$ , i.e. $2f(x)=f(x^2+f(1))-f(x^2)-f(1)$ . The right side depends on $x$ only through $x^2$ , which is unchanged when $x$ is replaced by $-x$ ; hence $f(x)=f(-x)$ for all $x$ .