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 8 of 9: Determine f on the positive reals

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x>0,\ m=f(x)\ \Longrightarrow\ f(m/x^2)=f(1)=1\ \Longrightarrow\ m/x^2=1\ \Longrightarrow\ f(x)=x^2

Detailed analysis

Fix $x>0$ and let $m=f(x)>0$ (step 5). By step 1, $f(m)=f(f(x))=f(x^2)$ ; by the lemma (step 3) with multiplier $1/x^2$ , $f(m/x^2)=f(x^2/x^2)=f(1)=1$ (step 7). Since $m/x^2>0$ , step 6 applied to $k=m/x^2$ gives $m/x^2=1$ , i.e. $m=x^2$ . Hence $f(x)=x^2$ for every $x>0$ .