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 6 of 9: Solving f(k)=1 forces k=1

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k>0,\ f(k)=1\ \Longrightarrow\ f(1/k)=f(k)=1,\ \text{WLOG}\ k\ge1;\qquad f(k^2-1)=-2f(k\sqrt{k^2-1})\le0\ \Longrightarrow\ k=1

Detailed analysis

Let $k>0$ satisfy $f(k)=1$ . By step 1, $f(k^2)=f(f(k))=f(1)$ , so by the lemma (step 3) with multiplier $1/k$ , $f(k)=f(1/k)$ ; hence $f(1/k)=1$ too, so we may assume $k\ge1$ . Using the full form of $(*)$ with $x=\sqrt{k^2-1},y=k$ : $f(k^2-1+f(k))=f(k^2-1)+f(k^2)+2f(k\sqrt{k^2-1})$ , i.e. $f(k^2)=f(k^2-1)+f(k^2)+2f(k\sqrt{k^2-1})$ (using $f(k)=1$ ), so $f(k^2-1)=-2f(k\sqrt{k^2-1})\le0$ by step 5's non-negativity. Since also $f(k^2-1)\ge0$ , we get $f(k^2-1)=0$ , so by step 4, $k^2-1=0$ , i.e. $k=1$ .