Problem 5

Suppose $f(s)=-t^2$ for some $s,t\ne0$. Using $(*)$ with $x=s,y=t$: $f(s^2+f(t))=f(t^2+f(s))=f(0)=0$, so (step 4) $s^2+f(t)=0$, i.e. $f(t)=-s^2$; then $f(t^2)=f(-t^2)=f(f(s))=f(s^2)$ (evenness, then $f(s)=-t^2$, then step 1). Applying the full form of $(*)$ with $x=\sqrt{s^2+t^2},y=s$ and again with $y=t$ gives $f(s^2+t^2)+2f(s\sqrt{s^2+t^2})=0=f(s^2+t^2)+2f(t\sqrt{s^2+t^2})$, so $f(s\sqrt{s^2+t^2})=f(t\sqrt{s^2+t^2})$; by the lemma (step 3) with multiplier $1/\sqrt{s^2+t^2}$, $f(s)=f(t)=-s^2$. Taking $x=y=s$ in the full form of $(*)$ now gives $f(s^2+f(s))=4f(s^2)$, while the left side is $f(s^2-s^2)=f(0)=0$; so $f(s^2)=0$ with $s^2\ne0$, contradicting step 4. Hence no such $s,t$ exist, and $f(x)\ge0$ for every real $x$; combined with step 4, $f(x)>0$ for $x\ne0$.