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 4 of 9: Rule out the trivial solution's zeros elsewhere

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f(a)=0,\ a\ne0\ \Longrightarrow\ f\equiv0;\qquad \text{assume}\ f\not\equiv0,\ \text{so}\ f(a)=0\Leftrightarrow a=0

Detailed analysis

If $f(a)=0$ for some $a\ne0$ , apply the lemma of step 3 with $b=0$ (so $f(a)=f(0)=0$ ): $f(ax)=f(0\cdot x)=f(0)=0$ for every $x$ ; taking $x=t/a$ shows $f(t)=0$ for every real $t$ , i.e. $f\equiv0$ . This constant function trivially satisfies the original equation. From now on we look for solutions with $f\not\equiv0$ , for which $f(a)=0$ forces $a=0$ .