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 3 of 9: A scaling lemma

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f(a)=f(b)\ \Longrightarrow\ f(ax)=f(bx)\ \text{for every real}\ x\qquad(1)

Detailed analysis

Suppose $f(a)=f(b)$ . Then $f(a^2)=f(f(a))=f(f(b))=f(b^2)$ by step 1. Writing the original equation with $y=a$ and with $y=b$ gives $f(x^2+f(a))=f(f(x))+f(a^2)+2f(xa)$ and $f(x^2+f(b))=f(f(x))+f(b^2)+2f(xb)$ . The left sides are equal (since $f(a)=f(b)$ ) and so are $f(a^2),f(b^2)$ ; subtracting the two identities leaves $2f(xa)=2f(xb)$ , i.e. $f(ax)=f(bx)$ for every real $x$ .