Problem 2

Let

\mathbb{R}

be the set of real numbers. Determine all functions

f:\mathbb{R}\rightarrow\mathbb{R}

such that, for all real numbers

x

and

y

f(f(x)f(y)) + f(x+y) = f(xy).

Step 3 of 6: Locating the unique zero of

f

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f(z)=0 \iff z=1,\qquad k=f(0)=1

Detailed analysis

Setting $x=y=0$ in the original equation gives $f(k^2)+k=k$ , i.e. $f(k^2)=0$ ; since $k\ne 0$ , $f$ has a zero. Now suppose $f(z)=0$ for some $z\ne 1$ . Because $z\ne 1$ , the numbers $x=z$ and $y=z/(z-1)$ satisfy $x+y=xy$ (both equal $z^2/(z-1)$ ), so $f(x+y)=f(xy)$ and the original equation forces $f(f(x)f(y))=0$ ; but $f(x)=f(z)=0$ , so $f(x)f(y)=0$ and hence $f(0)=0$ , contradicting $k\ne 0$ . So $z=1$ is the only zero of $f$ . Applying this to $f(k^2)=0$ gives $k^2=1$ , and since $k>0$ , $k=1$ ; in particular $f(0)=1$ and $f(1)=0$ .