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 2 of 6: Setting

y=0

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k:=f(0),\qquad f(kf(x))+f(x)=k

Detailed analysis

Write $k=f(0)$ . Substituting $y=0$ gives $f(f(x)f(0))+f(x)=f(0)$ , i.e. $f(kf(x))+f(x)=k$ for every $x$ — call this relation $(\ast)$ . If $k=0$ , relation $(\ast)$ reads $f(0)+f(x)=0$ , so $f(x)=-f(0)=0$ for every $x$ ; this is the solution $f\equiv 0$ . From now on assume $k\ne 0$ (and, by the sign symmetry of Step 1, $k>0$ ).