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 6 of 6: Finishing the computation

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f(x)=1-x

Detailed analysis

Apply the identity $f(f(x))=1-f(x)$ at the point $f(x)$ itself: $f(f(f(x)))=1-f(f(x))=1-(1-f(x))=f(x)$ . On the other hand, substituting the identity directly, $f(f(f(x)))=f(1-f(x))$ . So $f(1-f(x))=f(x)$ , and injectivity gives $1-f(x)=x$ , i.e. $f(x)=1-x$ for every $x$ . Undoing the sign reduction of Step 1 (every solution is $f$ or $-f$ for some $f$ with $f(0)\ge 0$ ), the complete list of solutions is $f\equiv 0$ , $f(x)=1-x$ , and $f(x)=x-1$ .