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 1 of 6: Trivial solutions and a sign symmetry

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f\equiv 0,\quad f(x)=x-1,\quad f(x)=1-x

Detailed analysis

Direct substitution checks that $f\equiv 0$ , $f(x)=x-1$ , and $f(x)=1-x$ all satisfy $f(f(x)f(y))+f(x+y)=f(xy)$ for every $x,y$ . Also, if $f$ is a solution then so is $-f$ (replacing $f$ by $-f$ turns the equation into $-f((-f(x))(-f(y)))-f(x+y)=-f(xy)$ , exactly the same equation for $-f$ , using $(-f(x))(-f(y))=f(x)f(y)$ ). So it suffices to find all solutions with $f(0)\ge 0$ , and every other solution is the negative of one of these.