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 4 of 6: Two consequences:

f\circ f

and the shift by

1

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f(f(x))=1-f(x),\qquad f(x+1)=f(x)-1

Detailed analysis

With $k=1$ , relation $(\ast)$ becomes $f(f(x))=1-f(x)$ for every $x$ . Separately, setting $y=1$ in the original equation and using $f(1)=0$ gives $f(f(x)f(1))+f(x+1)=f(x)$ , i.e. $f(0)+f(x+1)=f(x)$ , so $f(x+1)=f(x)-1$ . By induction, $f(x+n)=f(x)-n$ for every integer $n$ (positive or negative).