leannan-capall wrote: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ with the property that $$P(x, y): f(x)f(y) = (xy - 1)^2f\left(\frac{x + y - 1}{xy - 1}\right)$$ for all real numbers $x, y$ with $xy \neq 1$. $P(x,0) \land P(1-x, 0) \implies f(x)(f(0)^2-1)=0$ $\forall x \in \mathbb{R}$. If $f(0)^2 \neq 1$, then $\boxed{f=0}$ which is a valid solution. If $f(0)^2=1$, then $P(0, 0): f(1)=1$. Suppose there exists $t \neq 0, 1$ such that $f(t)=0$, then $P(t, \frac{t}{t-1}) \implies 1=f(1)=0$ which is a contradiction. So $f(x) \neq 0$ for all $x$. $P\left(x,\frac{1+\sqrt{x^2-x+1}}{x}\right) \implies f(x)=(x^2-x+1)^2$ $\forall x \neq 0$. So for all $x \neq 0, 1$, $f(1-x)=f(x)$. But $P(x, 0) \implies f(x)f(0)=f(1-x)$, which means $f(0)=1$. If follows that $\boxed{f(x)=(x^2-x+1)^2}$ $\forall x$ which is indeed a solution.