IRL_ABooklet_2023

Letting $x=0,y=1$ we get $f(1)f(0)=f(0)$, so $f(0)=0$ or $f(1)=1$.

Case 1: $f(0)=0$ Letting $y=0$ and $x=1-t$ we get $f(t)=0$ in this case for all $t$. Indeed, this is a valid solution for our equation.

Case 2: $f(0)\ne 0$ In this case we have $f(1)=1$. We first show that $f(x)\ne 0$ for all $x$. Suppose $f(x)=0$ for some $x\ne 0,1$, then $$f\left(\frac{x+y-1}{xy-1}\right)=0\text{for all }y\ne \frac{1}{x}.(6)$$ But note that for any $t\ne 1/x$, $$t=\frac{x+y-1}{xy-1}⟺y=\frac{x+t-1}{xt-1}$$ so by giving this value of $y$ in equation (6) we get $f(t)=0$ for all $t\ne 1/x$ provided that $\frac{x+t-1}{xt-1}\ne \frac{1}{x}$. In particular, when $t=1$ we have $t\ne 1/x$ and $\frac{x}{x-1}\ne \frac{1}{x}$ because $x^2-x+1=(x-\frac{1}{2}{)}^2+\frac{3}{4}>0$, hence $f(1)=0$, which is not true in the current case. Therefore $f(x)\ne 0$ for all $x$. Letting $y=0$ in the original equation we get $f(x)f(0)=f(1-x)$. Hence for $x=\frac{1}{2}$ we get $f(\frac{1}{2})f(0)=f(\frac{1}{2})$ and so $f(0)=1$ since $f(\frac{1}{2})\ne 0$.

One way to find an explicit expression for $f$ starts with the observation that $f(0)=1$ and $f(x)f(0)=f(1-x)$ imply $f(x)=f(1-x)$ for all $x$. For any $x$ we let $y=1-x$ in the original equation, which is possible since $xy-1=-(x^2-x+1)\ne 0$, and get $$f(x{)}^2=(x^2-x+1{)}^2$$ which means that $f(x)=\pm (x^2-x+1)$ for all $x$, where the choice of sign may depend on $x$. If $x\ne \pm 1$ then letting $y=x$ in the original equation we get $$f(x{)}^2=(x^2-1{)}^{2}f\left(\frac{2x-1}{x^2-1}\right)\text{hence}f\left(\frac{2x-1}{x^2-1}\right)\ge 0\forall x\ne \pm 1.$$ We then note that $y=\frac{2x-1}{x^2-1}$ is equivalent to $x^{2}y-2x-y+1=0$, or $x=\frac{1\pm \sqrt{y^2-y+1}}{y}$. Notice that for every $y$, we have $y^2-y+1>0$ and for each $y\ne 0$ there is an $x\ne \pm 1$ satisfying the relation above. Hence $f(y)\ge 0$ for all $y\ne 0$, which implies $f(y)=y^2-y+1$.

A second way to find this formula for $f$ starts with any $x\ne \pm 1$ so that we can define $$y=\frac{2x-1}{x^2-1}.(7)$$ We then have $y(x^2-1)=2x-1$ which implies $y^2{x}^2-2xy=y^2-y$ after multiplication by $y$. This can be rewritten as $$(xy-1{)}^2=y^2-y+1.(8)$$ Moreover, using expression (7) for $y$ we obtain $$\frac{x+y-1}{xy-1}=\frac{(x-1)(x^2-1)+2x-1}{x(2x-1)-(x^2-1)}=\frac{x(x^2-x+1)}{x^2-x+1}=x$$ because $x^2-x+1>0$ for all $x$. Therefore, for any $x\ne \pm 1$ and $y$ given by (7), the original equation becomes $$f(x)f(y)=(y^2-y+1)f(x).$$ Since $f(x)\ne 0$, we obtain $f(y)=y^2-y+1$ for each $y$ of the form (7). Finally, given $y\ne 0$ we can use (8) to find $$x=\frac{1\pm \sqrt{y^2-y+1}}{y}$$ such that $x$ and $y$ are related by (7). This means that each non-zero $y$ can be expressed in the form (7), hence $f(y)=y^2-y+1$ for all $y\ne 0$. As we know that $f(0)=1$, this equation holds for all $y$. Finally we check that the function $f(x)=x^2-x+1$ satisfies the original functional equation. $$ $$\begin{array}{rl}f(x)f(y)& =(x^2-x+1)(y^2-y+1)\\ & =(xy-1{)}^2-(xy-1)(x+y-1)+(x+y-1{)}^2\\ & =(xy-1{)}^2\left({\left(\frac{x+y-1}{xy-1}\right)}^2-\left(\frac{x+y-1}{xy-1}\right)+1\right)\\ & =(xy-1{)}^{2}f\left(\frac{x+y-1}{xy-1}\right).\end{array}$$