## Task 3 - 211233

Determine all triples $(a, b, c)$ of real numbers for which the following holds:

The function $f$ defined for all real numbers $x \neq -c$ by $f(x) = \frac{a x + b}{x + c}$ satisfies the following conditions:

(1) There exist real numbers $x$ for which $f(x)$, $f(f(x))$, and $f(f(f(x)))$ are defined.

(2) For every such number $x$ with $x \neq -1$, it holds that $f(f(f(x))) = \frac{x - 1}{x + 1}$.
I. Suppose a triple $(a, b, c)$ has the required properties. Then:

The function $f$ defined by $f(x)=\frac{a x+b}{x+c}$ is not constant; for if $f(x)=d \neq -c$ for all $x \neq -c$, then also $f(f(f(x)))=d$ for all $x \neq -c$, so (2) would not be satisfied, since $\frac{x-1}{x+1}=d$ can hold for at most one $x$; in the case $f(x)=-c$ for all $x \neq -c$, however, there would be no $x$ for which $f(x)$ and $f(f(x))$ are defined, contradicting (1).

Consequently, for each $d$, the equation $a x+b=x d+c d$ does not have all $x \neq -c$ as solutions. Thus, as a linear equation, it has at most one solution.

Therefore, there is at most one $x$ such that $f(x)=-c$ and at most one $x$ such that $f(f(x))=-c$. Hence, and because of (2), there are infinitely many numbers $x \neq 1$ for which $f(x)$ and

$$
\begin{gathered}
f(f(x))=\frac{\left(a^{2}+b\right) x+(a+c) b}{(a+c) x+\left(b+c^{2}\right)} \quad \text { and } \\
f(f(f(x)))=\frac{\left(a^{3}+2 a b+b c\right) x+\left(a^{2}+a c+b+c^{2}\right) b}{\left(a^{2}+a c+b+c^{2}\right) x+\left(a b+2 b c+c^{3}\right)}
\end{gathered}
$$

exist and for which

$$
\left(\left(a^{3}+2 a b+b c\right) x+\left(a^{2}+a c+b+c^{2}\right) b\right)(x+1)=\left(\left(a^{2}+a c+b+c^{2}\right) x+\left(a b+2 b c+c^{3}\right)\right)(x-1)
$$

holds. From the equality of these two polynomials for infinitely many $x$, the coefficient equalities follow:

$$
\begin{aligned}
a^{3}+2 a b+b c & =a^{2}+a c+b+c^{2} \\
a^{3}+2 a b+b c+\left(a^{2}+a c+b+c^{2}\right) b & =-\left(a^{2}+a c+b+c^{2}\right)+a b+2 b c+c^{3} \\
\left(a^{2}+a c+b+c^{2}\right) b & =-\left(a b+2 b c+c^{3}\right)
\end{aligned}
$$

From (4), (5), (6) it follows by addition that both sides of (5) are equal to 0; from this and (4) we get

$$
a^{3}+2 a b+b c=a b+2 b c+c^{3} \quad \text { so } \quad a\left(a^{2}+b\right)=c\left(b+c^{2}\right)
$$

and by adding $a c(a+c)$

$$
a\left(a^{2}+a c+b+c^{2}\right)=c\left(a^{2}+a c+b+c^{2}\right)
$$

If $a^{2}+a c+b+c^{2}=0$, then (3) would not be defined for any $x$. Therefore,

$$
a^{2}+a c+b+c^{2} \neq 0
$$

and from (7) it follows that $a=c$. Hence, and from (4), (2) we have

$$
\left(a^{2}+a c+b+c^{2}\right) b=-\left(a^{2}+a c+b+c^{2}\right)
$$

because of (8) thus $b=-1$. With this, (4) becomes

$$
a^{3}-3 a^{2}-3 a+1=0
$$

One solution of this is $a=-1$ (or after the known factorization of the factor $a+1$ from $a^{3}+1$ and from $3 a^{2}+3 a$) follows $(a+1)\left(a^{2}-4 a+1\right)=0$ and from this

$$
a=-1 \quad \text { or } \quad a=2 \pm \sqrt{3}
$$

Thus, only the functions $f$ with

$$
\begin{array}{r}
f(x)=\frac{-x-1}{x-1} \quad \text { for all } x \neq 1 \\
f(x)=\frac{(2 \pm \sqrt{3}) x-1}{x+(2 \pm \sqrt{3})} \quad \text { for all } x \neq-(2 \pm \sqrt{3})
\end{array}
$$

(always with the upper or always with the lower sign) can satisfy the conditions (1), (2), i.e., only the triples

$$
(-1,-1,-1) \quad ; \quad(2+\sqrt{3},-1,2+\sqrt{3}) \quad ; \quad(2-\sqrt{3},-1,2-\sqrt{3})
$$

can have the required properties.

II. For the triple $(-1,-1,-1)$, there exist

$$
f(f(x))=\frac{-x-1}{x-1}=-\frac{1}{x}
$$

for all $x$ with $x \neq 1, x \neq 0$ and

$$
f(f(f(x)))=f\left(-\frac{1}{x}\right)=\frac{x-1}{x+1}
$$

for all $x$ with $x \neq 1, x \neq 0, x \neq-1$. For the other two triples, there exist

$$
f(f(x))=f\left(\frac{(2 \pm \sqrt{3}) x-1}{x+(2 \pm \sqrt{3})}\right)=\frac{(3 \pm 2 \sqrt{3}) x-(2 \pm \sqrt{3})}{(2 \pm \sqrt{3}) x+(3 \pm 2 \sqrt{3})}
$$

for all $x$ with $x \neq-(2 \pm \sqrt{3}), x \neq \mp \sqrt{3}$ and

$$
f(f(f(x)))=f\left(\frac{(3 \pm 2 \sqrt{3}) x-(2 \pm \sqrt{3})}{(2 \pm \sqrt{3}) x+(3 \pm 2 \sqrt{3})}\right)=\frac{(5 \pm 3 \sqrt{3})(x-1)}{(5 \pm 3 \sqrt{3})(x+1)}
$$

for all $x$ with $x \neq-(2 \pm \sqrt{3}), x \neq \mp \sqrt{3}, x \neq-1$.

Therefore, and because $5 \pm 3 \sqrt{3} \neq 0$, these functions satisfy the conditions (1), (2). Consequently, exactly the triples

$$
(-1,-1,-1) \quad ; \quad(2+\sqrt{3},-1,2+\sqrt{3}) \quad ; \quad(2-\sqrt{3},-1,2-\sqrt{3})
$$

have the required properties.