A rational function

The function $f(z) = p(z)/q(z)$ extends to a meromorphic function on the complex numbers with $1$ removed, or to a well defined analytic function on the extended (with $\infty$ ) complex numbers. Since $f(f(f(x)))=x$ for all real $x\neq1$ , we deduce that $f(f(f(z)))=z$ for all $z$ . Thus $f(z)$ is a bijection of the extended complex plane.

Suppose that either $p(z)$ or $q(z)$ has degree greater than $1$ . If $p(z)$ and $q(z)$ have distinct degrees, choose a nonzero complex number $t$ . If $p(z)$ and $q(z)$ have the same degree, choose a complex number $t$ such that $a - tb \neq 0$ where $a\,,\,b$ are the leading coefficients of $p(z)\,,\,q(z)$ respectively. Then $p(z) - tq(z)$ is a polynomial of degree at least $2$ . If there are two distinct roots of the equation $p(z) - tq(z) = 0$ , then there are two distinct solutions of the equation $f(z) = t$ in the extended complex plane, and hence $f$ is not a bijection. Thus all the roots of $p(z) - tq(z) = 0$ must be the same. If this root is $w$ , then $p(z) = tq(z) + a(z-w)^n$ for some nonzero complex number $a$ and integer $n\ge2$ . But then $f'(w) = 0$ , which again contradicts the bijectivity of $f$ .

Thus we deduce that both $p(z)$ and $q(z)$ have degree no greater than $1$ . Since $f(x)$ is defined for all real numbers except $1$ , $q(x)$ is either constant or a multiple of $x-1$ .

If $q(x)$ is constant, then $f(x) = cx+d$ for real constants $c,d$ . Then $f(f(f(x))) \; = \; c^3x + (c^2+c+1)d$ and hence $c=1,d=0$ , so that $f(z) = x$ . In this case $f(0)=0$ is not positive.

If $q(x)$ is not constant, then we deduce that $f(z) \; = \; \frac{cz - d}{z-1} \qquad \qquad c \neq d$ for real constants $c,d$ . But $f(1)=\infty$ , $f(\infty) = c$ , and hence we must have $f(c)=1$ , so we deduce that $c^2-d=c-1$ , or $d = c^2-c+1$ . The requirement that $c \neq d$ tells us that $c \neq 1$ . We see that $f(f(z)) \; = \; \frac{z-d}{z-c} \qquad \qquad f(f(f(z))) \;=\; z$ as required. But now $f(0) = d = c^2-c+1 = (c-1/2)^2 + 3/4 \ge 3/4$ , giving us $a+b=7$ .

We begin by proving some preliminary lemmas about the form of our function $f(x)$ . Our goal will be to show that both $p(x)$ and $q(x)$ have degree at most 1 (assuming, of course, that the polynomials $p(x)$ and $q(x)$ are relatively prime).

Lemma 1 : If $f(f(f(x))) = x$ for all real $x$ for which $f(f(f(x)))$ is defined, then $f(f(f(x))) = x$ for all complex values of $x$ for which $f(f(f(x)))$ is defined.

Proof : Note that $f(f(f(x)))$ is a rational function of $x$ , so we can write $f(f(f(x)) = \frac{P(x)}{Q(x)}$ for some $P(x)$ and $Q(x)$ . Then, the statement $f(f(f(x))) = x$ becomes the polynomial identity $P(x) = xQ(x)$ . Since this is true for infinitely many real values (i.e. all real values $x$ where $f(f(f(x)))$ is defined), it must be true for all complex values $x$ .

Lemma 2 : $p(x)$ must have degree 0 or 1.

Proof : Assume that $p(x)$ has degree at least 2. Then we claim that there is some value $\lambda$ such that $p(x) = \lambda q(x)$ has at least two distinct roots which are not roots of $q(x)$ . To see this, first note that since $p(x)$ has degree at least 2, $p(x) - \lambda q(x)$ will also almost always have degree at least 2 (with the possible exception of when $\lambda$ times the leading term of $q(x)$ equals the leading term of $p(x)$ ), so $p(x) - \lambda q(x)$ will almost always have at least two roots. Now, if $r$ was a root both of $p(x) - \lambda q(x)$ and of $q(x)$ , it would also be a root of $p(x)$ ; but this implies $x-r$ divides both $p(x)$ and $q(x)$ , contradicting that $p(x)$ and $q(x)$ are relatively prime. Therefore, it also follows that these two roots will not be roots of $q(x)$ . Finally, to make sure that these roots are distinct, we can simply perturb $\lambda$ slightly if they aren't.

Call these two roots $r_1$ and $r_2$ . Note that $\frac{p(r_1)}{q(r_1)} = \frac{p(r_2)}{q(r_2)} = \lambda$ , so in particular, $f(f(f(r_1))) = f(f(f(r_2))) = f(f(\lambda))$ . But since $f(f(f(x))) = x$ , this implies $r_1 = r_2$ , a contradiction. Therefore our initial assumption was wrong, and $p(x)$ has degree at most 1.

Lemma 3 : $q(x)$ must have degree 0 or 1.

Proof : Similar to the proof of lemma 2 above (we again consider roots of $p(x) = \lambda q(x)$ and use the lack of injectivity to our favor).

We have now therefore shown that we can write $f(x)$ as

$f(x) = \frac{ax + b}{cx + d}$

These are known as Mobius transformations and importantly, compositions of Mobius functions are given by multiplication of the corresponding 2-by-2 matrices (this is straightforward to check). Therefore, in order for $f(f(f(x))) = x$ , we must have:

$\begin{bmatrix} a & b \\ c & d \end{bmatrix}^3 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Equivalently, all eigenvalues of this matrix (which we will henceforth call $M$ ) must be cube roots of unity. In order for $M$ to have real entries, its eigenvalues can either both be 1 (giving rise to the identity matrix and the function $f(x) = x$ ) or they can be $e^{\pm 2\pi i /3}$ . In this case, the characteristic polynomial of our matrix will be $x^2 + x + 1$ , so we must have $\mathrm{tr}(M) = -1$ and $\det(M) = 1$ . This gives rise to the system

$a + d = -1$ $ad - bc = 1$

Moreover, since we know $f(x)$ is defined everywhere except for $x = 1$ , we know that $c = -d$ . Substituting this and $a = -(d+1)$ into the second equation, we have that

$b = -\frac{d^2 + d + 1}{d}$

Finally, $f(0)$ = $\frac{b}{d}$ , which equals

$\frac{b}{d} = \frac{d^2 + d + 1}{d^2}$

This is minimized at when $d = -2$ , where $\frac{b}{d} = \frac{3}{4}$ . Substituting back, we see this minimum is attained by the function

$f(x) = \frac{2x - 3}{4(x-1)}$

Our desired answer is therefore $3+4=7$ .