IMC 2005 · Day 2 · P12

First consider the case when $q = 0$ and $r$ is rational. Choose a positive integer $t$ such that $r^2 t$ is an integer and set $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 + rt & -r^2 t \\ t & 1 - rt \end{pmatrix}.$$ Then $$\det \begin{pmatrix} a & b \\ c & d \end{pmatrix} = 1 \quad \text{and} \quad \frac{ar + b}{cr + d} = \frac{(1 + rt) r - r^2 t}{tr + (1 - rt)} = r.$$ Now assume $q \ne 0$. Let the minimal polynomial of $r$ in $\mathbb{Z}[x]$ be $u x^2 + v x + w$. The other root of this polynomial is $\bar{r} = p - q\sqrt{7}$, so $v = -u(r + \bar{r}) = -2up$ and $w = u r \bar{r} = u(p^2 - 7q^2)$. The discriminant is $v^2 - 4uw = 7 \cdot (2uq)^2$. The left-hand side is an integer, implying that also $\Delta = 2uq$ is an integer.

The equation $\frac{ar+b}{cr+d} = r$ is equivalent to $c r^2 + (d - a) r - b = 0$. This must be a multiple of the minimal polynomial, so we need $$c = ut, \quad d - a = vt, \quad -b = wt$$ for some integer $t \ne 0$. Putting together these equalities with $ad - bc = 1$ we obtain that $$(a + d)^2 = (a - d)^2 + 4ad = 4 + (v^2 - 4uw) t^2 = 4 + 7 \Delta^2 t^2.$$ Therefore $4 + 7 \Delta^2 t^2$ must be a perfect square. Introducing $s = a + d$, we need an integer solution $(s, t)$ for the Diophantine equation $$\tag{1} s^2 - 7 \Delta^2 t^2 = 4$$ such that $t \ne 0$.

The numbers $s$ and $t$ will be even. Then $a + d = s$ and $d - a = vt$ will be even as well and $a$ and $d$ will be really integers.

Let $(8 \pm 3\sqrt{7})^n = k_n \pm l_n \sqrt{7}$ for each integer $n$. Then $k_n^2 - 7 l_n^2 = (k_n + l_n\sqrt{7})(k_n - l_n\sqrt{7}) = ((8 + 3\sqrt{7})^n (8 - 3\sqrt{7}))^n = 1$ and the sequence $(l_n)$ also satisfies the linear recurrence $l_{n+1} = 16 l_n - l_{n-1}$. Consider the residue of $l_n$ modulo $\Delta$. There are $\Delta^2$ possible residue pairs for $(l_n, l_{n+1})$ so some are the same. Starting from such two positions, the recurrence shows that the sequence of residues is periodic in both directions. Then there are infinitely many indices such that $l_n \equiv l_0 = 0 \pmod{\Delta}$.

Taking such an index $n$, we can set $s = 2 k_n$ and $t = 2 l_n / \Delta$.

Remarks. 1. It is well-known that if $D > 0$ is not a perfect square then the Pell-like Diophantine equation $$x^2 - D y^2 = 1$$ has infinitely many solutions. Using this fact the solution can be generalized to all quadratic algebraic numbers.

2. It is also known that the continued fraction of a real number $r$ is periodic from a certain point if and only if $r$ is a root of a quadratic equation. This fact can lead to another solution.