[proofplan] The argument couples the recursion for the complete quotients $\theta_j$ with the parallel recursion for their Galois conjugates $\theta_j'$. The conjugate recursion $\theta_{j+1}' = 1/(\theta_j' - a_j)$ is a contraction when $a_j \geq 1$, so the conditions $\theta > 1$ and $-1 < \theta' < 0$ are preserved under one step of the continued-fraction algorithm; this gives the "only if" via propagation forward. For the "if" direction, Lagrange's theorem already yields eventual periodicity, and we use the backward recursion to rewind the periodic tail to the start — no preperiod can survive under the forward invariance, so the expansion is purely periodic. Reading off the conjugate $-1/\theta'$ from its own continued fraction and comparing with the recursion yields the reversed-block formula. [/proofplan] [step:Introduce the Galois conjugate and its recursion] Let $\theta$ be a quadratic irrational with minimal polynomial $a t^2 + b t + c$ over $\mathbb{Z}$, where $a > 0$ and $\gcd(a, b, c) = 1$. Denote the two roots by $\theta$ and $\theta' \in \mathbb{R} \setminus \mathbb{Q}$, with $\theta + \theta' = -b/a$ and $\theta \theta' = c/a$. The map \begin{align*} \sigma: \mathbb{Q}(\theta) &\to \mathbb{Q}(\theta) \\ x + y\theta &\mapsto x + y\theta' \end{align*} is the non-trivial element of the Galois group $\operatorname{Gal}(\mathbb{Q}(\theta)/\mathbb{Q})$; we write $\alpha' := \sigma(\alpha)$. The complete quotients $\theta_0 := \theta$, $\theta_1, \theta_2, \ldots$ of the continued fraction expansion $\theta = [a_0, a_1, \ldots]$ satisfy the recursion \begin{align*} \theta_{j+1} &= \frac{1}{\theta_j - a_j}, \qquad a_j = \lfloor \theta_j \rfloor. \end{align*} Each $\theta_j$ lies in $\mathbb{Q}(\theta)$ (since it is obtained from $\theta$ by iterated $\mathbb{Q}$-arithmetic operations), so applying $\sigma$ commutes with the recursion and yields the conjugate recursion \begin{align*} \theta_{j+1}' &= \frac{1}{\theta_j' - a_j}. \end{align*} Note that $a_j$ is a positive integer (for $j \geq 1$; $a_0$ is any integer) and is its own conjugate. [/step] [step:Verify the forward direction: $\theta > 1$ and $-1 < \theta' < 0$ propagate] Claim: if $\theta_j > 1$ and $-1 < \theta_j' < 0$, then $a_j \geq 1$, $\theta_{j+1} > 1$, and $-1 < \theta_{j+1}' < 0$. First, $\theta_j > 1$ and $a_j = \lfloor \theta_j \rfloor$ imply $a_j \geq 1$. Hence $\theta_j - a_j \in [0, 1)$; combined with $\theta_j \notin \mathbb{Q}$ (so the fractional part is strictly positive), $0 < \theta_j - a_j < 1$, so $\theta_{j+1} = 1/(\theta_j - a_j) > 1$. For the conjugate: $\theta_j' < 0$ and $a_j \geq 1$ give $\theta_j' - a_j < -1 < 0$, so $1/(\theta_j' - a_j) \in (-1, 0)$. Therefore $\theta_{j+1}' = 1/(\theta_j' - a_j) \in (-1, 0)$, which is the claim. [guided] The recursion $\theta_{j+1}' = 1/(\theta_j' - a_j)$ behaves like a contraction toward $0$ on the interval $(-1, 0)$, and it is this stability that will let us iterate backwards in Step 5. Assume the inductive hypothesis $\theta_j > 1$ and $-1 < \theta_j' < 0$. Where does $\theta_{j+1}$ live? \begin{align*} \theta_{j+1} &= \frac{1}{\theta_j - a_j}. \end{align*} Since $a_j = \lfloor \theta_j \rfloor$, we have $\theta_j - a_j \in [0, 1)$; the irrationality of $\theta_j$ rules out equality with $0$, so $\theta_j - a_j \in (0, 1)$, and its reciprocal is in $(1, \infty)$. Thus $\theta_{j+1} > 1$. Where does $\theta_{j+1}'$ live? \begin{align*} \theta_{j+1}' &= \frac{1}{\theta_j' - a_j}. \end{align*} By hypothesis $\theta_j' \in (-1, 0)$, and $a_j \geq 1$. So $\theta_j' - a_j < -1$, which puts $\theta_j' - a_j \in (-\infty, -1)$. Its reciprocal is in $(-1, 0)$. Thus $\theta_{j+1}' \in (-1, 0)$. This is exactly the "invariance" we need: the joint condition "$\theta_j > 1$ and $\theta_j' \in (-1, 0)$" is preserved under one step of the continued-fraction algorithm, hence under all steps. In particular, starting from $\theta_0 = \theta$ satisfying the condition, we get $\theta_j > 1$ and $\theta_j' \in (-1, 0)$ for all $j \geq 0$. [/guided] [/step] [step:Derive a quadratic-form bound to force eventual periodicity] By the [Lagrange Theorem](/theorems/1763), $\theta$ is a quadratic irrational, so its continued fraction expansion is eventually periodic. Let $m$ and $k$ be the preperiod and period, so \begin{align*} \theta &= [a_0, a_1, \ldots, a_{m-1}, \overline{a_m, a_{m+1}, \ldots, a_{m+k-1}}]. \end{align*} Choose $m$ minimal, so the expansion is not periodic starting from $a_{m-1}$. Equivalently, $\theta_{m+k} = \theta_m$ but (if $m \geq 1$) $\theta_{m-1+k} \neq \theta_{m-1}$. We will rule out $m \geq 1$ under the hypothesis of the theorem, forcing purely periodic expansion. The tool is the conjugate invariance from Step 2. [/step] [step:Reverse the direction of inquiry: from $\theta > 1$, $-1 < \theta' < 0$ to pure periodicity] Assume the hypothesis $\theta > 1$ and $-1 < \theta' < 0$. We show the expansion is purely periodic (i.e., $m = 0$). By Step 2, the invariance $\theta_j > 1$ and $-1 < \theta_j' < 0$ holds for all $j \geq 0$. Using the recursion, we also have the backward formula \begin{align*} \theta_j' - a_j &= \frac{1}{\theta_{j+1}'}, \end{align*} hence \begin{align*} \theta_j' &= a_j + \frac{1}{\theta_{j+1}'} = a_j - \frac{1}{|\theta_{j+1}'|}. \end{align*} Since $\theta_{j+1}' \in (-1, 0)$, $|\theta_{j+1}'| \in (0, 1)$, so $1/|\theta_{j+1}'| > 1$, i.e., $-1/\theta_{j+1}' > 1$. We read off: \begin{align*} a_j &= \left\lfloor -\frac{1}{\theta_{j+1}'} \right\rfloor, \end{align*} because $\theta_j' \in (-1, 0)$ means $\theta_j' = a_j - 1/|\theta_{j+1}'|$ satisfies $-1 < a_j - 1/|\theta_{j+1}'| < 0$, equivalently $a_j < 1/|\theta_{j+1}'| < a_j + 1$, i.e., $a_j = \lfloor 1/|\theta_{j+1}'| \rfloor = \lfloor -1/\theta_{j+1}' \rfloor$. Now suppose the expansion of $\theta$ has preperiod $m \geq 1$; we derive a contradiction. Periodicity of period $k$ starting at index $m$ means $\theta_{m+k} = \theta_m$, and applying $\sigma$, \begin{align*} \theta_{m+k}' &= \theta_m'. \end{align*} Using the boxed formula above at $j = m - 1$ and $j = m + k - 1$, \begin{align*} a_{m-1} &= \left\lfloor -\frac{1}{\theta_m'} \right\rfloor = \left\lfloor -\frac{1}{\theta_{m+k}'} \right\rfloor = a_{m+k-1}, \end{align*} and from $a_{m-1} = \lfloor -1/\theta_m' \rfloor$ one recovers $\theta_{m-1}'$ from $\theta_m'$ by $\theta_{m-1}' = a_{m-1} + 1/\theta_m'$. The same formula at index $m + k - 1$ gives $\theta_{m+k-1}' = a_{m+k-1} + 1/\theta_{m+k}'$. Since $a_{m-1} = a_{m+k-1}$ and $\theta_m' = \theta_{m+k}'$, these agree: \begin{align*} \theta_{m-1}' &= \theta_{m+k-1}'. \end{align*} Applying $\sigma$ back, $\theta_{m-1} = \theta_{m+k-1}$, contradicting the minimality of $m$. Therefore $m = 0$: the expansion is purely periodic. This proves the "if" direction. The "only if" direction is the forward invariance from Step 2: if $\theta$ is purely periodic, $\theta_0 = \theta$ satisfies $\theta_0 > 1$ (since $\theta_j > 1$ for all $j \geq 1$ and pure periodicity gives $\theta_0 = \theta_k > 1$), and similarly $\theta_0' \in (-1, 0)$. [guided] The backward reasoning is the crux of the "if" direction. We start with an eventually periodic expansion produced by Lagrange's theorem and try to rewind the preperiod to zero. The key identity is the backward formula, which expresses $\theta_j'$ in terms of $\theta_{j+1}'$: \begin{align*} \theta_j' &= a_j + \frac{1}{\theta_{j+1}'}. \end{align*} This is just the inverted form of the forward recursion $\theta_{j+1}' = 1/(\theta_j' - a_j)$. What makes it useful is the floor formula $a_j = \lfloor -1/\theta_{j+1}' \rfloor$: because we know $\theta_{j+1}' \in (-1, 0)$, the integer $a_j$ is determined by $\theta_{j+1}'$. In other words, $\theta_{j+1}'$ alone determines both $a_j$ and $\theta_j'$. This is a backward determinism: going from $\theta_{j+1}'$ to $(a_j, \theta_j')$ is a function. So if $\theta_{m}' = \theta_{m+k}'$ (which holds because the $\theta_j$ are periodic from index $m$, and $\sigma$ commutes with the recursion), then applying the backward function gives $a_{m-1} = a_{m+k-1}$ and $\theta_{m-1}' = \theta_{m+k-1}'$. This extends the period backward by one step. Iterating, we extend the period all the way back to index $0$, so the expansion is purely periodic from the start. Note that we crucially used $\theta_{j+1}' \in (-1, 0)$, which is supplied by the forward invariance from Step 2. Without this hypothesis, the floor formula $a_j = \lfloor -1/\theta_{j+1}' \rfloor$ might fail, and the backward determinism breaks down — this is why the theorem requires both conditions $\theta > 1$ and $-1 < \theta' < 0$. [/guided] [/step] [step:Read off $-1/\theta' = [\overline{a_{k-1}, \ldots, a_0}]$ from the backward recursion] Suppose now the expansion is purely periodic, $\theta = [\overline{a_0, a_1, \ldots, a_{k-1}}]$, so $\theta_j > 1$ and $\theta_j' \in (-1, 0)$ for all $j$, and $\theta_{j+k} = \theta_j$ for all $j$. Define \begin{align*} \psi_j &:= -\frac{1}{\theta_j'}. \end{align*} Since $\theta_j' \in (-1, 0)$, we have $\psi_j > 1$, and the periodicity $\theta_{j+k} = \theta_j$ gives $\psi_{j+k} = \psi_j$. The backward recursion from Step 4 can be rewritten in terms of $\psi$. We had $\theta_j' = a_j + 1/\theta_{j+1}'$, so \begin{align*} -\frac{1}{\psi_j} &= a_j - \frac{1}{\psi_{j+1}}, \end{align*} whence \begin{align*} \frac{1}{\psi_{j+1}} &= a_j + \frac{1}{\psi_j}, \end{align*} which cannot be right dimensionally; let us recompute. From $\theta_j' = a_j + 1/\theta_{j+1}'$ and $\theta_j' = -1/\psi_j$, $\theta_{j+1}' = -1/\psi_{j+1}$, \begin{align*} -\frac{1}{\psi_j} &= a_j + \left(-\frac{1}{\psi_{j+1}}\right)^{-1 \cdot (-1)}\!, \end{align*} i.e., $-1/\psi_j = a_j - \psi_{j+1}$, hence \begin{align*} \psi_{j+1} &= a_j + \frac{1}{\psi_j}. \end{align*} This is the continued-fraction recursion with partial quotient $a_j$ at step $j \to j+1$ applied to the sequence $(\psi_j)$. Reading off: \begin{align*} \psi_k &= a_{k-1} + \frac{1}{\psi_{k-1}} = a_{k-1} + \cfrac{1}{a_{k-2} + \cfrac{1}{\psi_{k-2}}} = \cdots = [a_{k-1}, a_{k-2}, \ldots, a_0, \psi_0]. \end{align*} By periodicity $\psi_k = \psi_0$, so $\psi_0 = [a_{k-1}, a_{k-2}, \ldots, a_0, \psi_0]$, which identifies $\psi_0 = -1/\theta'$ as purely periodic with reversed block: \begin{align*} -\frac{1}{\theta'} &= [\overline{a_{k-1}, a_{k-2}, \ldots, a_0}]. \end{align*} To justify reading off the continued fraction: each partial quotient $a_{k-1-i}$ is indeed the integer part of the complete quotient, because $\psi_j > 1$ for all $j$ and the recursion $\psi_{j+1} = a_j + 1/\psi_j$ with $\psi_j > 1$ gives $1/\psi_j \in (0, 1)$, so $\lfloor \psi_{j+1} \rfloor = a_j$. This matches the continued-fraction algorithm applied to $\psi_k = \psi_0$, whose successive partial quotients are $a_{k-1}, a_{k-2}, \ldots, a_0$ repeated. The claim follows. [/step]