[proofplan] We prove the complete quotient formula by induction on $n$, strengthening the induction hypothesis with the auxiliary bound $-\sqrt{D}<m_n<\sqrt{D}$. The floor identity is proved by writing the integer remainder of $a_0+m_n$ modulo $d_n$; this avoids the invalid estimate that $x<\alpha_n<x+1/d_n$ alone controls the floor. The induction step then subtracts $a_n$, inverts, rationalises the denominator, and uses congruence modulo $d_n$ to prove divisibility of the next denominator. The auxiliary bound gives positivity of $D-m_{n+1}^2$ without invoking any unproved external facts about the square-root recurrence. [/proofplan]

[step:Establish the initial quotient formula and divisibility] Since $D$ is not a square, $\sqrt{D} \notin \mathbb{N}$, and therefore \begin{align*} a_0 < \sqrt{D} < a_0 + 1. \end{align*} By definition, $m_0 = 0$ and $d_0 = 1$, so \begin{align*} \frac{\sqrt{D} + m_0}{d_0} = \frac{\sqrt{D} + 0}{1} = \sqrt{D} = \alpha_0. \end{align*} Also $d_0 = 1 > 0$, and $1$ divides $D - m_0^2 = D$. Since $m_0=0$ and $D\in\mathbb{N}$ is nonsquare, we also have the auxiliary bound \begin{align*} -\sqrt{D}<m_0<\sqrt{D}. \end{align*} Finally, \begin{align*} a_0 = \lfloor \sqrt{D} \rfloor = \lfloor \alpha_0 \rfloor. \end{align*} Thus all asserted properties hold for $n=0$. [/step]

[step:Show that the recurrence floor agrees with the complete quotient floor]Assume that, for some $n \ge 0$, \begin{align*} \alpha_n = \frac{\sqrt{D} + m_n}{d_n}, \end{align*} with $d_n$ a positive integer. Define the integer remainder $r_n$ by \begin{align*} r_n := a_0+m_n-d_n a_n. \end{align*} Since the recurrence defines \begin{align*} a_n = \left\lfloor \frac{a_0 + m_n}{d_n} \right\rfloor, \end{align*} the defining property of the floor gives \begin{align*} 0\le r_n<d_n. \end{align*} Using the induction formula for $\alpha_n$ and the definition of $r_n$, we compute \begin{align*} \alpha_n-a_n=\frac{\sqrt{D}+m_n}{d_n}-a_n. \end{align*} Combining the two terms over the common positive denominator $d_n$ gives \begin{align*} \alpha_n-a_n=\frac{\sqrt{D}+m_n-d_n a_n}{d_n}. \end{align*} Since $r_n=a_0+m_n-d_n a_n$, this becomes \begin{align*} \alpha_n-a_n=\frac{\sqrt{D}-a_0+r_n}{d_n}. \end{align*} Because $a_0=\lfloor\sqrt{D}\rfloor$ and $D$ is nonsquare, $0<\sqrt{D}-a_0<1$. Together with $0\le r_n<d_n$, this gives \begin{align*} 0<\sqrt{D}-a_0+r_n<d_n. \end{align*} Dividing by the positive integer $d_n$ yields \begin{align*} 0<\alpha_n-a_n<1. \end{align*} Equivalently, \begin{align*} a_n<\alpha_n<a_n+1, \end{align*} and therefore \begin{align*} a_n=\lfloor\alpha_n\rfloor. \end{align*}[/step]

[guided]The quantity that controls the floor is not merely the size of the interval between $(a_0+m_n)/d_n$ and $\alpha_n$; it is the fractional part of $(a_0+m_n)/d_n$. We encode that fractional part by defining the integer remainder \begin{align*} r_n := a_0+m_n-d_n a_n. \end{align*} This is an integer because $a_0$, $m_n$, $d_n$, and $a_n$ are integers. Since \begin{align*} a_n=\left\lfloor \frac{a_0+m_n}{d_n}\right\rfloor \end{align*} and $d_n>0$, the floor inequality \begin{align*} a_n\le \frac{a_0+m_n}{d_n}<a_n+1 \end{align*} is equivalent, after multiplying by $d_n$, to \begin{align*} 0\le a_0+m_n-d_n a_n<d_n. \end{align*} Thus \begin{align*} 0\le r_n<d_n. \end{align*} Now use the induction hypothesis for the complete quotient: \begin{align*} \alpha_n=\frac{\sqrt{D}+m_n}{d_n}. \end{align*} Subtracting $a_n$ gives \begin{align*} \alpha_n-a_n=\frac{\sqrt{D}+m_n}{d_n}-a_n. \end{align*} Combining the two terms over the common positive denominator $d_n$ gives \begin{align*} \alpha_n-a_n=\frac{\sqrt{D}+m_n-d_n a_n}{d_n}. \end{align*} Because $r_n=a_0+m_n-d_n a_n$, equivalently $m_n-d_n a_n=r_n-a_0$, we obtain \begin{align*} \alpha_n-a_n=\frac{\sqrt{D}-a_0+r_n}{d_n}. \end{align*} The hypothesis that $D$ is nonsquare matters here: it implies $\sqrt{D}$ is not an integer, so from $a_0=\lfloor\sqrt{D}\rfloor$ we get the strict bound \begin{align*} 0<\sqrt{D}-a_0<1. \end{align*} Combining this with $0\le r_n<d_n$ gives \begin{align*} 0<\sqrt{D}-a_0+r_n<1+(d_n-1)=d_n. \end{align*} After division by the positive integer $d_n$, we obtain \begin{align*} 0<\alpha_n-a_n<1. \end{align*} Hence \begin{align*} a_n<\alpha_n<a_n+1. \end{align*} This proves that $a_n$ is the unique integer immediately below $\alpha_n$, so \begin{align*} a_n=\lfloor\alpha_n\rfloor. \end{align*}[/guided]

[step:Rationalise the next complete quotient] The previous step gives $a_n<\alpha_n<a_n+1$, so $\alpha_n-a_n>0$. Hence the next complete quotient is well-defined by \begin{align*} \alpha_{n+1}=\frac{1}{\alpha_n-a_n}. \end{align*} Substituting the induction hypothesis gives \begin{align*} \alpha_{n+1}=\frac{1}{\frac{\sqrt D+m_n}{d_n}-a_n}. \end{align*} Combining the denominator over $d_n$ gives \begin{align*} \alpha_{n+1}=\frac{d_n}{\sqrt D+m_n-d_n a_n}. \end{align*} By the recurrence definition $m_{n+1}=d_n a_n-m_n$, the denominator is $\sqrt D-m_{n+1}$, and therefore \begin{align*} \alpha_{n+1}=\frac{d_n}{\sqrt D-m_{n+1}}. \end{align*} Since $m_{n+1}$ is an integer and $\sqrt{D}$ is irrational because $D$ is nonsquare, $\sqrt{D}-m_{n+1}\neq 0$. Thus multiplying numerator and denominator by $\sqrt{D}+m_{n+1}$ is valid and yields \begin{align*} \alpha_{n+1}=\frac{d_n(\sqrt D+m_{n+1})}{D-m_{n+1}^2}. \end{align*} Once $d_{n+1}=(D-m_{n+1}^2)/d_n$ is shown to be a positive integer in the next step, this identity becomes \begin{align*} \alpha_{n+1}=\frac{\sqrt D+m_{n+1}}{d_{n+1}}. \end{align*} [/step]

[step:Propagate divisibility, positivity, and the auxiliary bound to the next denominator] First prove divisibility independently of the definition of $d_{n+1}$. The induction hypothesis gives $d_n\mid D-m_n^2$, and the recurrence gives \begin{align*} m_{n+1}=d_n a_n-m_n. \end{align*} Therefore $m_{n+1}\equiv -m_n \pmod{d_n}$, so $m_{n+1}^2\equiv m_n^2 \pmod{d_n}$. It follows that \begin{align*} D-m_{n+1}^2\equiv D-m_n^2\equiv 0 \pmod{d_n}. \end{align*} Thus $d_n$ divides $D-m_{n+1}^2$, and the recurrence formula \begin{align*} d_{n+1}=\frac{D-m_{n+1}^2}{d_n} \end{align*} defines an integer. It remains to prove that this integer is positive. From the previous step, $a_n<\alpha_n<a_n+1$. Substituting \begin{align*} \alpha_n=\frac{\sqrt{D}+m_n}{d_n} \end{align*} and multiplying by the positive integer $d_n$ gives \begin{align*} d_n a_n<\sqrt{D}+m_n<d_n(a_n+1). \end{align*} Subtracting $m_n$ from the left inequality gives \begin{align*} m_{n+1}=d_n a_n-m_n<\sqrt{D}. \end{align*} For the lower bound, the same floor identity gives $a_n=\lfloor\alpha_n\rfloor$. Since $\alpha_n>0$, we have $a_n\ge 0$. The strengthened induction hypothesis gives $m_n<\sqrt{D}$, and hence \begin{align*} m_{n+1}=d_n a_n-m_n\ge -m_n> -\sqrt{D}. \end{align*} Therefore \begin{align*} -\sqrt{D}<m_{n+1}<\sqrt{D}. \end{align*} This auxiliary bound implies both factors in \begin{align*} D-m_{n+1}^2=(\sqrt D-m_{n+1})(\sqrt D+m_{n+1}) \end{align*} are positive, so \begin{align*} D-m_{n+1}^2>0. \end{align*} Since $d_n>0$, the quotient \begin{align*} d_{n+1}=\frac{D-m_{n+1}^2}{d_n} \end{align*} is a positive integer. Substituting this definition into the rationalised identity from the previous step gives \begin{align*} \alpha_{n+1}=\frac{\sqrt D+m_{n+1}}{d_{n+1}}. \end{align*} Consequently the strengthened induction hypotheses are valid at index $n+1$. [/step]

[step:Conclude the strengthened induction] The initial step proves the desired assertions for $n=0$, including the auxiliary bound $-\sqrt{D}<m_0<\sqrt{D}$. The induction steps show that, whenever the strengthened assertions hold at index $n$, they also hold at index $n+1$. Therefore, for every $n \ge 0$, \begin{align*} \alpha_n = \frac{\sqrt{D} + m_n}{d_n}, \end{align*} with $d_n \in \mathbb{N}$, $d_n > 0$, $d_n$ dividing $D-m_n^2$, and \begin{align*} a_n = \lfloor \alpha_n \rfloor. \end{align*} Discarding the auxiliary bound, these are exactly the asserted complete quotient properties for the continued fraction expansion of $\sqrt{D}$. [/step]