\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We show this by induction on \(n\).
We first consider the base case where \(n = 1\). Notice \(\LHS = x_1 = a\), and \[ \RHS = 2 + 4^{1 - 1} (a - 2) = 2 + (a - 2) = a. \]
Hence, \(\LHS \geq \RHS \) is true.
Now, assume that the original statement \[ x_n \geq 2 + 4^{n - 1} (a - 2) \] is true for some \(n = k\).
Consider the case where \(n = k + 1\). We first notice that since \(a > 2\), we must have \[ x_n \geq 2 + 4^{n - 1} (a - 2) > 0. \]
Hence, we have \begin {align*} \LHS & = x_{k + 1} \\ & = x_{k}^2 - 2 \\ & \geq \left (2 + 4^{k - 1} (a - 2)\right )^2 - 2 \\ & = 4 + 4^{2k - 2}(a - 2)^2 + 4 \cdot 4^{k - 1}(a - 2) - 2 \\ & = 2 + 4^{k} (a - 2) + 4^{2k - 2} (a - 2)^2 \\ & > 2 + 4^{(k + 1) - 1}(a - 2) \\ & = \RHS , \end {align*}
and this shows that the original statement is true for the case \(n = k + 1\) as well.
Hence, the original statement is true for the base case \(n = 1\), and given it holds for \(n = k\), it holds for \(n = k + 1\). By the principle of mathematical induction, it must hold for all integers \(n \geq 1\) given \(a > 2\), as desired.
Only-if direction. We attempt to prove the contrapositive of the only-if direction, i.e. given that \(\abs *{a} \leq 2\), we want to show that \(x_n\) does not diverge to \(\infty \).
We would like to show that \(\abs *{x_n} \leq 2\) for all \(n \in \NN \).
The base case where \(n = 1\) is true, since \(0 \leq a \leq 2\).
Now, assume that this is true for some \(n = k\), i.e. \[ \abs *{x_n} \leq 2 \iff -2 \leq x_n \leq 2 \iff 0 \leq x_n^2 \leq 4. \]
For \(n = k + 1\), \[ x_n = x_{k + 1} = x_k^2 - 2, \] and hence \[ -2 \leq x_{k + 1} \leq 2 \iff \abs *{x_{k + 1}} \leq 2. \]
So this statement is true for the base case where \(n = 1\), and given it holds for some \(n = k\) it holds for the case \(n = k + 1\). Hence, by the principle of mathematical induction, this statement is true for all \(n \in \NN \).
This means that \(x_n\) is bounded above and below, and hence it cannot diverge to infinity. This proves the contrapositive of the only-if direction, and hence the only-if direction is true.
In conclusion, we have shown that \(x_n \to \infty \) as \(n \to \infty \) if and only if \(\abs *{a} > 2\).
If this is true for all \(n \geq 1\), then this is true for \(n = 1\). On one hand, \[ y_{1} = \frac {A x_1}{x_2} = \frac {A a}{a^2 - 2}, \] and on the other hand \[ y_{1} = \frac {\sqrt {x_2^2 - 4}}{x_2} = \frac {\sqrt {(a^2 - 2)^2 - 4}}{a^2 - 2} = \frac {\sqrt {a^4 - 4a^2}}{a^2 - 2} = \frac {a \sqrt {a^2 - 4}}{a^2 - 2}. \]
Hence, we must have \begin {align*} A & = \sqrt {a^2 - 4} \\ A^2 & = a^2 - 4 \\ a^2 & = A^2 + 4 \\ a & = \sqrt {A^2 + 4}, \end {align*}
since \(a > 2\).
We still have to show that this \(a\) gives the desired relation for every \(n \geq 1\).
Notice that by definition, \begin {align*} y_{n + 1} & = \frac {A \prod _{i = 1}^{n + 1} x_i}{x_{n + 2}} \\ & = \frac {A \prod _{i = 1}^{n}}{x_{n + 1}} \cdot \frac {x_{n + 1}^2}{x_{n + 2}} \\ & = y_{n} \cdot \frac {x_{n + 1}^2}{x_{n + 2}}. \end {align*}
We aim to show this by induction on \(n\). The base case where \(n = 1\) is shown above.
Now, assume that \[ y_n = \frac {\sqrt {x_{n + 1}^2 - 4}}{x_{n + 1}} \] for a certain value of \(n = k\).
For \(n = k + 1\), \begin {align*} y_n & = y_{k + 1} \\ & = y_k \cdot \frac {x_{n + 1}^2}{x_{n + 2}} \\ & = frac{\sqrt {x_{n + 1}^2 - 4}}{x_{n + 1}} \cdot \frac {x_{n + 1}^2}{x_{n + 2}} \\ & = \frac {\sqrt {x_{n + 1}^2 - 4} x_{n + 1}}{x_{n + 2}} \\ & = \frac {\sqrt {x_{n + 1}^4 - 4 x_{n + 1}^2}}{x_{n + 2}} \\ & = \frac {\sqrt {\left (x_{n + 1}^2 - 2\right )^2 - 4}}{x_{n + 2}} \\ & = \frac {\sqrt {x_{n + 2}^2 - 4}}{x_{n + 2}}, \end {align*}
which is precisely the original statement for \(n = k + 1\).
By the principle of mathematical induction, for \(a = \sqrt {A^2 + 4}\), we have shown that this desired statement holds for the base case \(n = 1\), and given that it holds for some \(n = k\), we can show it holds for \(n = k + 1\). Hence, by the principle of mathematical induction, we have that \[ y_n = \frac {\sqrt {x_{n + 1}^2 - 4}}{x_{n + 1}} \] for every value of \(n \geq 1\) for this certain value of \(a = \sqrt {A^2 + 4}\).
Hence, for the value \(a = \sqrt {A^2 + 4}\), we have the statement holds for all \(n \geq 1\). We have also shown that if the statement holds for all \(n \geq 1\), it must be the case that \(a = \sqrt {A^2 + 4}\). Hence, for precisely this value of \(a = \sqrt {A^2 + 4}\), we have \[ y_n = \frac {\sqrt {x_{n + 1}^2 + 4}}{x_{n + 1}}. \]
For this value of \(a > 2\), we have \(x_n \to \infty \) as \(n \to \infty \). Hence, \[ y_n = \frac {\sqrt {x_{n + 1}^2 + 4}}{x_{n + 1}} = \sqrt {1 + \frac {4}{x_{n + 1}^2}} \] converges to \(1\) as \(n \to \infty \).