\(% 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}}\)
The base case is when \(n = 2\), and we have \[ T_2 = (\sqrt {a + 1} + \sqrt {a})^2 = (2a + 1) + 2 \cdot \sqrt {a (a + 1)}. \]
We therefore have \(A_2 = 2a + 1\) and \(B_2 = 2\), and we verify that \[ a(a + 1) B_2^2 + 1 = a (a + 1) \cdot 2^2 + 1 = 4a^2 + 4a + 1 = (2a + 1)^2 = A_2^2, \] as desired, and the statement holds for the base case when \(n = 2\).
Now, assume that this statement is for some even \(n = k\), i.e. \[ T_k = A_k + B_k \sqrt {a (a + 1)} \] where \(A_k\) and \(B_k\) are both integers, and \(A_k^2 = a(a + 1) B_k^2 + 1\).
Notice that \begin {align*} T_{k + 2} & = T_k \cdot \left (\sqrt {a + 1} + \sqrt {a}\right )^2 \\ & = \left (A_k + B_k \sqrt {a(a + 1)}\right ) \cdot \left (2a + 1 + 2 \sqrt {a(a + 1)}\right ) \\ & = A_k \cdot (2a + 1) + B_k \cdot 2 \cdot a (a + 1) + 2A_k \sqrt {a(a + 1)} + (2a + 1) B_k \sqrt {a(a + 1)} \\ & = \left [(2a + 1)A_k + 2a(a + 1)B_k\right ] + \left [2 A_k + (2a + 1)B_k\right ] \sqrt {a(a + 1)}. \end {align*}
Now let \(A_{k + 2} = (2a + 1) A_k + 2a (a + 1) B_k\), and \(B_{k + 2} = 2 A_k + (2a + 1)B_k\). Since \(a\) is a positive integer, and \(A_k\) and \(B_k\) are both integers, we must have \(A_{k + 2}\) and \(B_{k + 2}\) are both integers. Furthermore, \begin {align*} & \phantom {=} A_{k + 2}^2 - \left [a (a + 1) B_{k + 2}^2 + 1\right ] \\ & = \left [(2a + 1) A_k + 2a (a + 1) B_k\right ]^2 - \left [a (a + 1) \left (2 A_k + (2a + 1)B_k\right )^2 + 1\right ] \\ & = \left [(2a + 1)^2 - 4 a (a + 1)\right ] A_k^2 \\ & \phantom {=} + \left [2 \cdot (2a + 1) \cdot 2a (a + 1) - 2 \cdot a (a + 1) \cdot 2 \cdot (2a + 1)\right ] A_k B_k \\ & \phantom {=} + [(2a (a + 1))^2 - a (a + 1) (2a + 1)^2] B_k - 1 \\ & = A_k^2 - a (a + 1) B_k^2 - 1 \\ & = 1 - 1 \\ & = 0, \end {align*}
and hence \[ A_{k + 2}^2 = a (a + 1) B_{k + 2}^2 + 1. \]
So the original statement holds for \(n = k + 2\).
By the principle of mathematical induction, the original statement must hold for all even integers \(n\).
If \(n\) is odd, then we have \begin {align*} T_n & = (\sqrt {a + 1} + \sqrt {a}) T_{n - 1} \\ & = (\sqrt {a + 1} + \sqrt {a}) (A_{n - 1} + B_{n - 1} \sqrt {a(a + 1)}) \\ & = A_{n - 1} \sqrt {a + 1} + A_{n - 1} \sqrt {a} + B_{n - 1} (a + 1) \sqrt {a} + B_{n - 1} a \sqrt {a + 1} \\ & = (A_{n - 1} + a B_{n - 1}) \sqrt {a + 1} + (A_{n - 1} + (a + 1) B_{n - 1})\sqrt {a}. \end {align*}
Now, consider \(C_n = A_{n - 1} + a B_{n - 1}\), and \(D_n = A_{n - 1} + (a + 1) B_{n - 1}\). Since \(a\) is a positive integer, and \(A_{n - 1}\) and \(B_{n - 1}\) are integers, we must have \(C_n\) and \(D_n\) are integers as well. Furthermore, \begin {align*} & \phantom {=} (a + 1) C_n^2 - (a D_n^2 + 1) \\ & = (a + 1) \left (A_{n - 1} + a B_{n - 1}\right )^2 - \left [a \left (A_{n - 1} + (a + 1) B_{n - 1}\right )^2 + 1\right ] \\ & = \left [(a + 1) - a\right ] A_{n - 1}^2 + \left [(a + 1) \cdot 2 \cdot a - a \cdot 2 \cdot (a + 1)\right ] A_{n - 1} B_{n - 1} \\ & \phantom {=} + [(a + 1) a^2 - a (a + 1)^2] B_{n - 1}^2 - 1 \\ & = A_{n - 1}^2 - a(a + 1) B_{n - 1}^2 - 1 \\ & = 1 - 1 \\ & = 0, \end {align*}
and hence \[ (a + 1) C_n^2 = a D_n^2 + 1. \]
This shows precisely the original statement.