Let
\[
T _n =
\left( \sqrt{a+1} + \sqrt a\right)^n\,,
\]
where \(n\) is a positive integer and \(a\) is any given positive integer.
In the case when \(n\) is even, show
by induction that
\(T_n\) can be written in the form
\[
A_n +B_n \sqrt{a(a+1)}\,,
\]
where
\(A_n\) and \(B_n\) are integers (depending on \(a\) and \(n\))
and \(A_n^2 =a(a+1)B_n^2 +1\).
In the case when \(n\) is odd, show by considering
\((\sqrt{a+1} +\sqrt a)T_m\) where \(m\) is even, or otherwise,
that \(T_n\)
can be written in the form
\[
C_n \sqrt {a+1} + D_n \sqrt a \,,
\]
where \(C_n\) and \(D_n\) are integers (depending on \(a\) and \(n\)) and
\( (a+1)C_n^2 = a D_n^2 +1\,\).
Deduce that, for each \(n\), \(T_n\) can be written
as the sum of the square roots of two consecutive integers.
Solution:
Claim: For all \(n \geq 1\) \(T_{2n} = A_{2n} + B_{2n}\sqrt{a(a+1)}\) where \(A_{2n}, B_{2n}\) are integers and \(A_{2n}^2 = a(a+1)B_{2n}^2+1\)
Proof: (By induction)
Base case: \(n =1\).
\begin{align*}
&& T_2 &= (\sqrt{a+1}+\sqrt{a})^2 \\
&&&= a+1+2\sqrt{a(a+1)}+a \\
&&&=2a+1+2\sqrt{a(a+1)} \\
\Rightarrow && A_2 &= 2a+1 \\
&& B_2 &= 2 \\
&& A_2^2 &= 4a^2+4a+1 \\
&& a(a+1)B_1^2 + 1 &= 4a^2+4a+1
\end{align*}
Therefore our base case is true. Suppose it is true for some \(n = k\) then consider \(n = k+1\) we must have \(T_{2k} = A_{2k}+B_{2k}\sqrt{a(a+1)}\)
\begin{align*}
&& T_{2(k+1)} &= T_{2k} (\sqrt{a+1}+\sqrt{a})^2 \\
&&&= \left (A_{2k}+B_{2k}\sqrt{a(a+1)} \right)\left (2a+1+2\sqrt{a(a+1)} \right) \\
&&&= (2a+1)A_{2k}+2a(a+1)B_{2k} + (2A_{2k}+(2a+1)B_{2k})\sqrt{a(a+1)} \\
\Rightarrow && A_{2(k+1)} &= (2a+1)A_{2k}+2a(a+1)B_{2k} \in \mathbb{Z} \\
&& B_{2(k+1)} &= 2A_{2k}+(2a+1)B_{2k} \in \mathbb{Z} \\
&& A_{2(k+1)}^2 &= \left ( (2a+1)A_{2k}+2a(a+1)B_{2k} \right)^2 \\
&&&= (2a+1)^2A_{2k}^2+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k} \\
&&&= a^2(a+1)^2(2a+1)^2B_{2k}^2+(2a+1)^2\\
&&&\quad\quad+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k}\\
&& a(a+1)B_{2(k+1)}^2 + 1 &= a(a+1)\left ( 2A_{2k}+(2a+1)B_{2k} \right)^2 \\
&&&= 4a(a+1)A_{2k}^2 + 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1\\
&&&= 4a^2(a+1)^2B_{2k}^2+4a(a+1) + \\
&&&\quad\quad+ 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1
\end{align*}
So our relation holds ad therefore by induction we're done.
For even \(n\) \(T_n = \sqrt{a(a+1)B_{2n}^2+1} + \sqrt{a(a+1)B_{2n}^2}\)
For odd \(n\), \(T_n = \sqrt{aD_n^2+1}+ \sqrt{aD_n^2}\) therefore it is always the sum of the square root of two consecutive integers.