Let $a_1=a_2=97$ and $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$ for $n>1$. Prove that

(a) $2+2a_n$ is a perfect square, and (b) $2+\sqrt{2+2a_n}$ is a perfect square.

I changed the given recursive formula by squaring, and the result was as follows: $$(a_{n+1}^2+a_n^2+a_{n-1}^2)-2a_{n+1}a_na_{n-1}-1=0$$ $\Rightarrow$ $$(a_{n+1}+a_n+a_{n-1})^2-2(a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1})-1=0$$ $\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(a_{n+1}+a_n+a_{n-1}+a_{n+1}a_n+a_na_{n-1}+a_{n-1}a_{n+1}+a_{n+1}a_na_{n-1}+1)-2=0$$ $\Rightarrow$ $$(1+a_{n+1})^2+(1+a_n)^2+(1+a_{n-1})^2-2(1+a_{n+1})(1+a_n)(1+a_{n-1})-2=0$$ And consequently, I lost the way :(

I thought of proving in a inductive way, that is : $$2+2a_1=196=14^2$$ When we set $2+2a_n=k^2$, $2+2a_{n-1}=l^2$, $$ 2+2_{n+1}=\frac{(k^2-2)(m^2-2)}{4}+\sqrt{\left(\left(\frac{k^2-2}{2}\right)^2-1\right)\left(\left(\frac{m^2-2}{2}\right)^2-1\right)}$$ As a result, I again lost the way :/

I think I sill have not got the main points. Could you give me some clues about this problem? Thanks.

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    $\begingroup$ b implies a. That is $2+\sqrt{2+2a_n}$ is perfect square implies $\sqrt{2+2a_n}$ is integer. $\endgroup$ – Baby desta Feb 10 at 14:44
  • $\begingroup$ @Prof.Shanku induction? !!! that is what he is doing. are you suggesting any other way? $\endgroup$ – Baby desta Feb 10 at 15:15
  • $\begingroup$ Through AoPS, I got an idea which is to displace $a_n$ into $cosh(t_n)$. Then $\sqrt{a_n^2-1}$ becomes $sinh(t_n)$. Thanks for giving comments to the problem! $\endgroup$ – ToBY Feb 13 at 10:43

$ $(Essentially taken from Sequence problem on AoPS.)

The key point is to recognize the connection between the recurrence formula $$ a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)} $$ and the addition formula for the inverse hyperbolic cosine: $$ \DeclareMathOperator{\arcosh}{arcosh} \arcosh u + \arcosh v=\arcosh \left(uv + {\sqrt {(u^{2}-1)(v^{2}-1)}}\right) \, . $$

First note that $a_n > 1$ for all $n$, so that the sequence is well-defined, and we can set $x_n = \arcosh a_n$. Then $$ x_{n+1} = x_n + x_{n-1} \, , $$ which together with $x_1 = x_2$ implies that $(x_n)$ is a multiple of the Fibonacci sequence: $x_n = F_n x_1$. (In the following we need only that $x_n$ is an integer multiple of $x_1$.)

So we have $$ a_n = \cosh x_n = \cosh \left( F_n x_1 \right) \, , $$ and using the half-angle formula for $\cosh$ we conclude that $$ 2 + 2a_n = \left( 2 \cosh \left( \frac{F_n x_1}{2} \right)\right)^2 $$ and then $$ 2+\sqrt{2+2a_n} = \left( 2 \cosh \left( \frac{F_n x_1}{4} \right)\right)^2 $$

It remains to show that $$ y_n = 2 \cosh \left( \frac{F_n x_1}{4} \right) = e^{F_n x_1/4} + e^{-F_n x_1/4} = \alpha^{F_n} + \left(\frac{1}{\alpha}\right)^{F_n} $$ with $\alpha = e^{x_1/4}$ is an integer for all $n$. Setting $n=1$ allows to determine $\alpha$: $$ y_1 = \sqrt{2+\sqrt{2+2 \cdot 97}} = 4 = \alpha + \frac 1\alpha $$ has the solution $$ \{ \alpha, \frac 1\alpha \} = \{ 2 + \sqrt 3, 2 - \sqrt 3 \} \, . $$ So finally we get $$ y_n = \left( 2 + \sqrt 3\right)^{F_n} + \left( 2 - \sqrt 3\right)^{F_n} $$ and that is indeed an integer for all $n$ (see for example The number $(3+\sqrt{5})^n+(3-\sqrt{5})^n$ is an integer).

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    $\begingroup$ Nice work, starting with the recognition of the cosh identity. $\endgroup$ – marty cohen Feb 13 at 3:04

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