# Solving the sequence $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$: proving that $2+2a_n$ is a perfect square

Question:

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.

• b implies a. That is $2+\sqrt{2+2a_n}$ is perfect square implies $\sqrt{2+2a_n}$ is integer. – Baby desta Feb 10 at 14:44
• @Prof.Shanku induction? !!! that is what he is doing. are you suggesting any other way? – Baby desta Feb 10 at 15:15
• 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! – ToBY Feb 13 at 10:43

## 1 Answer

(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).

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