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

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.

Answer 1

$ $(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).

Answer 2

The question is

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.

You were on the right track with squaring the given recursive formula. You just need a little bit more work to follow to get the final answer.

Define a sequence of numbers with initial values (note that $\;a_2 = 97$) and recursion

$$ a_0 = 1, a_1 = 97, \;\; \text{ and } \;\; a_{n+1} = a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}. \tag1 $$

Isolate the square root in $(1)$, square both sides and simplify to get

$$ (a_{n+1}^2+a_n^2+a_{n-1}^2)-2a_{n+1}a_na_{n-1}-1=0. \tag2 $$

Define another sequence of numbers with

$$ b_n := \sqrt{2+2a_n} \qquad \text{ or } \qquad a_n = (b_n^2-2)/2. \tag3 $$

Verify algebraically the identity

$$ 4((x+y+z)^2-2xyz-1) = (u^2+v^2+w^2-4)^2-(uvw)^2 = \\ (u^2+v^2+w^2-4 -uvw)(u^2+v^2+w^2-4 +uvw) \\ \text{where } \qquad u^2 = 2+2x,\;\; v^2 = 2+2y,\;\; w^2 = 2+2z. \tag4 $$

Combine $(2)$ and $(3)$ where $\,y=a_n,\, v=b_n\,$ and verify that the left factor is zero to get

$$ (b_{n+1}^2+b_n^2+b_{n-1}^2)- b_{n+1}b_n b_{n-1}-4=0. \tag5 $$

Given $\,b_n,b_{n-1},\,$ regard $(5)$ as a quadratic in $\,b_{n+1}\,$ and notice that $\,b_{n-2}\,$ is the other root.

This is a simple example of Vieta jumping. Use Vieta's formulas to get

$$ b_{n+1} + b_{n-2} = b_n b_{n-1} \qquad \text{ and } \qquad b_{n+1} b_{n-2} = b_n^2 + b_{n-1}^2 - 4. \tag6 $$

Rewrite the first formula to get the recursion

$$ b_{n+1} = b_n b_{n-1} - b_{n-2} \qquad \text{ while } \qquad b_0 = 2, b_1 = b_2 = 14 \tag7 $$

which implies that $\,b_n\,$ is an integer sequence. This proves part (a) of the question.

Define another sequence of numbers with

$$ c_n := \sqrt{2+b_n} \qquad \text{ or } \qquad b_n = c_n^2-2. \tag8 $$

A very similar argument to $(4),(5),(6),(7)$ implies the recursion

$$ c_{n+1} = c_n c_{n-1} - c_{n-2} \qquad \text{ while } \qquad c_0 = 2, c_1 = c_2 = 4 \tag9 $$

which implies that $\,c_n\,$ is an integer sequence. This proves part (b) of the question.

Both parts of the question are now proved but the sequences $\;a_n,\,b_n,\,c_n\;$ remain to be identified.

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Define $\,A(n) := T_n(2)\,$ where $\,T_n(x)\,$ is the Chebyshev polynomial of the second kind.

As usual, let $\,F_n\,$ be the Fibonacci sequence.

This is a small table of sequence values: $$\begin{array}{|c|c|c|c|c|} \hline n & a_n & b_n & c_n & A(n) \\ \hline 0 & 1 & 2 & 2 & 1 \\ \hline 1 & 97 & 14 & 4 & 2 \\ \hline 2 & 97 & 14 & 4 & 7\\ \hline 3 & 18817 & 194 & 14 & 26\\ \hline 4 & 3650401 & 2702 & 52 & 97 \\ \hline 5 & 137379191137 & 524174 & 724 & 362 \\ \hline \end{array}$$

Using properties of $\;F_n\;$ and $\;A(n)\;$ it can be proved that $$ a_n = A(4F_n), \qquad b_n = 2A(2F_n), \qquad c_n = 2A(F_n). \tag{10}$$

Answer 3

Simply redoint the argument of @Martin R: so I understand it better.

If $a = \frac{1}{2}(s +1/s)$ then $a^2-1= \left(\frac{1}{2}(s - 1/s)\right)^2$. Now for $a = \frac{1}{2}(s + 1/s)$, $b = \frac{1}{2}(t + 1/t)$ ( $s$, $t> 1$) we get

$$a b + \sqrt{(a^2-1)(b^2-1)}= \frac{1}{4} (s+1/s)(t+ 1/t) + \frac{1}{4} (s-1/s)(t- 1/t)=\\ = \frac{1}{2}( s t + 1/(s t) )$$