Prove the sequence is perfect square

Question

Question

Given a sequence $a_0 = a_1 = 97\sqrt{2}$ and $$\forall n \space \space a_{n+1}= \frac{1}{\sqrt{2}}[a_n a_{n-1} + \sqrt{(a_n^2-2)(a_{n-1}^2-2)}]$$ Proof that $$2+\sqrt{2+a_n\sqrt{2}}$$ is a perfect square

My approach

Use mathematical induction to prove that (base case are trivial, this is the inductive step)

$$2+\sqrt{2+a_na_{n-1}+\sqrt{(a_n^2-2)(a_{n-1}^2-2)}}$$ However, this does not hold true for all $a_n$, there seems to be an invariant from the initial value that makes it perfect square. Any hints would be appreciated.

Out of curiosity, where did you find this recurrence? Or did you come up with this? — Marco Bellocchi, Jun 13 '18 at 19:09
@Marco Bellochi I found it in one of Mathematics Olympiad questions — Clark Makmur, Jun 20 '18 at 05:51

score 1 · Accepted Answer · edited Jun 11 '18 at 10:04

1

This is not an answer, but I want to share some of my findings and I cannot do it in a comment. Let $$x_n=2+\sqrt{2+a_n\sqrt{2}}$$ for $n=0,1,...$. Now it is easy to prove that indeed $x_0,x_1,x_2,x_3$ are perfect squares. But there is a precise relationship between $x_n$ for $n>2$ and it is $$\sqrt{x_n}=\sqrt{x_{n-1}x_{n-2}}-\sqrt{x_{n-3}}$$ Now the base case for $k=3$ is trivial to verify. If by induction we suppose that then for $k=4,5,,...n$ we have $$\sqrt{x_k}=\sqrt{x_{k-1}x_{k-2}}-\sqrt{x_{k-3}}$$ and also $$x_k \text{ is a perfect square}$$ then it remains to show that $$\sqrt{x_{n+1}}=\sqrt{x_{n}x_{n-1}}-\sqrt{x_{n-2}}$$ which implies than that $x_{n+1}$ is a perfect square and by induction we are done. However the algebra is a bit messy, but I believe it could be a good line of attack.

edited Jun 11 '18 at 10:04

Mau

113

answered Jun 06 '18 at 19:16

Marco Bellocchi

686

This equation $\sqrt{x_n}=\sqrt{x_{n-1}x_{n-2}}-\sqrt{x_{n-3}}$ is not true. For example, $a_2 = (497^2 - 2)/\sqrt{2}$ and $a_3 = (897^3 - 6*97)/\sqrt{2}$ – sku Jun 07 '18 at 02:25
I specified for n>2 – Marco Bellocchi Jun 07 '18 at 02:54
Hence base case is for n=3. For n=0,1,2 it is easy to verify that xn is a perfect square. Then for n=3, which is my base case, the relationship holds I believe – Marco Bellocchi Jun 07 '18 at 02:57
Indeed we have $\sqrt{x_3}=52$, $\sqrt{x_2}=14$, $\sqrt{x_1}=4$, $\sqrt{x_0}=4$ and obviously $14*4-4=52$ – Marco Bellocchi Jun 07 '18 at 03:04
I am sorry. I mistook your $x_n$ for $a_n$. – sku Jun 07 '18 at 03:05

Prove the sequence is perfect square

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