Find squares in recurrence relation Find all squares in $\{a_n\}$ which obeys the following recurrence relation:
$a_0=a_1=1, a_{n+2}=6a_{n+1}-a_n$
I've tried to solve the equation $n^2+(n+1)^2=m^4$ and got this recurrence relation. 
(If and only if $m^2\in\{a_i\}_{i=2,3,\dots}$, $m$ satisfies this equation, I suppose.)
I'd be happy if you could share your ideas on how to find the all square numbers in $\{a_i\}_{i=2,3,\dots}=\{5,29,169,985,5741,\dots\}$ with me. There seems to be only one square in the first 30 terms, according to my computer. Interestingly, each term has very few divisors. I've read one of the proofs of square Fibonacci numbers, but failed to apply it to this problem.
Please give me some hints.
EDIT
Here is the page of this sequence.
Solution
Thanks to your help, I found the elementary solution here.
 A: COMMENT:
I worked many years ago on famous Euler equation $2y^4-1=z^2$. This is my experience:
Consider following sets their members are correspondingly make Pythagorean triples:
$a∈\{0, 3, 119, 4059, 137903, . . . \}$
$b∈\{1, 4, 120, 4060, 137904, . . . \}$
$c∈\{1, 5, 169, 5741, 195025, . . . \}$
Now take triple (119, 120, 169) we can write:
$(120+119)^2=120^2+119^2+2\times120\times 119$
$(120-119)^2=120^2+119^2-2\times120\times 119$
Summing these relations we get:
$239^2+1=2(120^2+119^2=169^2)$
Or:
$239^2+1=2\times 13^4$
Which gives $z=239$ and $y=13$
In set c which you have found its complete form only $1$ and $169$ can be perfect square.This was proved by Lungern on 1942.
A: You can directly obtain the solution to this, as it's a homogeneous recurrence in constant coefficients
The characteristic polynomial is $p(x) = x^2-6x+1$
The roots of the above are $r_{1,2} = 3 \pm 2\sqrt{2}$
Hence, we have
$$a_n = k_1(3+2\sqrt{2})^n + k_2(3-2\sqrt{2})^n$$
$$a_0 = 1 = k_1 + k_2$$
$$a_1 = 1 = 3(k_1 + k_2) + 2\sqrt2(k_1-k_2)$$
Solving for $k_1,k_2$ we have
$$k_1 = \frac{\sqrt{2} - 1}{2\sqrt2}, k_2 = \frac{\sqrt{2}+1}{2\sqrt{2}}$$
Now I don't know for sure how this would help finding perfect squares, but at least you have a way of determining the $n^{th}$ term directly
EDIT
Something interesting about $a_n$, you can rewrite it as follows
$$a_n = \frac{1}{2\sqrt2}\left((\sqrt{2}+1)^{2n}(\sqrt{2}-1) + (\sqrt{2}+1)(\sqrt{2}-1)^{2n}\right)$$
$$a_n = \frac{1}{2\sqrt2}\left((\sqrt{2}+1)^{2n-1} + (\sqrt{2}-1)^{2n-1}\right)$$
I'd expand the sum in terms of binomial theorem, terms would cancel out, maybe that might help?
