Find all positive integers $n$ for which $1 + 5a_n.a_{n + 1}$ is a perfect square. The sequence $a_1, a_2, \ldots $ is defined by the initial conditions $$a_1 = 20;  \quad a_2 = 30$$ and the recursion
$$a_{n+2} = 3a_{n+1} - a_n$$  and
for $n \geq 1$. Find all positive integers $n$ for which $1 + 5a_n * a_{n+1}$ is a perfect square.
I could only find the $n$-th term and don't know how to proceed further.pls help 

Note:Just a try. If I'm wrong feel free to comment. Maybe this approach will give someone an idea
I'll post a solution, or rather attempt to solve this problem. 
Let $1 + 5 \cdot a_n \cdot a_{n+1} = x^2$
First obviously every number in the sequence $a$ is multiple of $10$. So:
$$a_n \cdot a_{n+1} \equiv 0 \pmod {100}$$
$$5 \cdot a_n \cdot a_{n+1} \equiv 0 \pmod {100}$$
$$1 + 5 \cdot a_n \cdot a_{n+1} \equiv 1 \pmod {100}$$
$$x^2 \equiv 1 \pmod {100} \implies x \equiv \pm 1, \pm49 \pmod {100}$$
This means that $x$ can be written as: $ x = 50k \pm 1$
$$1 + 5 \cdot a_n \cdot a_{n+1} = (50k \pm 1)^2$$
$$1 + 5 \cdot a_n \cdot a_{n+1} = 2500k^2 \pm 100k + 1$$
$$5 \cdot a_n \cdot a_{n+1} = 100k(25k \pm 1)$$
Now we can introduce another sequence $b$, where $b_n = \frac{a_n}{10}$
$$5 \cdot b_n \cdot b_{n+1} = k(25k \pm 1)$$
The RHS need to be divisible by $5$, but that's only possible if $k$ is multiple of $5$. So we write $k=5l$
$$5 \cdot b_n \cdot b_{n+1} = 5l(25k \pm 1)$$
$$b_n \cdot b_{n+1} = l(125l \pm 1)$$
Now using the formula for the sequence we write $b_{n+1} = 3b_n - b_{n-1}$
$$b_n (3b_n - b_{n-1}) = l(125l \pm 1)$$
$$3b_n^2 - b_n \cdot b_{n-1} - l(125l \pm 1) = 0$$
We are now solving a quadratic equation for $b_n$
$$b_n = \frac{b_{n-1} \pm \sqrt{b_{n-1}^2 + 12l(125l \pm 1)}}{6}$$ 
But because $b_n$ can have one unique value it means that this equation has double root, implying that:
$$b_{n-1}^2 + 12l(125l \pm 1) = 0$$
But $12l(125l \pm 1) > 0$, which means that $b_{n-1}^2 < 0$, which is impossible. This leads to conclusion that an $1 + 5 \cdot a_n \cdot a_{n+1}$ can't be a perfect square
