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

-
do you mean $a_{n+2}=3a_{n+1}-a_n$ and $1+5a_n*a_{n+1}$? – kaine Sep 10 '13 at 17:17
Arshdeep Singh, welcome to Math.SE. Please check, if the edit I suggested reflects what you wanted to express. Also, please use MathJax for Maths in the future. What is you "." operation? Is it multiplication (usually denoted by "*" in plain text), or something special? – AlexR Sep 10 '13 at 17:18
Alex R -yes it is multiplication – Arshdeep Singh Sep 10 '13 at 17:58
@ArshdeepSingh, Have you tried en.wikipedia.org/wiki/Recurrence_relation#Solving to find $a_n$? – lab bhattacharjee Sep 10 '13 at 18:58
I can determine that if $a_{2j+1}=1250x^2+10x-30$ for some integer $x$ and the $j$ I used is $j=n-1$, then $1+5*a_j*a_{j+1}$ is a perfect square. I cannot, however, determine if any such points exist besides the obvious $j=2$ at this point in time. – kaine Sep 10 '13 at 20:15

The only such $n$ is $n=3$, with $$1 + 5 a_3 a_4 = 1 + 5 \cdot 70 \cdot 180 = 63001 = 251^2.$$ Let $b_n = a_n/10 = 2, 3, 7, 18, 47, \ldots$ for $n=1,2,3,4,5,\ldots$ . These are sums of consecutive odd-order Fibonacci numbers: $2 = 1+1$ (with the first $1$ being $F_{-1}$), $3 = 1+2$, $7 = 2+5$, $18 = 5+13$, $47 = 13+34$, etc. by induction. It soon follows that $b_n b_{n+1} = 5 F^2 + 1$ where $F$ is the Fibonacci number common to $b_n$ and $b_{n+1}$: $$2\cdot 3 = 5 \cdot 1^2 + 1,\phantom{M} 3\cdot 7 = 5 \cdot 2^2 + 1,\phantom{M} 7\cdot 18 = 5 \cdot 5^2 + 1,\phantom{M} 18\cdot 47 = 5 \cdot 13^2 + 1,$$ etc. So we're looking to make $$1 + 5 a_n a_{n+1} = 1 + 500 b_n b_{n+1} = 2500 F^2 + 501$$ a square, and it's easy to see that $F = 5$ is the only positive integer that makes this happen even without the hypothesis that $F$ be a Fibonacci number. (For instance, if $2500 F^2 + 501 = y^2$ with $y>0$, we may factor $501 = y^2 - 2500F^2 = (y-50F) (y+50F)$, or bound $y$ between $50F$ and $50F+1$ once $F>5$, or use the technique I described in this Mathoverflow answer.) Therefore $n=3$ is the unique answer as claimed.

-

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

-
If $x=50k\pm 1$, $x^2 = 2500k^2\pm 100k+1$, so $x^2 = 1 (mod 100)$. In other words, there are other solutions besides $\pm 1$. – marty cohen Sep 11 '13 at 4:02
I checked and it means that $x \equiv \pm 1, \pm 49 \pmod {100}$. So x can be written as $x= 50k \pm 1$. Again I used the same method and I come to the same conclusion that solution doesn't exist. One question is it the general idea good? Is it enough to prove that there aren't solution? – Stefan4024 Sep 11 '13 at 10:24