Find n that : $1+5u_nu_{n+1}=k^2, k \in N$

Question

Let ${u_n}$ be such that: $$\begin{cases}u_1=20;\\u_2=30;\\ u_{n+2}=3u_{n+1}-u_{n},\; n \in \mathbb N^*.\end{cases}$$ Find $n$ such that: $$1+5u_nu_{n+1}=k^2,\; k \in \mathbb N.$$

Answer 1

Here is a way that you could imagine proving $n=2$ is the only solution. Notice that $$u_n^2 - 3 u_n u_{n+1} + u_{n+1}^2 = -500$$ for all $n$. So we are searching for $(x,y, z)$ solving $$x^2-3xy+y^2 = -500$$ $$z^2 = 1+5xy$$ That intersection is an elliptic curve; now see this question.

Answer 2

First, we solve the recurrence relation with its initial conditions

$$ \begin{cases}u_0=20;\\u_1=30;\\ u_{n+2}=3u_{n+1}-u_{n},\; n \in \mathbb N^*.\end{cases} \,.$$

The solution is given by

$$ u(n) = 10\, \left( \frac{\sqrt {5}}{2}+\frac{3}{2} \right) ^{n}+10\, \left( \frac{3}{2}-\frac{\sqrt {5}}{2}\,\right)^{n} \,,\quad n \geq 0 \,. $$

Now, you need the above solution to solve for $n$ $$ 1+5u_nu_{n+1}=k^2,\; k \in \mathbb N. $$

Substituting the solution in the above equation and simplifying, we have

$$ \frac{10^2}{2^{2n+1}} \left(\left( 3+\sqrt{5} \right)^{n}+ \left( {3}-{\sqrt {5}}\,\right)^{n}\right)\left(\left( 3+\sqrt{5} \right)^{n+1}+ \left( {3}-{\sqrt {5}}\,\right)^{n+1}\right) = k^2 - 1 \,.$$

I will leave it for you to finish the solution of your problem (the solution is $n = 2$).