Find n that : $1+5u_nu_{n+1}=k^2, k \in N$ 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: $$1+5u_nu_{n+1}=k^2,\; k \in \mathbb N.$$
 A: 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.
A: 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 $n = 2$). 
