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

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It really helps us take questions more seriously when you include your thoughts on a solution, what you've tried, and don't simply copy homework questions and post them hoping for help. We can help best when we better understand where, exactly, you are "stuck"...Your questions are more likely to receive votes as a "good question" when you show some work. Just some suggestions ;-) – amWhy Sep 20 '12 at 11:21
    
sorry because of my bad English :( – LevanDokite Sep 20 '12 at 12:48
up vote 2 down vote accepted

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.

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Noam Elkies gives a much better answer here math.stackexchange.com/a/490225/448 – David Speyer Sep 26 '13 at 12:37

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 $n = 2$).

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@labbhattacharjee:Here $ n\geq 0$. You can shift $n$. – Mhenni Benghorbal Sep 20 '12 at 16:11
    
thanks you so much, but I know your way, I need the solution with easy caculating. I wonder is it exist? – LevanDokite Sep 20 '12 at 16:17
    
@MhenniBenghorbal, could you please share why $n$ is only $2$ and nothing else? – lab bhattacharjee Sep 22 '12 at 12:10

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