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: $1+5u_nu_{n+1}=k^2,\; k \in \mathbb N.$
Find n that : $1+5u_nu_{n+1}=k^2, k \in N$
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0sorr$y$ because of my bad English :( – 2012-09-20
2 Answers
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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0Noam Elkies gives a much better answer here http://math.stackexchange.com/a/490225/448 – 2013-09-26
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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0@MhenniBenghorbal, could you please share why $n$ is only $2$ and nothing else? – 2012-09-22