Solving Recursive Sequence Homework

[HallsofIvy]

Homework Statement

Sequence of integers a1, a2, a3, . . . is fixed by:
a1 =1,
a2 =2,
a_{n}={3a}_{n-1}+5a_{n-2} for n=3,4,5, . . . .
Decide if exisist such an integer k\geq2, that {a}_{k+1}\bullet {a}_{k+1} is divisible by {a}_{k}.

Homework Equations

don't care about latex error.

 

[HallsofIvy]

I edited to try to correct the Latex and accidently deleted the entire thread. Sorry about that. I don't know why the
a_n= 3a_{n-1}+ 5a_{n-2}
still won't come out properly.
To make up for that, I'll try to help with the question!

Before you can prove anything, you need to decide if it is true or false! The question asks b if there exist an integer larger than 1 such that a_{k+1}^2 is divisible by a_k.

From the recursion formula, a_{k+1}= 3a_k+ 5a_{k-1}.
(a_{k+1})^2= 9a_k^2+ 30a_ka_{k-1}+ 25a_{k-1}
Since the first two terms are obviously divisble by a_k, the whole thing is if and only if 25a_{k-1}^2 is divisible by a_k.

That's as far as I can get. Don't know if it helps.

 

[Feldoh]

Find a summation then use induction^^

 

[dalle]

Notice that if a_{k+1}^2 is divisible by a_k then there is a prime p such that both a_{k+1} and a_k are divisible by p. in other words
a_{k+1} = 0 modulo p
a_k = 0 modulo p.

for a fixed prime q define the sequence of integers b_1,b_2, \dots by:
b_{1} = 1

b_{2} = 2

b_{n} = 3 b_{n-1} + 5 b_{n-2} modulo q

Show: there is no integer k>=2 such that both b_k and b_{k+1} are 0.