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$$\geq$$2, 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.