We have a integer sequence: $\displaystyle a_{1}$, $\displaystyle a_{2}$, $\displaystyle a_{3}$, .... and we have
$\displaystyle a_{1}=1$
$\displaystyle a_{2}=2$.
$\displaystyle a_{n}=3a_{n - 1}+5a_{n - 2}$
$\displaystyle n=3,4,5....$
Does exist integer number $\displaystyle x\geqslant 2$ such that $\displaystyle a_{x}$ is a divisor of $\displaystyle a_{x + 1}a_{x + 2}$ ??

2. Originally Posted by Ununuquantium
We have a integer sequence: $\displaystyle a_{1}$, $\displaystyle a_{2}$, $\displaystyle a_{3}$, .... and we have
$\displaystyle a_{1}=1$
$\displaystyle a_{2}=2$.
$\displaystyle a_{n}=3a_{n - 1}+5a_{n - 2}$
$\displaystyle n=3,4,5....$
Does exist integer number $\displaystyle x\geqslant 2$ such that $\displaystyle a_{x}$ is a divisor of $\displaystyle a_{x + 1}a_{x + 2}$ ??
You can find the general expression for $\displaystyle a_n$.

Define the generating function,
$\displaystyle f(x) = a_1+a_2x+a_3x^2+...$
Now consider,
$\displaystyle x^2f(x) - 3xf(x) -5x^2f(x) = 0$
And find the form of $\displaystyle a_n$ ...

3. General expression doesn't work

4. Originally Posted by Ununuquantium
We have a integer sequence: $\displaystyle a_{1}$, $\displaystyle a_{2}$, $\displaystyle a_{3}$, .... and we have
$\displaystyle a_{1}=1$
$\displaystyle a_{2}=2$.
$\displaystyle a_{n}=3a_{n - 1}+5a_{n - 2}$
$\displaystyle n=3,4,5....$
Does exist integer number $\displaystyle x\geqslant 2$ such that $\displaystyle a_{x}$ is a divisor of $\displaystyle a_{x + 1}a_{x + 2}$ ??
Suppose (in order to get a contradiction) that $\displaystyle a_n$ divides $\displaystyle a_{n + 1}a_{n + 2}$.

Let p be a prime divisor of $\displaystyle a_n$. Then p must divide one or other of $\displaystyle a_{n + 1}$ and $\displaystyle a_{n + 2}$. But this means that p divides two of the three terms in the relation $\displaystyle a_{n+2}=3a_{n + 1}+5a_{n}$ and therefore it divides the third one as well. However, $\displaystyle a_{n+1}=3a_{n}+5a_{n - 1}$, and since p divides $\displaystyle a_{n + 1}$ and $\displaystyle a_{n}$ it must also divide $\displaystyle a_{n-1}$. Continuing in this way by "reverse induction", you see that p must divide $\displaystyle a_{n-2}$, $\displaystyle a_{n-3}$, ..., and eventually that p divides $\displaystyle a_{1}=1$.

That contradiction shows more than was asked for, namely that none of the prime divisors of $\displaystyle a_{n}$ can divide $\displaystyle a_{n+1}$ or $\displaystyle a_{n+2}$.

Edit: That proof doesn't quite work. It's okay unless p = 3 or 5. But those two primes will require some other argument. I don't have time to think about that just now.

Second edit: No real problem with p = 3 or 5. Suppose that p = 3. The recurrence relation $\displaystyle a_{n}=3a_{n - 1}+5a_{n - 2}$ says that if $\displaystyle a_n$ is a multiple of 3 then so is $\displaystyle a_{n-2}$. The "reverse induction" procedure then tells us that either $\displaystyle a_2$ or $\displaystyle a_1$ is a multiple of 3, which is a contradiction.

In the same way, if p = 5 and $\displaystyle a_n$ is a multiple of 5 then the recurrence relation shows that $\displaystyle a_{n-1}$ is a multiple of 5. Continuing down the sequence to $\displaystyle a_1$, we again get a contradiction.

5. Yes:
I made it also by induction, it is similar:
First I prove that every number from the sequence is not divisible by $\displaystyle 5$:

INDUCTION BASE: I see that $\displaystyle 5$ doesn't divide $\displaystyle a_{k}$, where $\displaystyle k=3$.
Induction step: I will prove for arbitrary $\displaystyle k\geqslant 3$(integer)
that if $\displaystyle a_{k}$ is not divisible by $\displaystyle 5$ therefore $\displaystyle a_{k+1}$ is also not divisible by $\displaystyle 5$.

I do similar with division by $\displaystyle 3$.

And then, also proving by induction I say:

Every three consecutive numbers in the sequence $\displaystyle a_{k}=3a_{k - 1}+5a_{k - 2}$ ($\displaystyle k\geqslant$) are relatively prime [we also know now that every number from the sequence is not divisible by $\displaystyle 5$ and $\displaystyle 3$]:.
INDUCTION BASE: Let $\displaystyle k=3$ - so $\displaystyle a_{1}, a_{2}, a_{3}$ are relatively prime.
Induction step: I will prove that if $\displaystyle a_{k}=3a_{k - 1}+5a_{k - 2}$ therefore $\displaystyle a_{k+1}=3a_{k}+5a_{k - 1}$. So, we see that $\displaystyle a_{k}, a_{k - 1}$ are reletively prime - so $\displaystyle a_{k+1}$ is also relatively prime with $\displaystyle a_{k}, a_{k - 1}$ - it is easy to deduce.

Taht's why there does not exist such $\displaystyle x\geqslant 2$ (integer of course) that $\displaystyle a_{x}$ is divisor of $\displaystyle a_{x + 1}a_{x + 2}$

Can anybody check my proof, please - I am not a specialist