Proving divisibility properties of Pell companion sequence V_n

[T.Rex]

Hi, I need some help about properties of Pell numbers:

$$U_n = 2 U_{n-1} + U_{n-2} , \text{ with: } U_0=0 \text{<br /> and: } U_1=1$$

$$V_n = 2 V_{n-1} + V_{n-2} , \text{ with: } V_0=2 \text{<br /> and: } V_1=2$$


I have a proof for:

$$\frac{V_{\displaystyle 2^{\scriptstyle n}}}{2}<br /> \ = \ 1 + 4 \prod_{i=0}^{n-2}V_{\displaystyle 2^{\scriptstyle<br /> i}}^{\scriptstyle 2} \ \equiv \ 1 \pmod{2^{\scriptstyle 2n}} \<br /> \text{\ \ \ \ (for } n \geq 2 )$$

But I have no proof for C1:

$$\frac{V_{\displaystyle 2^{\scriptstyle n}+1}}{2}<br /> \ \equiv \ 1 \pmod{2^{\scriptstyle n+1}} \ \text{\ \ \ \ (for<br /> } n \geq 2 )$$

and C2:

$$p,q \ \text{ odd primes }, \ \ p \mid V_{\displaystyle q} \ \ \Longrightarrow \ \ p \equiv 1 \pmod{2q}$$

and C3:

$$p \ \text{ odd prime }, \ \ p \mid V_{\displaystyle 2^{\scriptstyle i}} \ \ \Longrightarrow \ \ p \equiv 1 \pmod{2^{\scriptstyle i+2}}$$

and C4 (a guess):
$$p \ \text{ odd prime }, \ \ p \mid V_{\displaystyle 2^{\scriptstyle i}} \ <br /> \text{ and } \ p = 1 + {(2^{\scriptstyle i}\alpha)}^2 \ \ <br /> \Longrightarrow \ \ \alpha = \prod_{j=0}^{k} F_j \ \text{ or } \<br /> \alpha = 1$$.

Can you help ?
Thanks,

Tony

 

[T.Rex]

Any idea ?

OK, I've got a proof for C1. It was not so difficult.

About C4, there was a mistake:
$$p \ \text{ odd prime }, \ \ p \mid V_{\displaystyle 2^{\scriptstyle n}} \ <br /> \text{ and } \ p = 1 + {(2^{\scriptstyle n}\alpha)}^2 \ \ <br /> \Longrightarrow \ \ \alpha = \prod_{j} F_j$$
where $$F_j$$ are Fermat prime numbers.

Any idea ?
tony

 

[M98Ranger]

Hi Tony,

I would be happy to help you with your questions about properties of Pell numbers. Firstly, let's review some basic properties of Pell numbers:

1. The Pell numbers are a sequence of numbers that start with 0 and 1, and each subsequent number is the sum of twice the previous number and the number before it. This can be represented by the formula U_n = 2U_{n-1} + U_{n-2}.

2. Similarly, there is a second sequence of numbers called the Companion Pell numbers that start with 2 and 2, and each subsequent number is the sum of twice the previous number and the number before it. This can be represented by the formula V_n = 2V_{n-1} + V_{n-2}.

3. The Pell numbers and Companion Pell numbers have a special relationship, where V_n = U_{n+1}. This means that the nth Pell number is equal to the (n+1)th Companion Pell number.

Now, let's address your specific questions:

C1: This statement is actually a special case of the general formula you provided in your question. If we let n = 1, then we have V_{2} = 2V_{1-1} + V_{1-2} = 2V_0 + V_{-1}. Since V_0 = 2 and V_{-1} = 2, we can plug in these values to get V_2 = 2(2) + 2 = 6. Dividing this by 2, we get 3, which is equivalent to 1 (mod 2^2). This is the same as saying V_{2+1}/2 \equiv 1 (mod 2^{2+1}).

C2: To prove this statement, we can use the fact that V_n = U_{n+1}. Let's assume that p is an odd prime and p divides V_q. This means that p divides U_{q+1}. Using the formula for Pell numbers, we can see that U_{q+1} = 2U_q + U_{q-1}. Since p divides both U_q and U_{q-1}, it must also divide U_{q+1}. This means that p divides U_{q+2}, and so on. Eventually, we will reach a point where p divides