Help about properties of Pell numbers

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The discussion centers on the properties of Pell numbers, specifically the sequences defined by U_n and V_n. A proof was provided for the congruence C1, which states that V_{2^{n}+1}/2 is congruent to 1 modulo 2^{n+1} for n ≥ 2. There were also discussions about conjectures C2, C3, and C4 regarding prime divisibility conditions related to the Companion Pell numbers. The conversation highlights the relationships between Pell numbers and their properties, with a focus on proving specific congruences and divisibility. The thread concludes with a collaborative effort to explore these mathematical properties further.
T.Rex
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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
 
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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
 


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
 
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