Pell number is a term of the sequence $\{P_n\}$ determined by a recurrence relation
$$P_{n+2}=2P_{n+1}+P_n, P_0=0, P_1=1.$$
Let $v_p(x)$ be the $p$-adic valuation of an integer $x$ (the number of times $x$ is divisible by $p$). Is the following conjecture correct?
(I) $v_2(P_n)=v_2(n)$,
(II) For any natural number $m$, if a Pell number $P_n$ is divisible by an odd prime $p$,
$$v_p(P_{mn})=v_p(P_n)+v_p(m).$$
The reason for making this conjecture is that I learned the following facts about the Fibonacci number $F_n$ determined by
$$F_{n+2}=F_{n+1}+F_n, F_0=0, F_1=1.$$
(I) If $F_n$ is even: $v_2(F_{2n})=v_2(F_n)+1$ if $n$ is even, and $v_2(F_{2n})=v_2(F_n)+2$ if $n$ is odd.
(II) For any natural number $m$, if a Fibonacci number $F_n$ is divisible by an odd prime $p$,
$$v_p(F_{mn})=v_p(F_n)+v_p(m).$$
Let $\mathcal{O}$ be the valuation ring of the local field $K=\mathbb{Q}_p(\sqrt{5})$ and $\mathfrak{p}$ be the maximal ideal of $\mathcal{O}$.
It is known that (II) can be proved using ''$\mathfrak{p}$-adic log'' of $K$.
$$\phi=\dfrac{1+\sqrt{5}}{2},\ \overline{\phi}=\dfrac{1-\sqrt{5}}{2},$$
$$v_{\mathfrak{p}}(F_n)=v_{\mathfrak{p}}(n)+v_{\mathfrak{p}}(\log(\phi\overline{\phi}^{-1}))-v_{\mathfrak{p}}(\sqrt{5}),$$
$$v_{\mathfrak{p}}(F_{mn})=v_{\mathfrak{p}}(mn)+v_{\mathfrak{p}}(\log(\phi\overline{\phi}^{-1}))- v_{\mathfrak{p}}(\sqrt{5}).$$
Let $e$ be the ramification index of $\mathfrak{p}$. We get (II) because $v_{\mathfrak{p}}=ev_p$.
I don't know enough about ''$\mathfrak{p}$-adic log'', so I don't know if similar proposition for Pell numbers can be done in the same way.
Please let me know if there is no problem with the following intuitive transformation of formulas, and if there is a problem, how to fix it?
Pell number $P_n$ is given by
$$\phi=1+\sqrt{2},\ \psi=1-\sqrt{2},$$
$$P_n=\dfrac{\phi^n-\psi^n}{2\sqrt{2}}.$$
Let $\mathcal{O}$ be the valuation ring of $K=\mathbb{Q}_p(\sqrt{2})$ and $\mathfrak{p}$ be the maximal ideal of $\mathcal{O}$ , can $v_p$ be extended to $v_{\mathfrak{p}}$?
$\sqrt{2}$ is the square root of $2$ of $K$.
$v_{\mathfrak{p}}(P_n)=v_{\mathfrak{p}}(\phi^n-\psi^n)-v_{\mathfrak{p}}(2\sqrt{2})$
$=v_{\mathfrak{p}}(\phi^n\psi^{-n}-1)-v_{\mathfrak{p}}(2\sqrt{2})$ ($\psi\in\mathcal{O}^{\times}$ correct?)
$=v_{\mathfrak{p}}(\log(\phi^n\psi^{-n}))-v_{\mathfrak{p}} (2\sqrt{2})$
$=v_{\mathfrak{p}} (n\log(\phi\psi^{-1}))- v_{\mathfrak{p}} (2\sqrt{2})$
$=v_{\mathfrak{p}}(n)+v_{\mathfrak{p}}(n\log(\phi\psi^{-1}))-v_{\mathfrak{p}}(2\sqrt{2}) $
Similarly
$$v_{\mathfrak{p}}(P_{mn})=v_{\mathfrak{p}}(mn)+v_{\mathfrak{p}}(n\log(\phi\psi^{-1}))-v_ {\mathfrak{p}}(2\sqrt{2})$$
is correct?
If the above is correct we get
$$ v_{\mathfrak{p}}(P_{mn})=v_{\mathfrak{p}}(P_n)+ v_{\mathfrak{p}}(m).$$
Perhaps the situation changes depending on whether $p=2$ or $p\neq2$, but I don't have the knowledge to understand this.
Best Answer
Let me advertise the following version of the Lifting The Exponent Lemma. As setup, let $K$ be any characteristic zero field with a nonarchimedean $p$-adic valuation $v_p: K \rightarrow \mathbb R \cup \{\infty\}$ with $v_p(p)=1$ (i.e. the restriction of $v_p$ to $\mathbb Q$ is the standard $p$-adic valuation). Then for any two elements $x,y \in K$, and all $n \in \mathbb N$, the following holds:
A proof goes like this: Via multiplying through with $y^{-n}$ (note that the condition implies $v_p(x) = v_p(y)$), WLOG we are in the special case with $y=1$ i.e. $x = 1+r$ for some $r \in K$ with $v_p(r) > \dfrac{1}{p-1}$.
Then $x^n-y^n = (1+r)^n – 1 = \sum_{k=1}^n \binom{n}{k} r^k$ and it suffices to consider the two cases
Case A. $n=p$, where $v_p (\binom{p}{k} r^k) \begin{cases} =v_p(r) + 1 \qquad \quad \text{ for } k=1 \\ \ge kv_p(r) +1 \qquad \text{ for } 2 \le k \le p-1 \\ = pv_p(r) \qquad \qquad\text{ for }k=p \end{cases}$
By the assumption $v_p(r) > \dfrac{1}{p-1}$ the first one is smaller than all the others, hence $=v_p((1+r)^n – 1)$.
Case B. $\gcd(p,n) =1$, where $v_p (\binom{n}{k} r^k) \begin{cases} = v_p(r) \text{ for } k=1 \\ \ge kv_p(r) \text{ for } 2 \le k \le n \end{cases}$
and the assumption ($v_p(r) >0$ would suffice) again guarantees the first one is smaller than all the others, and hence $=v_p((1+r)^n – 1)$. $\square$
This implies both your assertions about the $P_n = \dfrac{\phi^n - \psi^n}{2\sqrt{2}}$, as follows.
First note that $v_p(\phi) = v_p(\psi) = 0$ for all $p$ (e.g. since $\phi, \psi$ are roots of $X^2+2X-1$), so we are in the "Special Case".
In case $p$ odd we have $v_p(2\sqrt{2}) = 0$, hence $v_p(P_n) = v_p(\phi^n-\psi^n)$ for all $n$. Now your assertion (II) is the lemma applied to $x=\phi^n, y=\psi^n$ (and $n=m$, sorry), noting that the hypothesis of that assertion is $v_p(P_n) \ge 1 >\dfrac{1}{p-1}$.
In case $p=2$ we have $v_2(2\sqrt{2}) = \frac32$, hence $0=v_2(P_1) = v_2(\phi-\psi) – v_2(2\sqrt{2})$ i.e. $v_2(\phi-\psi) = \frac32 > \dfrac{1}{2-1}$. So $$v_p(P_n) = v_p(\phi^n - \psi^n) - \frac32 \stackrel{LTE}= \frac32 + v_p(n) -\frac32 = v_p(n)$$ for all $n$.
Now, your proposed proof works just as well, and is more conceptual by using the $p$-adic logarithm. In fact, I recommended that line of thought here. However, you have to be careful about two things (which, furthermore, might interfere with each other):
Normalizing the extended valuation $v_{\mathfrak p}$, which is $e\cdot v_p$ for the ramification index of the field: That might cause some case distinctions which ultimately don't matter but make a proof cumbersome.
The two facts $$\log(x^n) = n \log(x)$$ and $$v_p(\log(x)) = v_p(x-1)$$ are not as trivial as they are beautiful. In fact, the first one does hold as long as $v_p(x) > 0$, but the second one is valid in general only if $v_p(x-1) > \dfrac{1}{p-1}$ (as it should, because otherwise why would that assumption be needed in our version of LTE). Cf. Corollary 8.6 and Theorem 8.7 in K. Conrad's superb summary of $p$-adic series. This ultimately relies on the same computations as the straightforward algebra above.
In fact, the number $\dfrac{1}{p-1}$ (which annoyingly is $<1$ for odd $p$ but $=1$ for $p=2$) occurs again and again in this theory, as it is the break point where the two operations "raising to the $p$-th power" and "multiplying with $p$" switch places (as to which one has more effect on the valuation).