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TL;DR How to prove the eight congruences at the end of this post?

Remark. My number theory is rusty and I'm trying to prove the following observations.

Motivation: This result easily implies that $3^n$ does not divide $8^n+1$ for $n\geq 4$, for example, as shown in my suggested answer on the linked question.

Definition. $\mathbb v_p(a)$ = The $p$-adic order of number $a$ is the number of times $p$ divides $a$.

When I start observing congruence classes, it is easy for me to see the initial observations:

$$\begin{array}{} \mathbf v_3(2^n-1) =\begin{cases} \mathbf v_3(n) + 1, & n\text{ even}\\0, & n\text{ odd} \end{cases} \\ \mathbf v_3(2^n+1) =\begin{cases}0, & n\text{ even}\\ \mathbf v_3(n) + 1, & n\text{ odd} \end{cases} \end{array}$$

In other words, I've actually observed that:

$$\begin{array}{} \mathbb v_3(2^n-1)=\begin{cases} 0, & n\equiv1\pmod{2}\\ 1, & n\equiv2,4\pmod{6}\\ 2, & n\equiv6,12\pmod{18}\\ 3, & n\equiv18,36\pmod{54}\\ \dots\\ k, & n\equiv2\cdot3^{k-1},4\cdot 3^{k-1}\pmod{2\cdot 3^{k}}\\ \dots \end{cases} \\ \mathbb v_3(2^n+1)=\begin{cases} 0, & n\equiv0\pmod{2}\\ 1, & n\equiv1,5\pmod{6}\\ 2, & n\equiv3,15\pmod{18}\\ 3, & n\equiv9,45\pmod{54}\\ \dots\\ k, & n\equiv1\cdot3^{k-1},5\cdot 3^{k-1}\pmod{2\cdot 3^{k}}\\ \dots \end{cases} \end{array}$$

Where notice that when we realize $k-1 = \mathbf v_3(n)$, the initial observations follow.

My question is, how would we formally and rigorously prove these observations?

Below is a possible starting point for the rigorous proof:

It seems to me like a starting point would be to notice that by definition:

$\space\space\space\space$ 1) $k\le\mathbf v_3(2^n\mp1)$ $\iff$ $2^n \pm1\equiv 0\pmod{3^{k}}$

$\space\space\space\space$ 2) $\mathbf v_3(2^n\mp1)\le k$ $\iff$ ${2^n}\pm1\not\equiv 0\pmod{3^{k+1}}$

Where depending on "$\mp$" we have:

$\space\space\space\space$ a) $ n\equiv2\cdot 3^{k-1}\pmod{2\cdot 3^{k}}$ or $ n\equiv4\cdot3^{k-1}\pmod{2\cdot 3^{k}}$

$\space\space\space\space$ b) $ n\equiv1\cdot 3^{k-1}\pmod{2\cdot 3^{k}}$ or $n\equiv5\cdot3^{k-1}\pmod{2\cdot 3^{k}}$

Now we need to prove that:

  • a) implies the RHS of 1) and RHS of 2) for "$-$"

  • b) implies the RHS of 1) and RHS of 2) for "$+$"

After proving that, we would have the implications: LHS $\iff$ RHS.

Finally, LHS of 1) combined with LHS of 2) would imply the observations.

To state the implications:

We can use a corollary of Euler's Theorem to obtain the implications:

a) implies the RHS of 1) and RHS of 2) for "$-$":

$$\begin{array}{} n \equiv 2\cdot3^{k-1} \quad(\bmod 2\cdot3^{k}) &\implies 2^{n} \equiv 2^{2\cdot3^{k-1}} \quad(\bmod 3^{k+1}) \\ n \equiv 4\cdot3^{k-1} \quad(\bmod 2\cdot3^{k}) &\implies 2^{n} \equiv 2^{4\cdot3^{k-1}} \quad(\bmod 3^{k+1}) \end{array}$$

b) implies the RHS of 1) and RHS of 2) for "$+$":

$$\begin{array}{} n \equiv 1\cdot3^{k-1} \quad(\bmod 2\cdot3^{k}) &\implies 2^{n} \equiv 2^{1\cdot3^{k-1}} \quad(\bmod 3^{k+1}) \\ n \equiv 5\cdot3^{k-1} \quad(\bmod 2\cdot3^{k}) &\implies 2^{n} \equiv 2^{5\cdot3^{k-1}} \quad(\bmod 3^{k+1}) \end{array}$$

To finish the proof, we have to prove:

$$\begin{array}{} 2^{2\cdot3^{k-1}} \not\equiv +1\quad(\bmod 3^{k+1}) \\ 2^{4\cdot3^{k-1}} \not\equiv +1 \quad(\bmod 3^{k+1}) \end{array}$$

$$\begin{array}{} 2^{1\cdot3^{k-1}} \not\equiv -1 \quad(\bmod 3^{k+1}) \\ 2^{5\cdot3^{k-1}} \not\equiv -1 \quad(\bmod 3^{k+1}) \end{array}$$

$$\begin{array}{} 2^{2\cdot3^{k-1}} \equiv +1\quad(\bmod 3^{k}) \\ 2^{4\cdot3^{k-1}} \equiv +1 \quad(\bmod 3^{k}) \end{array}$$

$$\begin{array}{} 2^{1\cdot3^{k-1}} \equiv -1 \quad(\bmod 3^{k}) \\ 2^{5\cdot3^{k-1}} \equiv -1 \quad(\bmod 3^{k}) \end{array}$$

But I'm not sure how to prove these eight congruences.

Vepir
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  • I think all this this follows quickly from the properties of the $p$-adic logarithm (for $p=3$) $log: (1+p\mathbb Z_p, \cdot) \simeq (\mathbb Z_p,+)$ and just observing in which filtration step of the $3$-adic unit group small powers of $2$ lie, but maybe you don't want to use such a big hammer? – Torsten Schoeneberg Apr 17 '20 at 02:27
  • Actually, one does not need the logarithm, just "well-known" results about how the $p$-power map operates on the higher principal unit groups. – Torsten Schoeneberg Apr 17 '20 at 03:06
  • @TorstenSchoeneberg I'm actually not familiar with p-adic number theory, but I did once read something little about it somewhere and remembered that my observation on these congruences actually represents the p-adic order. If you have any suggestions on reading material that would be cool. If not, still thank you for your comment for giving me some more context. – Vepir Apr 17 '20 at 11:09
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    Naturally, books on p-adic numbers contain discussions of this. Good short treatments might be in sources on local class field theory (where it's a preliminary; the goal is then to relate the filtration of the unit group to filtrations of Galois extensions), I remember a succinct one in Neukirch's "Class Field Theory". A decent free online source is https://www.maths.nottingham.ac.uk/plp/pmzibf/book/vol.pdf, chapter I sec. 5 and 6. Raising to the $p$-th power is discussed in 5.6 et seq., all far more general and hence a little abstract compared to what we need here. – Torsten Schoeneberg Apr 17 '20 at 16:37

2 Answers2

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Equivalences for Induction

Since $2^6\equiv1\pmod9$, we have $$ \begin{align} 2^{6m+1}+1&\equiv3\pmod9\tag1\\ 2^{6m+2}-1&\equiv3\pmod9\tag2\\ 2^{6m+4}-1&\equiv6\pmod9\tag3\\ 2^{6m+5}+1&\equiv6\pmod9\tag4\\ \end{align} $$

If $n$ is even, $n\in\{0,2,4\}\pmod6$, $2^{2n}+2^n+1\equiv3\pmod9$ and $$ \left(2^{2n}+2^n+1\right)\left(2^n-1\right)=\left(2^{3n}-1\right)\tag5 $$ If $n$ is odd, $n\in\{1,3,5\}\pmod6$, $2^{2n}-2^n+1\equiv3\pmod9$ and $$ \left(2^{2n}-2^n+1\right)\left(2^n+1\right)=\left(2^{3n}+1\right)\tag6 $$

Applying the Equivalences

Let $P(k)$ be the statement that $v_3(n)=k\implies v_3\!\left(2^n-(-1)^n\right)=k+1$.

$P(0)$ is verified by $(1)-(4)$.

$P(k)\implies P(k+1)$ is verified by $(5)-(6)$.

Therefore, $$ \bbox[5px,border:2px solid #C0A000]{v_3\!\left(2^n-(-1)^n\right)=v_3(n)+1}\tag7 $$

robjohn
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  • Thank you for your idea. But, I'm trying to directly prove $2^n\pm1$ is divisible by $3^k$ and is not divisible by $3^{k+1}$, which would directly imply $\mathbf v_3(2^n\pm1)=k=\mathbf v_3(n)+1$ for corresponding parity of $n$. That is, I've edited the post to hopefully make it clear that I want to know how to prove the eight congruences at the end of my post, if possible. – Vepir Apr 16 '20 at 21:59
  • I have added more of the induction. – robjohn Apr 17 '20 at 01:26
  • Ah I see it the induction is indeed a simpler and nicer proof than what I tried. I was hoping to learn how would one directly evaluate the eight congruences that appear in my brute force proof, but if that isn't given, I'll eventually accept your answer. – Vepir Apr 17 '20 at 10:59
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More generally,

$$ (p+(-1))^n-(-1)^n = \sum_{k=0}^{n-1}\binom{n}{k}(-1)^k p^{n-k} $$

and now note that the term for $k=n-1$ on the RHS has $p$-adic valuation $v_p(n\cdot p) = v_p(n)+1$, whereas for odd (!) primes $p$, all terms with $k \le n-2$ have strictly higher valuation. Consequently,

$$v_p \left[(p-1)^n - (-1)^n\right]=v_p(n)+1$$ for all odd primes $p$ and $n \ge 1$. Yours is the case $p=3$.

Torsten Schoeneberg
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    This does now remind me of LTE lemma that I've just now found out about. I appreciate your insight. – Vepir Apr 17 '20 at 11:22
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    (+1) The Lemma from this answer shows that for any prime $p\gt2$ and for $j\ge2$, $p^k,|,n\implies\left.p^{k-j+2},\middle|,\binom{n}{j}\right.$, which means that $\left.p^{k+2},\middle|,\binom{n}{j}p^j\right.$. This validates your claim regarding the valuation of the extra terms. – robjohn Apr 17 '20 at 12:22
  • Vepir, yes, this is a special case of the LTE lemma. @robjohn: Thanks for amending a validation for that. I got the idea for this answer from the final rephrasing of the statement in your answer, which I also upvoted. – Torsten Schoeneberg Apr 17 '20 at 16:28
  • @TorstenSchoeneberg: yes, writing it that way, I was thinking of adding an answer using the Binomial Theorem, but you wrote yours first. This is nice in its generality. – robjohn Apr 17 '20 at 17:24