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The 1991 USAMO Problem 3 offers a concise proof-by-contradiction that if $n$ is any positive integer, then the sequence $2, 2^2, 2^{2^2}, 2^{2^{2^2}}, \dots$ is eventually constant $\pmod n$. However, I don't fully understand the following step in that proof:

Claim: If the sequence $2, 2^2, 2^{2^2}, 2^{2^{2^2}},\dots$ is not eventually constant $\pmod n$, then the sequence of its exponents $1, 2, 2^2, 2^{2^2},\dots$ is not eventually constant $\pmod k$, where $k$ is the eventual period of $2^0,2^1,2^2,2^3,\dots\pmod n$.

Equivalently: If $1, 2, 2^2, 2^{2^2},\dots$ is eventually constant $\pmod k,$ then $2, 2^2, 2^{2^2}, 2^{2^{2^2}},\dots$ is eventually constant $\pmod n.$

Equivalently: If $2, 2^2, 2^{2^2}, 2^{2^{2^2}},\dots$ is eventually constant $\pmod k,$ then this same sequence is eventually constant $\pmod n.$

I'm sorry if this is something obvious/elementary, but would someone please explain why this is true when $n\gt 1?$ How exactly does the claim then depend on $k$ being the eventual period of $2^0,2^1,2^2,2^3,\dots\pmod n?$

r.e.s.
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  • If the exponents $e_i$ are eventually all of the form $km_i+r$ for fixed $r$ and some $m_i$, then $2^{e_i}=2^{km_i+r}=2^{k(m_i-1)+r}=...=2^{r}\pmod{n}$, where each equation is using that $k$ is the period modulo $n$. – plop May 10 '21 at 14:57
  • What I find confusing is whether the proof shows why such a $k$ would exist in the first place? If it is shown to exist, wouldn't that conclude the proof in itself since $2, 2^2, 2^{2^2}...$ is a subsequence of $2^0, 2^1, 2^2...$? – smalldog May 10 '21 at 15:02
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    @acephalous It's simply asserted that for all integers $b>1$, the sequence $2^0, 2^1, 2^2, \dots$ is eventually cyclic $\pmod b$, with $k$ defined as the corresponding period. (Of course a subsequence of such a periodic sequence need not be constant.) – r.e.s. May 10 '21 at 15:20
  • @plop Thank you -- I think that's all there is to it! We have $2^{km_i+r}=2^{k(m_i-1)+r}$, etc., just because $k$ is the eventual period of $2^0, 2^1, 2^2,2^3,\ldots\pmod n\ \ $ (don't know why I couldn't see that). – r.e.s. May 10 '21 at 17:07
  • I just discovered this may be a duplicate of https://math.stackexchange.com/q/1160795/16397 (voting to close) – r.e.s. May 10 '21 at 21:55

1 Answers1

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The power towers might distract your attention from what is really happening, which is just this:

If $a_1, a_2, a_3, \ldots$ is strictly increasing and eventually constant modulo $k$, then $2^{a_1}, 2^{a_2}, 2^{a_3},\ldots$ is eventually constant modulo $n$.

Proof. By definition $k$ has the property that $(2^i \bmod n)_i$ eventually has period $k$, which is the same as saying that if $A$ is large enough we will have $2^A \equiv 2^{A+k} \pmod n$.

But that means that if we go far enough in the increasing sequence $a_i$ we eventually reach a point where $a_i$ is large enough to this for happen. And if we go far enough for this to happen and for $(a_i)$ to become constant modulo $k$, we will forever have $a_{i+1}=a_i+jk$ for some $j$, and therefore $$ 2^{a_i} \equiv 2^{a_i+k} \equiv 2^{a_i+2k} \equiv \cdots \equiv 2^{a_i+jk} = 2^{a_{i+1}} \pmod n$$

Troposphere
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