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if $3^k \equiv \ 1\pmod{2^n}$ then $2^{n-2} | k$. I got this Number Theory problem and I'm stuck. The solution has to be proven without the use of Hensel's Lemma, and might require the LTE lemma. But I'm not familiar with those tools.

Problemsolved
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    What's the LTE lemma? – Dylan Moreland Jun 20 '12 at 19:27
  • Perhaps some kind of "L"ifting lemma? It would be related to Hensel's. – Patrick Da Silva Jun 20 '12 at 19:32
  • This is a nice theorem though. It would say that the order of $3$ modulo $2^n$ is always $2^{n-2}$ for $n > 2$, since the order of $3$ divides $\phi(2^n) = 2^{n-1}$, and since there is no primitive root modulo $2^n$, the order of $3$ can't be $2^{n-1}$. – Patrick Da Silva Jun 20 '12 at 19:34
  • @Pippo : Note that you can't use the LTE lemma here, at least not directly : the prime is $p=2$ and $4 , \nmid , 3-1$. – Patrick Da Silva Jun 20 '12 at 19:48
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    Ah, there's an article on AoPS that explains the name. I use Hensel's lemma quite a bit and had never heard this abbreviation before, so I don't think it's wise to assume that people know it. – Dylan Moreland Jun 20 '12 at 20:04
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    This is a standard simple proof, e.g. see Theorem $4.2'\ p. 43$ in Ireland and Rosen, where they do it for $5$ vs. $3$, but the proof is the same. The arithmetic amounts to the same as that in Andre's answer, but there is more conceptual background in Ireland and Rosen (see also the LTE link in Dylan's comment). These ideas will become clearer once you learn about Hensel lifting. – Bill Dubuque Jun 20 '12 at 20:11
  • A near duplicate of this question. The answers given there work here as well with the minor exception of $n=2$. – Jyrki Lahtonen Jun 20 '12 at 20:57

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Make a separate calculation for $n=2$ and $n=3$.

We will prove that if $n \ge 4$, then $$3^{2^{n-3}}\equiv 1+2^{n-1} \pmod{2^n}.\tag{$1$}$$
This implies that the order of $3$ modulo $2^n$ is greater than $2^{n-3}$. Since the order of $3$ divides $\varphi(2^n)$, it is at least $2^{n-2}$. And since $3^k\equiv 1\pmod{2^n}$, the order of $3$ modulo $2^n$ must divide $k$. It follows that $2^{n-2}$ divides $k$.

The proof of $(1)$ is by induction on $n$. The base case $n=4$ is easy to verify, since $3^{2^1}\equiv 1+8\pmod{16}$.

Suppose now that $3^{2^{n-3}}\equiv 1+2^{n-1} \pmod{2^n}$. So $$3^{2^{n-3}}=1+2^{n-1}+q2^n$$ for some integer $q$. Square both sides. On the left we get $3^{2^{n-2}}$. On the right, we get $1+2^n$ plus terms divisible by $2^{n+1}$. This completes the induction step.

André Nicolas
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