1

Let $n,k,l\in\mathbb{N}_{\geq 1}$. I am interested in knowing that is known about the congruence of $2^k$ mod $3^n$, and specifically what sufficient conditions for

$$2^k \equiv 2^l \mod 3^n$$

other than having $k\equiv l \mod \phi(3^n)$ for the Euler's Totient function (see this Wikipedia article). Actually in this case to my knowledge $\phi(3^n) = 3^n - 3^{n-1} = 2\cdot 3^{n-1}$ (since there are $3^{n-1}$ multiples of $3$ between $1$ and $3^n$). I am asking this question on the basis of the interesting article: https://proofwiki.org/wiki/Congruence_Modulo_3_of_Power_of_2 in ProofWiki. I don't have much background in number theory/modular arithmetic, so I can't answer this question myself.

Bill Dubuque
  • 282,220
Epsilon Away
  • 1,070
  • 1
    The condition is necessary and sufficient. See https://math.stackexchange.com/questions/594782/2-is-a-primitive-root-mod-3h-for-any-positive-integer-h – lhf May 03 '24 at 15:26
  • @lhf Could you explain why the condition is necessary? Sufficient part is covered by the Wikipedia article. So we know that $2$ is primitive root of every $3^n$ and so by definition for any coprime $a$ w.r.t. $3^n$, e.g. $a = 2^k, k \geq 1$, we know that $2^l \equiv 2^k \mod 3^n$ for some $l$. At least I don't see from this why necessarily $l \equiv k \mod \phi(3^n)$, since in principle $l$ could be $k$. – Epsilon Away May 03 '24 at 15:40

2 Answers2

2

In general: when $\gcd(a,q)=1$, set $m$ to be the multiplicative order of $a$ modulo $q$. Then the congruence $a^k\equiv a^\ell\pmod q$ is equivalent to the congruence $k\equiv \ell\pmod m$. The equivalence can be quickly proved from the definition of multiplicative order.

In this special case, all that remains is to verify that $2$ has order $\phi(3^n)$ modulo $3^n$ for every $n\ge1$. (lhf has pointed out that this was done in another question on this site)

Greg Martin
  • 92,241
0

Necessity follows from the fact that $2$ is primitive.

If not, we get $1\equiv 2^{k-l}\equiv 2^{(k-l)\pmod {\varphi (3^n)}}\pmod {3^n},$ with $0\lt (k-l)\pmod {\varphi (3^n)}\lt\varphi(3^n),$ which is a contradiction.

suckling pig
  • 1