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I'm facing with the following problem:

Find all $(h,k)$ such that $$2^h \equiv 1 ~(\text{mod}~ 3^k) ~~~~~~~~(1)$$ and $$2^h \geq 3^k+1 ~~~~~~~~(2).$$

I'm just able to prove that the $(1)$ holds for all $(h,k) = (2a, 1)$, where $a$ is an integer. Indeed:

$$2^{2a} \equiv 1 ~(\text{mod}~ 3^1) \Rightarrow 4^a \equiv 1 ~(\text{mod}~ 3) \Rightarrow $$ $$\Rightarrow 4^{a-1} + 3 \cdot 4^{a-1} \equiv 1 ~(\text{mod}~ 3) \Rightarrow 4^{a-1} \equiv 1 ~(\text{mod}~ 3) \Rightarrow $$ $$\Rightarrow 4^{a-2} + 3 \cdot 4^{a-2} \equiv 1 ~(\text{mod}~ 3) \Rightarrow 4^{a-2} \equiv 1 ~(\text{mod}~ 3) \Rightarrow $$ $$\ldots$$ $$\Rightarrow 4^0 + 3 \cdot 4^0 \equiv 1 ~(\text{mod}~ 3) \Rightarrow 4^0 = 1 \equiv 1 ~(\text{mod}~ 3)$$

In this case, the $(2)$ is always satisfied since:

$$2^{2a} = 4^a \geq 3^1 + 1 = 4.$$

How can I proceed in order to find other $(h,k)$ couples? Or maybe, are these couples the only ones that satisfy my problem?

the_candyman
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  • Pick $k = 2$. Find out when $2^h \equiv 1 \pmod{9}$. Pick $k = 3$. Find when $2^h \equiv 1 \pmod{27}$. Try to spot a pattern. – Daniel Fischer Oct 18 '14 at 13:16

1 Answers1

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HINT:

Let $2^h=1+a3^k$ where $(a,3)=1$

$(2^h)^2=(1+a3^k)^2=1+2a(3^k)+a(3^k)^2\not\equiv1\pmod{3^{k+1}}$

$(2^h)^3=(1+a3^k)^3=a^33^{3k}+3(a3^k)^2+3(a3^k)+1\equiv1\pmod{3^{k+1}}$ for $2k\ge k+1\iff k\ge1$

See also : Prove that , any primitive root $r$ of $p^n$ is also a primitive root of $p$

ord$_{(3^k)}2=h\implies$ ord $_{(3^{k+1})}2=3h$

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lab bhattacharjee
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