Prove that: $$2^{n} \equiv 1 \pmod {9} \implies 2^{n} \equiv 1 \pmod {7}$$ Please a hint and a help
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No we did not study it – user315918 Apr 12 '16 at 16:33
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Note that $7.9 = 2^6 - 1$. – anomaly Apr 12 '16 at 16:49
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@anomaly That's a weird way of writing multiplication, looks more like a decimal marker to me. – Alice Ryhl Apr 12 '16 at 16:53
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@KristofferRyhl: It's very common in number theory. – anomaly Apr 12 '16 at 17:58
4 Answers
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Observe that $2^3\equiv-1\pmod9\implies$ord$_92=6\implies6\mid n$
If $n=6m,$ $$2^{6m}-1$$ is divisible by $$2^6-1$$ which is always divisible by $7$
lab bhattacharjee
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Raise 2 to successive powers and check against $2^n \equiv 1$ (mod 9). You will find the least such $n$ is ... (fill in with your answer). Then show that if $2^n \equiv 1$ (mod 9) then $n$ must be divisible by that "least such $n$." Finally, show that this implies $2^n\equiv 1$ (mod 7) as well.
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Put $n=6k+a, a \in \{0,1,..,5\}$. In this case,
$$2^n=2^{6k+a}=64^k.2^a \equiv 2^a \equiv 1 (\text{mod 9})$$
Imply that $a=0$. Then:
$$2^n=2^{6k}=64^k \equiv 1 (\text{mod 7}).$$
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