Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$.

Question

Prove or disprove that there isn't any positive integer $n>7$ such that the linear congruence below is true.

$ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $

score 3 · Accepted Answer · edited Mar 02 '15 at 07:36

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Consider $2^{3^4}$, and it's easy to show $1\equiv2^{3^4}=2^{81} \pmod{511}$. Therefore, take $n=511$, $2^{3^{4^{.^{.^{.^{n-1}}}}}}-1$ is apparently a multiple of $2^{3^4}-1$ since $3^{4^{.^{.^{.^{n-1}}}}}$ is divisible by $3^4$ and $$2^{3^{4^{.^{.^{.^{510}}}}}}\equiv1 \pmod{511}$$ holds.

edited Mar 02 '15 at 07:36

rubik

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answered Mar 02 '15 at 07:30

Wei Zhan

740

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Note that if you don't want to do the calculation with $511$ you could just use $n=2^{81}-1$ instead since $2^{81}\equiv1$ is obvious that way. – Mario Carneiro Mar 02 '15 at 07:37
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@MarioCarneiro Right. Also $511$ is an obvious factor since it is $2^9-1$. – Wei Zhan Mar 02 '15 at 07:38
T. H. A. N. K. Y. O. U. – GohP.iHan Mar 02 '15 at 14:40

score -1 · Answer 2 · answered Mar 02 '15 at 07:35

-1

Note that $2$ mod $7$ has order $3$ (since $2^2 \equiv 4 \bmod 7$ and $2^3 \equiv 1 \bmod 7$).

So $2^{3^m} \equiv 1 \bmod 7$ for any $m\geq 1$.

answered Mar 02 '15 at 07:35

fretty

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-1 The modulus is not fixed at $7$. – Mario Carneiro Mar 02 '15 at 07:38
Oh my bad...I was reading the title of the question! – fretty Mar 02 '15 at 07:41
The title has been edited! – fretty Mar 02 '15 at 07:42
Yes, admittedly it was a bit confusing. Now it's better. – rubik Mar 02 '15 at 07:43

Can $ 2^{3^{4^{.^{.^{.^{n-1}}}}}}\equiv 1 \bmod {n} $ for some $n>7$.

2 Answers2