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Show that if $n\geq3$ is odd, then $2^n-1\equiv7\mod24$.

I tried solving this backwardly. We want to prove that $2^3(2^{n-3}-1)=2^n-2^3\equiv0\mod24$. Since $\frac{24}{2^3}=3$, this leaves us to prove that $3\mid2^{n-3}-1$. From here on I got stuck, although I feel like I am ignoring something very simple.

Once the result is proved, I need to find all $n\in\mathbb{Z}_{>0}$ for which $2^n-1$ divides $3^n-1$.

This problem is given in a chapter on the Legendre symbol, introducing the law of quadratic reciprocity.

Bill Dubuque
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bitsbit
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3 Answers3

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Remember $n$ is odd, so writing $n=2k+1$ gives $$2^{n-3}-1\equiv 2^{2k-2}-1\equiv (2^2)^{k-1}-1\equiv 1-1\equiv 0\pmod{3}$$

user10354138
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Hint $\,\ 2^{\overbrace{\large 3+2k}^{\LARGE n}}\bmod 24\, =\, 2^{\large 3}(\overbrace{\color{#c00}2^{\large\color{#c00} 2k}\!\bmod 3}^{\large \color{#c00}{2^{\LARGE 2}}\ \equiv\ 1})\, =\, 2^{\large 3}\ $

We used: $\,\ ab\bmod ac\, =\, a(b\bmod c)\,= $ mod Distributive Law to factor $\,a = 2^{\large 3}$ out of the mod.

Bill Dubuque
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  • Thank you for your answer, I now also learned a new law in modular arithmetic! – bitsbit Jun 21 '19 at 14:55
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    @bitsbit Great. MDL is usually the quickest way to do these type of problems (it yields an operational way to apply CRT, as I explain in the linked answer). The little time spent learning it will reap much greater time savings in future analogous calculations. – Bill Dubuque Jun 21 '19 at 15:16
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In one hand, since $n\geq 3$ then $8|2^n$, i.e. $2^n\equiv 0\bmod 8$. Hence $$2^n-1\equiv 7\mod 8$$

On the other hand, since $n$ is odd, we have $2^n\equiv(-1)^n=-1\bmod 3$. Hence, $2^n-1\equiv -2\bmod 3$. In particular $$2^n-1\equiv 7\mod 3.$$

So, we conclude $2^n-1\equiv 7\mod 24.$

Julian Mejia
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