Show that the order of 5 mod $2^k$ is $2^{k-2}$

Asked Oct 16 '13 at 11:16

Active Oct 16 '13 at 11:16

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I have that $5^{2^{k-2}}\equiv 1 \mod 2^{k}$ but I am unsure how to show that $2^{k-2}$ is the least integer that has this property.

I thought perhaps I could show that $2^{k-2}|\operatorname{ord}_{2^k}(5)$ since I already have (from above) $\operatorname{ord}_{2^k}(5)|2^{k-2}$ but I have been stuck with this.

Any help appreciated :)

asked Oct 16 '13 at 11:16

hotstuff69

Show inductively that $5^{2^e} \equiv 1 + 2^{e+2} \pmod{2^{e+3}}$. – Daniel Fischer Oct 16 '13 at 11:18
http://math.stanford.edu/~ksound/Math152A10/Solution3.pdf – lab bhattacharjee Oct 16 '13 at 11:37

Show that the order of 5 mod $2^k$ is $2^{k-2}$

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