Determine all k ∈ N such that $n^{k}$ ≡ n (mod 7) for all integers n.

Asked Jul 25 '22 at 17:44

Active Jul 25 '22 at 18:02

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How would I prove this statement? This is what I have tried, but it seems to be incorrect:

writing the congruence as 7 | n(1-$n^{k-1}$)
then, either 7 | n or 7 | 1-$n^{k-1}$, none of which cannot be true for all n regardless of the k value

I'm not sure where I went wrong, are there any other approaches I should try?

edited Jul 25 '22 at 18:02

Bill Dubuque

asked Jul 25 '22 at 17:44

user

Your second bullet point is either a typo or a misstep in logic. – David P Jul 25 '22 at 17:47
Specialize $,n=7,$ in the Theorem in the linked dupe. – Bill Dubuque Jul 25 '22 at 18:00
Why not just try small $k$ to see if you find a pattern? Certainly you only need to try $7$ values for $n$. – Ross Millikan Jul 25 '22 at 18:01

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