I noticed that :

\[\Large{\boxed{a^{n^{n^{...^{\text{m times}}}}} \equiv a^{n^{n^{...^{\text{m-1 times}}}}} \mod (n+m-1)}}\]

\[\text{is always true}\]

\(\text{for all}\) \(m\) , \(\text{and}\) \(n\) \(\text{is a prime}\) \(> 7\)

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## Comments

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TopNewestDo you mean "For some m" or do you mean "For every m"?

In the former, that is obvious since the sequence is eventually repeating.

For the latter, that isn't true. E.g. \( a = n , m \neq kn + 1 \).

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sorry for all m then...

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Try \( a = 7, n = 7, m = 2 \).

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@Calvin Lin

Now it is true right ?Log in to reply

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I currently do not understand the statement enough to comment if it is true or false.

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@Harsh Shrivastava you may try to prove this..

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@Otto Bretscher Dont forget to see the other note and suggest....

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