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. \( a = n , m \neq kn + 1 \). – Calvin Lin Staff · 1 year, 4 months ago

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– Chinmay Sangawadekar · 1 year, 4 months ago

sorry for all m then...Log in to reply

– Calvin Lin Staff · 1 year, 4 months ago

Try \( a = 7, n = 7, m = 2 \).Log in to reply

@Calvin Lin – Chinmay Sangawadekar · 1 year, 4 months ago

Now it is true right ?Log in to reply

– Chinmay Sangawadekar · 1 year, 4 months ago

Should it be then , anot equal to n ? also n> 7 ?Log in to reply

I currently do not understand the statement enough to comment if it is true or false. – Calvin Lin Staff · 1 year, 4 months ago

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– Chinmay Sangawadekar · 1 year, 4 months ago

Yes adding value of m-1 to n is what I wanted .... Please help me to rephrase...Log in to reply

@Harsh Shrivastava you may try to prove this.. – Chinmay Sangawadekar · 1 year, 4 months ago

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@Otto Bretscher Dont forget to see the other note and suggest.... – Chinmay Sangawadekar · 1 year, 4 months ago

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