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Another Interesting fact

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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Note by Chinmay Sangawadekar
8 months, 3 weeks ago

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Do 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 · 8 months, 2 weeks ago

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@Calvin Lin sorry for all m then... Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Chinmay Sangawadekar Try \( a = 7, n = 7, m = 2 \). Calvin Lin Staff · 8 months, 2 weeks ago

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@Calvin Lin Now it is true right ? @Calvin Lin Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Calvin Lin Should it be then , anot equal to n ? also n> 7 ? Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Chinmay Sangawadekar Do you mean \( \mod { n + m-1} \)? If not, the problem would be interpreted as adding \(m-1 \) to the value of it mod \(n\).

I currently do not understand the statement enough to comment if it is true or false. Calvin Lin Staff · 8 months, 2 weeks ago

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@Calvin Lin Yes adding value of m-1 to n is what I wanted .... Please help me to rephrase... Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Harsh Shrivastava you may try to prove this.. Chinmay Sangawadekar · 8 months, 2 weeks ago

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@Otto Bretscher Dont forget to see the other note and suggest.... Chinmay Sangawadekar · 8 months, 2 weeks ago

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