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# Another Interesting fact

I noticed that :

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

$\text{is always true}$

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

You may also see Interesting fact

Note by Chinmay Sangawadekar
6 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. $$a = n , m \neq kn + 1$$. Staff · 6 months, 2 weeks ago

sorry for all m then... · 6 months, 2 weeks ago

Try $$a = 7, n = 7, m = 2$$. Staff · 6 months, 2 weeks ago

Now it is true right ? @Calvin Lin · 6 months, 2 weeks ago

Should it be then , anot equal to n ? also n> 7 ? · 6 months, 2 weeks ago

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. Staff · 6 months, 2 weeks ago

Yes adding value of m-1 to n is what I wanted .... Please help me to rephrase... · 6 months, 2 weeks ago

@Harsh Shrivastava you may try to prove this.. · 6 months, 2 weeks ago