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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
1 year 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 · 1 year ago

sorry for all m then... · 1 year ago

Try $$a = 7, n = 7, m = 2$$. Staff · 1 year ago

Now it is true right ? @Calvin Lin · 1 year ago

Should it be then , anot equal to n ? also n> 7 ? · 1 year 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 · 1 year ago

Yes adding value of m-1 to n is what I wanted .... Please help me to rephrase... · 1 year ago

@Harsh Shrivastava you may try to prove this.. · 1 year ago