# 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

2 years, 2 months 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 - 2 years, 2 months ago

sorry for all m then...

- 2 years, 2 months ago

Try $$a = 7, n = 7, m = 2$$.

Staff - 2 years, 2 months ago

Now it is true right ? @Calvin Lin

- 2 years, 2 months ago

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

- 2 years, 2 months 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 - 2 years, 2 months ago

- 2 years, 2 months ago

@Harsh Shrivastava you may try to prove this..

- 2 years, 2 months ago

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

- 2 years, 2 months ago