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I tried to find a proof for fermat's little theorem, any thoughts?

4 Answers

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Nice proof! Especially the first part flows very nicely. I had some trouble understanding the last part (after you introduce b, it wasn't clear to me what you were trying to do), but I figured it out after re-reading it a few times. I would try to rewrite that last part to make it a bit more rigorous: for example, can you deduce that n_1 - n_2 divides p-1 by writing down equations instead of appealing to the reader's intuition?

Anyway, great job producing a proof like this when you're still in high school :)
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Maybe I'm not reading this correctly, but it seems like there are two kinds of sequences invoked:

1. the sequence ( a^n ) for a given a,

2. the sequence ( b*a^n ) for the given a, b.

Are you sure you're not conflating these two? I don't really follow the reasoning at the end.

This looks like a great attempt either way!
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Well done!

When you have a multi-part proof it helps (at least the reader…) to introduce each part with a kind of announcement so it's clear what's to come next:

We will first show that [blah blah]

the proof, part one

.

.

Now we can prove that [blah blah]

following part of the proof

.

.

&c.

-----------------------

A little problem which could interest you in the context.

The period of the decimal development of 1/p, p a prime other than 2 or 5 (which divide the base 10), has of course a length at most p-1. But some reach that maximal length, and others not:

1/3 = 0.**3**333… has period length 1 < p-1 = 3-1 = 2

1/7 = 0.**142857**142857… has maximal period length p-1 = 6

1/11 = 0.0**90**90… has period length 2

1/13 = 0.0**769230**769230… has period length 6

For what primes is the period length maximal? What about other bases, 2 for example?
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For the part where you deduce that the sequence length, don't you also need to show that all the sequences contain an equal amount of representation of each number?

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