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

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 :)
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!
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.

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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?
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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