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Message boards : General discussion : Why does p divide 2^(p-1)-1?

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Message 147019 - Posted: 24 Dec 2020 | 7:17:02 UTC

This might sound like a strange question and I am not looking for a mathmatical proof. But maybe there is some more accessible way to show why there is a relation at all? It seems so arbitrary to me, which, of course, it actually isn't.
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Message 147024 - Posted: 24 Dec 2020 | 10:25:03 UTC - in response to Message 147019.

This might sound like a strange question and I am not looking for a mathmatical proof. But maybe there is some more accessible way to show why there is a relation at all? It seems so arbitrary to me, which, of course, it actually isn't.

Not sure what kind of answer you are after, when not a proof.

This result is known as Fermat's little theorem (Wikipedia). I suggest you read about it. What they call a, is 2 in your example, so the result is valid if 2 does not divide the prime p.

A modern way of understanding the result, is that the (nonzero) integers modulo p form a group with multiplication, and in that group, 2 is a member; and the order of a member divides the order of the group which is p-1 (Lagrange's theorem).

/JeppeSN

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Message boards : General discussion : Why does p divide 2^(p-1)-1?

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