- #1

amcavoy

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Thanks for your help.

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- Thread starter amcavoy
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- #1

amcavoy

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Thanks for your help.

- #2

Muzza

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Let a be invertible mod p.

Consider the p - 1 numbers

1a, 2a, ..., (p - 1)a.

Since a was invertible, they are all distinct mod p. So we have p - 1 numbers which are distinct mod p, so they must be congruent to 1, 2, 3, ..., p - 1 in some order.

Thus

(1a)(2a)...((p - 1)a) = 1*2*...*(p - 1) (mod p)

<=>

(p - 1)! * a^(p - 1) = (p - 1)! (mod p).

- #3

amcavoy

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Since a was invertible, they are all distinct mod p. So we have p - 1 numbers which are distinct mod p, so they must be congruent to 1, 2, 3, ..., p - 1 in some order.

Can you explain this? I understand the rest of the proof except for this part.

Thanks again.

While I'm at it: I have asked this before, but received responses with suggestions for

Last edited:

- #4

robert Ihnot

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- #5

matt grime

Science Advisor

Homework Helper

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apmcavoy said:It seems like it would be true if the (mod p) was instead (mod a).

if it were mod a then the RHS would be identically zero, wouldn't it?

- #6

amcavoy

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Yes, right. I have it now thanks.

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