Okay, so the proof of the fact that the value of (p-1)! mod p is always p-1 is below:
If p is prime, the numbers {1,2,....p-1} form a group under multiplication modulo p. So for each element a in this group, there is an element b such that ab is congruent to 1 mod p.
If a is congruent to b mod p, then a^2 is congruent to 1 (mod p). This means that a^2 - 1 = (a+1)(a-1) = 0 (mod p). Since p is prime, this means a is congruent to 1 or -1 mod p (which means a = 1 or a = p-1).
Okay, so my question is, how did they get this part:
"If a is congruent to b mod p, then a^2 is congruent to 1 (mod p)."
2007-10-18
04:56:41
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