From: "Bill Bouris" <leavemsg1@...>

[a^ (p-2)] is equiv. to [p - (p-1)/a] (mod p)

Interesting? or not?

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It's trivial:

a^(p-1) = = 1

a^(p-2)*a = = 1

a^(p-2)*a = = p-p+1

a^(p-2)*a = = p-(p-1)

a^(p-2) = = [p-(p-1)]/a

a^(p-2) = = p/a-(p-1)/a

And we have that GCD(a,p)=1

p/a = = x (mod p)

a*x = = p = = 0 (mod p)

x = = 0 (mod p)

p/a = = 0 = = p (mod p).

So:

a^(p-2) = = p-(p-1)/a.

QED

Jose Brox