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Re: Lucky primes & omlets on my face...
> > Recall: x^p = x mod p therefore, x^(p-1) = 1 mod p. So what we need is:
> > (x^e)^d = x^ed = x^(p-1)*i+1 = x mod p.
> This would only be true for prime p, but with RSA we are dealing with
> composite moduli. What we want is ed=1 mod phi(n), where
> phi(n)=(p-1)(q-1). (Actually you want to use (p-1)(q-1)/gcd((p-1),(q-1)).
> I forget what that is called.)
"Least common multiple," or LAMBDA(n).
finger for PGP public key
D4 99 54 2A 98 B1 48 0C CF 95 A5 B0 6E E0 1E 1D