Re: RSA: more than one secret exponent d exists ???


bert wrote:
If n = p*q, phi(n) = (p-1)*(q-1) as I'm sure you already
know, but take lambda(n) = phi(n)/hcf(p-1, q-1), and
d = e^(-1) mod(lambda(n)). Then there at least two
exponents, d and d + lambda(n), which correctly
decrypt every message. In fact there are precisely
hcf(p-1, q-1) such exponents less than phi(n), only
one of which is e^(-1) mod(phi(n)).

very nice solution !! thanks !!
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