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Old 06-05-2016, 17:58
dila dila is offline
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In textbook RSA, you compute phi(n) from (p-1)(q-1), but that formula does not work if the modulus contains repeated prime factors. Here we have n = 2^64, so there are repeated 2's.

The requirement for modular exponentiation to be reversible is that gcd(e, phi(n)) = 1. If that turns out to be true, then a private exponent exists, which can "decrypt" the magic constant into the value of key.

@UniSoft: How does your solving code work??
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