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??
