Post by v***@chelnydom.rue=2
d=((p-1)*(q-1),e)=2
m=p*q-1
E(p*q-1)=1
D(1)=1
p*q-1 is not equal to 1.
This is cryptic (no pun intended) as far as I can see.
In most papers written about RSA, "p" represents a prime number, "q"
represents another prime number, "e" is the encryption exponent, and "d" is
the decryption exponent.
e is chosen such that it has no factors in common with (p-1)*(q-1)
...but, since p is prime, then p-1 is an even number, and
...since q is prime, then q-1 is an even number, therefore
...(p-1)*(q-1) = an even number and has 2 as one of its factors
Since e must have no factors in common with (p-1)*(q-1), then 2 cannot be a
factor of e. In other words, e must be an odd number.
Neil - Salem, MA USA