|visits||member for||2 years, 3 months|
|seen||Jul 11 '14 at 0:52|
|bio||website||karelbilek.com||visits||member for||2 years, 3 months|
|location||seen||Jul 11 '14 at 0:52|
Why do we need in RSA the modulus to be product of 2 primes?
Oh.... so because the attacker would know $\phi(N)$, he would be able to deduce $d$ from $e$ because $de=1$ in mod $\phi(N)$. I think I am starting to get it.