# Tag Info

Because RSA takes advantage of the fact that: $a^{\phi(n)} \equiv 1 \mod n$ Which means: $a^{\phi(n)+1} \equiv a \mod n$ So, in order to encrypt and decrypt using $e$ and $d$, we need: $ed \equiv 1 \mod \phi(n)$ Which can only be true if $gcd(e,\phi(n)) = gcd(d,\phi(n)) = 1$