Why RSA must make sure that $\gcd(e,\varphi(n)) = 1$ instead of $\gcd(e,n) = 1$?

Question

By Euler's theorem, if $\gcd(e,n) = 1$, then $e^{\varphi(n)} \equiv 1 \pmod n$. But why does RSA need to make sure that $\gcd(e,\varphi(n)) = 1$?

Answer 1

You want to be able to encrypt a message and decrypt it, that is translated to:

$m^{ed} \equiv m^1 \equiv m \pmod n$

But why does this work? Since $ed \equiv 1 \pmod{\varphi(n)}$

If $\gcd(e,d) \neq 1$ then $e$ and $d$ are not coprime then $ed \not \equiv 1 \pmod{\varphi(n)}$. So $e$ must be coprime with $\varphi(n)$ to have a modular multiplicative inverse.

Also $\gcd(e,n)=1$, since the public key is presented as $(e,n)$. Then you trivially can compute a factor of $n$ if $e$ is not coprime.