Is gcd(e,p−1)=1=gcd(e,q−1) similar to gcd(e,phi(n))=1?

Question

I wonder, is $\gcd(e,p−1)=1=\gcd(e,q−1)$ similar to $\gcd(e,\phi(n))=1$ ??

Answer 1

If $n=p\,q$, and $p$ and $q$ are prime, and $p\ne q$, then $\varphi(p\cdot q) = (p-1)(q-1)$, from which it follows that for any integer $e$, the propositions $\gcd(e,p-1)=1=\gcd(e,q-1)$ and $\gcd(e,\varphi(n))=1$ are equivalent.

Each of the conditions $n=p\,q$, $p$ and $q$ prime, and $p\ne q$, is necessary for the equivalence to hold for all $e$. These conditions are standard in RSA. For an example proving that $p\ne q$ is necessary, consider $p=q=e=3$: it holds $\gcd(e,p-1)=1=\gcd(e,q-1)$, but $\gcd(e,\varphi(9))=\gcd(e,6)=3$.

When generating an RSA key pair with a desired $e$ (which is common), $\gcd(e,p-1)=1=\gcd(e,q-1)$ is convenient, because it allows generating suitable $p$ and $q$ essentially independently. Also, when $e$ is prime (which is common: $e$ is typically a Fermat prime $F_i=2^{(2^i)}+1$ for some $i\in\{0,1,2,3,4\}\,$), this condition becomes $p\bmod e\ne1$ and $q\bmod e\ne1$, which are easy to check or ascertain by construction of $p$ and $q$.