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

Chris
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Nicha59
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    Welcome to Cryptography.se What is the origin of this question? do you know that $\varphi(p\cdot q) = (p-1)(q-1)$ and from there what can you conclude? – kelalaka Nov 12 '23 at 12:19
  • Thank you for your clues and the origin question is from this link @fgrieu – Nicha59 Nov 12 '23 at 12:52

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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 $q$.

fgrieu
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  • Thank you so much , Your answer is very useful for me – Nicha59 Nov 12 '23 at 12:54
  • I'm still expecting proof for the first part. – kelalaka Nov 14 '23 at 20:52
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    @kelalaka: the proof uses that for all integers $e,u,v\in\mathbb N^*$, it holds $\gcd(e,u,v)=1$ $\iff$ $(\gcd(e,u)=1$ and $\gcd(e,v)=1)$ [which itself follows from the fact that prime divisors of $u,v$ are the union of prime divisors of $u$ and prime divisors of $v$]. That's instantiated with $u=p-1$, $v=q-1$, $u,v=\varphi(p,q)=\varphi(n)$. – fgrieu Nov 15 '23 at 07:21