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In RSA ,the encryption, Can we choose the public exponent (e) greater than m (modulus) or e > φ(n) ? I wonder about choosing public key exponents (e) because the most information on the internet or articles has generated public keys from these 2 conditions :

  1. 1 < e < phi(n)

  2. e < m

So is it possible that we can choose to use e>m or e > phi(n) for the publickey encryption. And If e > m will there be any problems ? (other effects than make the calculation slow)

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    $\begingroup$ How does this differ from your previous question $\endgroup$
    – kelalaka
    Commented Nov 10, 2023 at 5:23
  • $\begingroup$ Again: the material linked as duplicate tells that yes, you can choose $e$ as large as you want, as long as: (1) You don't care for interoperability with standard implementations of RSA which assume an upper bound for $e$, like $e<n$ or $e<2^{256}$. (2) Slowness is not an issue. (3) $\gcd(e,p-1)=1=\gcd(e,q-1)$ holds. (4) $e$ is not chosen in a manner effectively helping towards the factorization of $n$, like it would be for $e$ close to $p^7$, and more generally for many ways to choose $e$ as a function of $p$ and $q$ beyond a check that $\gcd(e,p-1)=1=\gcd(e,q-1)$. $\endgroup$
    – fgrieu ♦
    Commented Nov 10, 2023 at 8:43
  • $\begingroup$ Sorry Could I asked , Is gcd(e,p−1)=1=gcd(e,q−1) similar to gcd(e,phi(n))=1 ?? $\endgroup$
    – Nicha59
    Commented Nov 12, 2023 at 9:21

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