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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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  • $\begingroup$ Hint: how does modular arithmetic over multiplicative group work? $\endgroup$
    – DannyNiu
    Commented Nov 9, 2023 at 7:48
  • $\begingroup$ $a^b \equiv a^{b \bmod \phi(n)} \pmod n $ by Euler's Theorem $\endgroup$
    – kelalaka
    Commented Nov 9, 2023 at 7:48
  • $\begingroup$ See e.g. this question and this question. In a nutshell: Mathematically we can choose $e$ as large as we want. Standards mandate various upper limit, with the most common being $e<n$, where $n$ is the public modulus, or $e<2^{256}$. Large $e$ (say $>2^{17}$) seems useful mostly as a deliberate way to slow down encryption. Large $e$ to allow small $d$ is not recommendable for security reasons. $\endgroup$
    – fgrieu ♦
    Commented Nov 9, 2023 at 7:51
  • $\begingroup$ Following edit of the question: $e≥n$ is incompatible with many deployed implementations. $e≥2^{256}$, with some. $e≥2^{32}$, with a few. $e≥2^{17}$ is unusual. Choosing $e$ with knowledge of factors of $n$ or knowledge of $φ(n)$ can lead to security issues, if that's beyond insuring $\gcd(φ(n),e)=1$. I think all this is addressed in answers to the two questions linked as duplicate. If supported by implementations, it's OK mathematically and security-wise to choose $e$ randomly in $[3,n^{100})$ and insure $\gcd(φ(n),e)=1$. If something is unclear, you are welcome to edit the question again. $\endgroup$
    – fgrieu ♦
    Commented Nov 9, 2023 at 8:32

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