# In RSA Encryption, Can the public exponent e > m (modulus) ? and can we choose any public key without paying attention to the conditions? [duplicate]

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)

• Hint: how does modular arithmetic over multiplicative group work? Nov 9, 2023 at 7:48
• $a^b \equiv a^{b \bmod \phi(n)} \pmod n$ by Euler's Theorem Nov 9, 2023 at 7:48
• 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.
– fgrieu
Nov 9, 2023 at 7:51
• 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.
– fgrieu
Nov 9, 2023 at 8:32