How does openssh actually choose the public exponent 'e' when generating a RSA key?

Question

What I've read:

Seems in practice, will first choose an e, then choose the 2 big primes p1 and p2.

What I've learned so far:

e is usually between (2^16, 2^32), e.g 2^16 + 1 = 65537. // TODO .. not sure
e should not be too large, otherwise the public-key operation is slow.

I've also try to read openssh source code, but can't figure out it yet.

Questions

How openssh actually choose e, when creating a RSA key?
If possible, can u point out relevant file/functions in the source code.
If a software choose a fixed e (e.g 65537) instead of a random value, will that make the RSA key less secure?

yyyyyyy · Accepted Answer · 2023-01-04 10:51:22Z

4

$e=65537$.
See here and here.
No, using fixed $e$ is the industry standard, and $e=65537$ is a particularly popular choice.

answered Jan 4, 2023 at 10:51

yyyyyyy

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3

$\begingroup$ Addition: for reasons why $e=65537$ is more popular than $e=3$, see this. Notice that the choice of $p$ and $q$ must be compatible with $e$, if that's chosen and frozen first. The reason why sometime it's enforced $e<2^{32}$ is that's there's no much reason to increase $e$ beyond $65537$, and an historical limitation of a Windows API. $\endgroup$
– fgrieu ♦
Commented Jan 4, 2023 at 10:58

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