Revision history [back]

$3e\equiv 1\pmod5$ means that $e=5k+2$ for some $k$. Then $e$ having $n=71$ bits means $e\in[2^{n-1},2^n-1]$, which translates into $k\in \big[\lceil (2^{n-1}-2)/5\rceil, \lfloor (2^n-3)/5\rfloor\big]$. Hence, it's enough to generate such a random integer $k$, and then compute $e$ out of it:

n = 71
L = ceil((2^(n-1)-2)/5)
U = (2^n-3)//5
import random
e = random.randint(L,U)*5 + 2