- #1
Firepanda
- 430
- 0
I started by saying
cince c = me mod n (1)
then me ≡ c mod n
Since this is true then
mek ≡ cek-1 mod n
and cek-1 = (me)ek-1 mod n, by (1)
= mek mod n
=> k is a generator of e
and so both are proved? Not sure..
Also I haven't a clue is it's dangerous or not for RSA.
Thanks for any help.