So, Let's assume we have
n which is made up of 2 strong primes which cannot be factored &
e which is textbook value of e that is 65537.
Now, I have a message1, first I convert that into decimal and then do
(msg1^e)%n and this turns out to be exactly same as
random1 . Now, I generated a lot of messages and performed the encryption same way and I found one of the message (let's call it msg2) generates same output after encryption as it is in plain.
So, usually RSA formula is
c=pow(m,e,n) . In my case
pow(msg2,e,n) gives output exactly same as
msg2. That implies for an edge case of a message the encryption provides same ciphertext as plaintext.
Does this means, that this RSA system for
n is broken/buggy?