What does it imply if in RSA pow(m, e, n) is same as the m So, Let's assume we have n which is made up of 2 strong primes which cannot be factored &amp; 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?
Not necessarily.  With the (not so strong) primes p=211 and q=311, we have n=65621.  Using e=157, I came up with the following messages that encrypt to themselves:

1
5908
5909
11197
11817
17105
17106
23013
23014
28922
36699
42607
42608
48515
48516
53804
54424
59712
59713
65620  

Finding such a message with strong primes would be extremely difficult, but I can't think of a reason it couldn't occur.

At this point I don't have an explanation for the consecutive numbers in the list.