shivajikobardan Posted August 28, 2022 Posted August 28, 2022 I'll present few cases of RSA encryption: [math]CT=(PT)^e mod \, n[/math] [math]PT=(CT)^d mod \, n[/math] CT= Cipher Text PT=Plain Text e*d/m should give remainder 1, where; m=(p-1)*(q-1); where p, q are 2 prime numbers. A) Sender takes p=3, q=11. The value of e=3, d=7 satisfied the remainder=1 condition. So, if sender wants to send "SELL": S=19, CT=28 E=5. CT=26 L=12, CT=12 L=12, CT=12(? What to do to not get the cipher text same as plain text without making things too complex and still being able to do it in paper manually?) B) p=2, q=11 e=3,d=7 I'll only write CT here(i.e S=17 means that 17 is a cipher text for S after RSA encryption): S=17 E=15 L=12 L=12 (Same here? why? Because of d,e being the same?) C) p=13, q=11 d=13, e=37 So, S=95 E=93 L=12 L=12 (Again got the same value, what is this? Even with different values of p,q,d,e!) Questions: a) Is there any restrictions if among p or q, anyone should be greater? b) Is there any restrictions like which of the d or e should be greater? I know e<m and e>1 And e should be relatively prime to m, i.e GCD(e,m)=1. GCD(13,120)=1 For d, de mod m=1 c) Would there be any cases, where while decryption, we would not be able to get the plain text due to some reasons (like if not choosing values properly for d,e,p,q in english alphabet encryption? This is the main confusion that is making me ask this )
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now