Let plaintext be written in the alphabet a, b, . . . , z and let ciphertext be written
in the alphabet A, B, . . . , Z. Both alphabets have N = 26 letters which are numbered
using the Standard Numbering.
By numbering digraphs as usual, we can use RSA encryption with public key
K = (589, 23) to encrypt plaintext single letters as ciphertext digraphs.
Find the formula for decryption and decrypt the message AE.
Now I believe i know how to find the formula for decryption.... i prime factorized 589 to get 31x19
Then i got (589) = (31-1)(19-1) = 540
Then i took the multiplicative inverse of 23(mod540) which = 47
So i now have Decryption formula, Dk = c^47 (mod589)
Now im not sure how to decrypt the message AE. I know A = 0 and E = 4. I dont understand the whole digraph thing in this instance!