# Hill cipher - two plaintext pairs encrypt to same ciphertext

1. Nov 18, 2012

### LANS

1. The problem statement, all variables and given/known data

Given the following key, find two plaintext pairs that encrypt to the same ciphertext.

$\begin{pmatrix} 9 & 5 \\ 7 & 3 \\ \end{pmatrix} (mod \ 26)$

2. Relevant equations

3. The attempt at a solution

Let the two plaintext pairs be (a, b) and (c, d). Let the ciphertext be (X, Y)

9a + 7b = X mod 26
9c + 7d = X mod 26
5a +3b = Y mod 26
5c + 3d = Y mod 26

I'm not sure where to go from here. Any help would be greatly appreciated.

2. Nov 18, 2012

### haruspex

You can add and subtract equations, and multiply by factors, in the usual way. The thing you have to be careful with is division. You can divide by any number that's prime to the base, 26, but it can be tricky working out the answer. E.g. 5/3 will be 19, because 5 = 5+2*26 = 57 (mod 26).
You can also cancel common factors that are shared with the base provided you cancel out in the base too:
4a + 6b = 8c (mod 26) implies
2a + 3b = 4c (mod 13)