Hill cipher - two plaintext pairs encrypt to same ciphertext

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SUMMARY

The discussion focuses on finding two plaintext pairs, (a, b) and (c, d), that encrypt to the same ciphertext using the Hill cipher with the key matrix \(\begin{pmatrix} 9 & 5 \\ 7 & 3 \end{pmatrix} \mod 26\). The equations derived from the encryption process are 9a + 7b = X \mod 26 and 5a + 3b = Y \mod 26. Participants emphasize the importance of careful manipulation of equations, particularly when dividing by numbers that are coprime to the modulus, 26. The discussion also highlights the concept of canceling common factors in modular arithmetic.

PREREQUISITES
  • Understanding of Hill cipher encryption techniques
  • Familiarity with modular arithmetic, particularly modulo 26
  • Knowledge of linear algebra concepts as applied to cryptography
  • Ability to manipulate equations in modular systems
NEXT STEPS
  • Study the properties of the Hill cipher and its encryption/decryption processes
  • Learn about modular arithmetic and its applications in cryptography
  • Explore examples of plaintext pairs that yield the same ciphertext in the Hill cipher
  • Investigate the implications of coprime numbers in modular division
USEFUL FOR

This discussion is beneficial for cryptography students, mathematicians interested in modular arithmetic, and anyone studying the Hill cipher's properties and applications in secure communications.

LANS
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Homework Statement



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

\begin{pmatrix}<br /> 9 &amp; 5 \\<br /> 7 &amp; 3 \\<br /> \end{pmatrix} (mod \ 26)<br />

Homework Equations



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.
 
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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)
 

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