Question

Let y = ax + b denote an affine cipher that encrypts a letter x in a plaintext into the letter y in the ciphertext. Assume that if we encrypt a plaintext with the affine cipher twice then the resultant ciphertext is the same as the original plaintext. Find the encryption key (a, b).

Answer 1

In an affine cipher where double encryption leads to the original text, the encryption key that satisfies this condition is a = 1 or a = -1 (for mod 26) and b = 0. This requirement is based on modular arithmetic and properties of the affine cipher.

Step-by-step explanation:

The student is asking about an affine cipher which is a type of substitution cipher used in cryptography. The cipher is defined by the encryption key given by the equation y = ax + b, where a and b are constants, x is the plaintext, and y is the ciphertext.If encrypting a plaintext with the affine cipher twice results in the same ciphertext as the original plaintext, it means that applying the affine cipher twice cancels out the encryption. This happens when 'a' has a multiplicative inverse modulo the size of the alphabet and 'b' is equal to zero.

If encrypting a plaintext letter twice results in the original plaintext, it implies that the cipher must effectively reverse itself on the second encryption. This requirement provides specific constraints on the values of a and b. For an affine cipher, such that encryption and its inverse (i.e., decrypting twice) is the identity, we typically need a such that a * a = 1 mod 26 and b such that 2b = 0 mod 26 (considering the English alphabet with 26 letters and working with modular arithmetic). The solutions to these constraints in such a modular system are a = 1 (or a = -1 for mod 26) and b = 0, since these are the values that satisfy the given equations in a system modulo 26.