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We first need to show that for each pair of plaintext-ciphertext letters ( x,y ), there are exactly 12 keys that encrypt x to y . For each choice of a , the key ( a,y − ax ) encrypts the plaintext letter x to the ciphertext letter y , since ax + ( y − ax ) = y . There are twelve possible choices for a so there are exactly twelve keys that map a given plaintext letter to a given ciphertext letter.

This is an easy question, I'm just trying to make sense of this statement found on page three here www.maths.uq.edu.au/courses/MATH3302/2010/files/cryptotute2.pdf

In the affine cipher how is their only 12 keys that map a given plaintext x to a given ciphertext y? I learned that the affinecipher has 286 nontrivial keys. So if thats the case how can only 12 of them map a given plaintext x to a given ciphertext y.

Temirzhan
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Assuming plaintext and ciphertext alphabets are $U=\{0,\ldots,25\}$, taken to be the ring of integers modulo 26, and the keyspace is $U^2$ with $$(a,b)\in U^2,$$ it is clear that for plaintext $x=2t$ corresponding to even numbers, $(a,b)$ and $(a+13,b)$ give the same $y$ violating the invertibility of the encryption. Moreover, if $b=0$ $a$ can't be zero for the same reason.

So we allow all $b$ but only $12$ nonzero $a$'s and can choose either $a$ or $a+13$ as $a$ ranges over $\{1,2,\ldots,12\}$.

kodlu
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