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Math Help - mod arithmetic

  1. #1
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    mod arithmetic

    Consider the affine enciphering transformation:

    f(x) = 9x + 4 mod 49

    on single letter message blocks of the ordinary English alphabet A - Z
    and the letter b for “space”.

    How many more letters can we add to our alphabet without any
    further changes to our encoding procedure?

    Would this just be 26 x 2 = 52
    52 - 49 = 3, hence 3 more letters could be added?
    Thanks in advance!
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  2. #2
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    Quote Originally Posted by hunkydory19 View Post
    Consider the affine enciphering transformation:

    f(x) = 9x + 4 mod 49

    on single letter message blocks of the ordinary English alphabet A - Z
    and the letter b for “space”.

    How many more letters can we add to our alphabet without any
    further changes to our encoding procedure?

    Would this just be 26 x 2 = 52
    52 - 49 = 3, hence 3 more letters could be added?
    Thanks in advance!
    Again I believe this is a number theoretic question.I have never had a formal course in cryptography, but I know a little bit of number theory.

    I think this question is essentially asking, find the largest number of x's such that x \rightarrow 9x \, mod \, 49 is a bijection. Since (9,49) = 1, 9\mathbb{Z}_{49} = \mathbb{Z}_{49}.
    This means there are 49 values of x which will map to a unique (9x mod 49).
    Since we have already added 27 letters, we can add 22(=49-27) more letters to the alphabet without any further changes to our encoding procedure.

    Again, feel free to point my mistakes. Since I am not entirely sure whether the utility of the affine enciphering stems from its bijective nature. It looks like common sense, but I am not sure
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