# mod arithmetic

• May 14th 2008, 02:36 AM
hunkydory19
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?
• May 14th 2008, 10:32 AM
Isomorphism
Quote:

Originally Posted by hunkydory19
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?

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 $\displaystyle x \rightarrow 9x \, mod \, 49$ is a bijection. Since $\displaystyle (9,49) = 1, 9\mathbb{Z}_{49} = \mathbb{Z}_{49}$.
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 :(