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Question 1183390: given set Z(26) = {0,1,2,3,...25}
Let a and b be integers. Consider the function f: Z(26) -> Z(26) defined by f(x) = ax + b (mod 26).
a)Prove that f(x) = 3x + b is bijective for any b.
b) find all the values of a for which the function f is bijective.
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website! .
Z(26) is the residue ring modulo 26.
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| Since you use this denoting, I assume that you are familiar |
| with the term "the residue ring modulo 26" |
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and with all other associated terms.
The map f : x ----> 3x + b is the bijective for any "b", because 3 mod 26 is an invertible element of the ring Z(26).
Indeed, in this ring 3*9 = 1 (since 3*9 = 27 = 1 mod 26); so, the element "9" of the ring Z(26) is inverse to the element "3".
It is the answer to question (a).
The answer to question (b) is "any number / (element of Z(26) ), which is mutually prime with /(or to) the number 26 ".
For example, the elements "1", "3", "5", "7", "9", "11" satisfy this criterion;
the elements "2", "4", "6", . . . , "13" do not satisfy the criterion.
Solved.
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I intently presented the solution in terms of the ring theory, because your post was formulated in these terms.
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