SOLUTION: 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

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.


    +------------------------------------------------------------------+
    |    Since you use this denoting, I assume that you are familiar   |
    |          with the term  "the residue ring modulo 26"             |
    +------------------------------------------------------------------+


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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