bijective function on Z/26Z

[Perlita]

Hello,

In cryptography, I need to prove that the function f(x)=ax+b (on Z/26Z) is bijective.
I started by proving that it's injective:
an+b=an'+b
a(n-n')=0
n-n'=0
n=n'
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• 
  [*=1]Is this correct?
  [*=1]I didn't know how to prove that it's surjective.

Can anyone help?
​

 

[HallsofIvy]

No, what you are doing is not at all correct. In fact, you have a very fundamental problem- what you are trying to prove is NOT TRUE!
There exist a and b such that ax+ b is neither injective nor surjective! 26= 2(13). Suppose a= 2, b= 0 so that f(x)= ax+ b= 2x. f(0)= 0 (mod 26) and f(13)= 26= 0 (mod 26) so f is NOT injective.

It is not surjective because f(x)= 2x is always even- there is no x such that f(x)= 3 (mod 26).