Mod. Arithmetic Proof: I don't see flaws in my logic, but it isn't working out.

[jdinatale]

1. Homework Statement .
1. Let [itex]a[/itex] and [itex]b[/itex] be constant integers with [itex]a \not = 0[/itex], and let the mapping [itex]f : Z \rightarrow Z[/itex] be defined by [itex]F(x) = ax + b[/itex]. Determine all values of [itex]a[/itex] such that f is a bijection. Prove that the aforementioned values are the only possible values resulting in a bijection.

The logic in my proof makes sense, but my conclusion that [tex] ax \cong 0 \mod a[/tex] doesn't make sense because that statement will always be true.

Homework Equations

N/A

The Attempt at a Solution

 

[HallsofIvy]

What's wrong with always being true? The only way f(x)= ax+ b will not be bijective is if there exist [itex]x_1[/itex] and [itex]x_2[/itex] such that [itex]x_1\ne x_2[/itex] but [itex]ax_1+ b= ax_2+ b[/itex]. What does that tell you about a?

 

[jdinatale]

What's wrong with always being true? The only way f(x)= ax+ b will not be bijective is if there exist [itex]x_1[/itex] and [itex]x_2[/itex] such that [itex]x_1\ne x_2[/itex] but [itex]ax_1+ b= ax_2+ b[/itex]. What does that tell you about a?

I understand that it will always be 1-1.
But my whole issue is that you could construct a [itex]f(x) = ax + b[/itex] that is not onto. For example, consider [itex]f(x) = 4x + 3[/itex]. There is no [itex]x \in \mathbf{Z}[/itex] such that [itex]f(x) = 5[/itex].

That's my whole problem.

 

[fzero]

You seem to have lost the [itex]b\mod a[/itex] term in your argument. Considering it should lead you to the correct condition.

 

[jdinatale]

You seem to have lost the [itex]b\mod a[/itex] term in your argument. Considering it should lead you to the correct condition.

Thanks for point that out, I did mess up there. It should be [itex]ax - b \cong b \mod a[/itex]. I don't quite see how this helps because that implies that [itex]a[/itex] must divide [itex]ax - 2b[/itex]. I don't see how that tells me anything about what a must be. I mean it tells us that [itex]a must divide [itex]2b[/itex] for this to be true.

But anyways, at the end of the day, we know [itex]a[/itex] must divide [tex]t - b[/itex]. The only number guaranteed to divided every integer [itex]k = t - b[/itex] for all integers [itex]t[/itex] and [itex]b[/itex] is 1. The problem is finding a mathematical argument for that. Given any other a, I could always find a number such that the function isn't a bijection.

 

[jdinatale]

I apologize for double posting, but it will no longer let me edit my last post. I believe I have found a solution, but I would greatly appreciate it if you could look through it and try to find errors.