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

[jdinatale]

1. Homework Statement .
1. Let $a$ and $b$ be constant integers with $a \not = 0$, and let the mapping $f : Z \rightarrow Z$ be defined by $F(x) = ax + b$. Determine all values of $a$ 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 $$ax \cong 0 \mod a$$ 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 $x_1$ and $x_2$ such that $x_1\ne x_2$ but $ax_1+ b= ax_2+ b$. 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 $x_1$ and $x_2$ such that $x_1\ne x_2$ but $ax_1+ b= ax_2+ b$. 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 $f(x) = ax + b$ that is not onto. For example, consider $f(x) = 4x + 3$. There is no $x \in \mathbf{Z}$ such that $f(x) = 5$.

That's my whole problem.

 

[fzero]

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

 

[jdinatale]

You seem to have lost the $b\mod a$ 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 $ax - b \cong b \mod a$. I don't quite see how this helps because that implies that $a$ must divide $ax - 2b$. I don't see how that tells me anything about what a must be. I mean it tells us that [itex]a must divide $2b$ for this to be true. <br /> <br /> But anyways, at the end of the day, we know $a$ must divide $$t - b$$[/itex][tex]. The only number guaranteed to divided every integer $k = t - b$ for all integers $t$ and $b$ 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.[/tex]

 

[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.