Q:

Prove by contradiction that if $a\geq 2$ and $b$ are integers, then $b$ or $b+1$ is not divisible by $a$

Prove by contradiction that if $a\ge2$ and $b$ are integers, then $a ∤ b$ or $a ∤ (b+1)$.
I have that assuming to the contrary $a∣b$ and $a∣(b+1)$ then $b=ax$ and $b+1=ax$, but I'm not sure where to go from there. 

A:

Assuming $a \mid b$ and $a \mid (b+1)$, then there exist two different integers $x_1$ and $x_2$ such that $b=ax_1$ and $(b+1)=ax_2$. Subtracting the two equations gives $1=a(x_2-x_1)$. Why is this a contradiction?

