Given distinct nonzero elements $a$ and $b$ in $A$…prove $A$ has characteristic $x$. I'm working on a problem from Pinter's Abstract Algebra and am wondering if someone can tell me if I'm on the right track.
Let $A$ be a finite integral domain. Prove that if there are distinct nonzero elements $a$ and $b$ in $A$ such that $125 \cdot a = 125 \cdot b$, then $A$ has characteristic 5.
Proof:
As this is a finite integral domain, it has nonzero characteristic. By Theorem 20.2, that characteristic must be a prime number. By Theorem 20.1, all nonzero elements in an integral domain have the same additive order $n$. So, unity (1) has additive order $n$, as do $a,b \in A$. 
Because $a,b$ have order $n$, $n \cdot a = 0$ and $n \cdot b = 0$. This implies that $n \cdot a = n \cdot b$. And since $125 \cdot a = 125 \cdot b$, and according to Theorem 10.5, 125 must be a multiple of $n$. The only prime factor of 125 is 5, so $char(A) = 5$.
Theorem 10.5: Suppose an element $a$ in a group has order $n$. Then $a^t = e$ iff $t$ is a multiple of $n$.
Theorem 20.1: All nonzero elements in an integral domain have the same additive order.
Theorem 20.2: In an integral domain with nonzero characteristic, the characteristic is a prime number.  
 A: The proof presented by our OP Alex Johnson seems fine to me, although I think there is a direct route to the desired result from first principles which avoids the need to develop the machinery of the stated theorems 10.5, 20.1, and 20.2, to wit:
It is a well-known and elementary result that a finite integral domain $A$ is in fact a field.  Indeed, this may easily be seen as follows:  for any $0 \ne a \in A$, we consider the function
$\theta_a:A \to A,\; \theta_a(r) = ar, \; \forall r \in A; \tag 1$
$\theta_a$ is easily seen to be injective, since for $r_1, r_2 \in A$ we have
$\theta_a(r_1) = \theta_a(r_2) \Longrightarrow ar_1 = ar_2 \Longrightarrow a(r_1 - r_2) = 0 \Longrightarrow r_1 - r_2 = 0 \Longrightarrow r_1 = r_2, \tag 2$
where we have used $a \ne 0$ in establishing this sequence of implications; now since
$\theta_a$ is injective and $A$ is finite, we have that $\theta_a$ is also surjective; therefore 
$\exists b \in A, \; \theta_a(b) = 1_A \Longrightarrow ab = ba = 1_A, \tag 3$
that is, $b$ is a multiplicative inverse of $a$; we have thus shown that every $a \in A$ has such an inverse, and thus that $A$ is indeed a field.
Now with 
$a \ne b, \tag 4$
we further have 
$a -b \ne 0, \tag 5$
so then
$125a = 125b \Longrightarrow 125(a - b) = 125a - 125b = 0$
$\Longrightarrow 5^3 = 125 = 0 \; \text{in} \; A \Longrightarrow 5 = 0 \; \text{in} \; A, \tag 6$
and now since $5$ is prime we have
$\text{char}A = 5, \tag 7$
as was to be shown.  $OE\Delta$.
