Nice proof for: In $\mathbb Z / m \mathbb Z$ associated implies strongly associated. I have stated the following problem:

Let $a,b,m \in \mathbb Z$ with $a | b$ and $b|a$ in $\mathbb Z / m \mathbb Z$. Then there exists a unit $\varepsilon \in \mathbb Z / m \mathbb Z^\times$ with $\varepsilon a = b$.

Since my proof might be a little bit hard to understand for undergrads (it is a little bit longer) I wonder if there is a nice way to show this.
You can see my proof below as an answer.
 A: For completeness the longer proof I have:
We already know that $a|b$ is equivalent to $\gcd(a,m) | b$ in $\mathbb Z$, hence we have $d:=\gcd(a,m)=\gcd(b,m)$ and wlog we can assume $b=d$.
Euclidean algorithm gives us the existence of $x \in \mathbb Z$ with $$xa\equiv d \quad \mod m.$$
Let $d':=\gcd(x,m)$. Since from above we can deduce $\gcd(x,\frac{m}{d})=1$ we have $d' | d$. Now we complete $d'$ with primes, namely for $d=p_1^{k_1}\cdot \ldots \cdot p_n^{k_n}$ as prime decomposition we define $$\tilde{d}:=\prod_{p_i | d'} p_i^{k_i} \text{ and }\tilde{x} = x- \frac{m}{\tilde{d}}$$
You can easily see, that still $\tilde xa\equiv d \mod m$ holds. 
Now we only need to show, that $\gcd(\tilde x,d)=1$ holds (for the same reason as above we have $\gcd(\tilde x,m)|d$). But you can easily check every prime factor of $d$ that it does not divide the sum $\tilde x =x -\frac{m}{\tilde{d}}$:
\begin{align*}
p|\tilde{d} &\Rightarrow p | d' \Rightarrow p | x \text{ and } p \not | \frac{m}{\tilde{d}} \Rightarrow p \not | \tilde x \\
p\not |\tilde{d} &\Rightarrow p \not | d' \text{ and } p | \frac{d}{\tilde{d}} \Rightarrow p \not | x \text{ and } p | \frac{m}{\tilde{d}}\Rightarrow p \not | \tilde x 
\end{align*}
