$s,t\mid k\Rightarrow st\mid k\,$ holds only when $s,t$ are coprime

Question

In the problem (below) we are asked to show: if $a \pmod {st} = b \pmod {st}$, then $a \pmod s = b \pmod s$ and $a \pmod t = b \pmod t$. I did this part of the question. Then we are asked, what makes the converse true? The answer is, that $\gcd (s, t) = 1$. How can this be shown?

Here is the actual problem (from Gallian, Contemporary Abstract Algebra 6E):

Let $a$, $b$, $s$, and $t$ be integers. If $a \pmod {st} = b \pmod {st}$, show that $a \pmod s = b \pmod s$ and $a \pmod t = b \pmod t$. What condition on $s$ and $t$ is needed to make the converse true?

My question is a little bit different than this one: x≡a (mod m) and x≡a(mod n) implies x≡a (mod mn) (and a few others like it). Usually in similar questions it is a supposition that $m$ and $n$ (in my case the $s$ and the $t$) are relatively prime. In my question, I am asking, how can we show that this is the case?

Here is my attempt at showing the converse (I don't think it is correct):

The converse would be that if $a \equiv b \pmod s$ and $a \equiv b \pmod t$, then $a \equiv b \pmod {st}$. Suppose that $a \equiv b \pmod s$ and $a \equiv b \pmod t$, then $s$ divides $a − b$ and $t$ divides $a − b$. There exist integers $m$ and $n$ such that $a − b = ms$ and $a − b = nt$. In order for $st$ to appear as a factor of $a − b$, $s$ and $t$ must be relatively prime. Or in other words, $\gcd (s, t) = 1$. Consider: \begin{align*} a - b & = ms \\ & = (nt)s \\ & = n(ts) \end{align*} Above, if $s$ divided $t$ or $t$ divided $s$, then they would not appear as factors alongside one another. We would not have $st$ as a factor.

EDIT:

My question was closed due to being similar to this question: $ d_1,d_2\mid n\iff {\rm lcm}(d_1,d_2)\mid n\,\ $ [LCM Universal Property or Definition]. My question is slightly different. This question is more open-ended. This question is seeking any satisfactory proof as an answer. In my question, I am asking about a specific condition: $\gcd (s, t) = 1$, and why it forces the result: $a \pmod {st} = b \pmod {st}$.

Answer 1

Note that $a\equiv b\ (\mathrm{mod}\ n)$ is the same as $n\mid a-b$, and so what you are asking is the same as $$s\mid x\ \mathrm{and}\ t\mid x\implies st\mid x$$ But if $s\mid x$ and $t\mid x$, then $x$ is a common multiple of $s$ and $t$, so it must be a multiple of $\operatorname{lcm}(s,t)$. If $s$ and $t$ are relatively prime, then the lcm is just the product, and that is the claim. But in general, you may or may not have it. You can always take $x$ to be the lcm itself, and then it won't hold. For a concrete example,$\newcommand{\Mod}[3]{#1\equiv#2\ (\mathrm{mod}\ #3)}$ $\Mod{12}04$ and $\Mod{12}06$ but $12\not\equiv0\ (\mathrm{mod}\ 24)$. You can always do this by replacing $4$ and $6$ with any not-coprime pair of integers.

Hope this helps. :)