If $x \equiv a \pmod{\!n}\,$ then $\,x \equiv a\,$ or $\, a + n \pmod{\!2n}$ My rationale was the following. If $n$ is even, then
$$n \mid x - a \implies n \mid x - a - n \implies n \mid \frac{1}{2} (x - a - n) \implies 2n \mid x - a - n  $$
so $x \equiv a + n \pmod{2n}$. On the other hand, if $n$ is odd, $(n:2)=1$. Hence the Chinese remainder theorem guarantees that the linear congruence system
$$\begin{cases} x \equiv a \pmod{n} \\ x \equiv a \pmod{2} \end{cases}$$
has a unique solution modulo $2n$. We know the solution is $x \equiv a(2+n) \mod{2n}$, which is congruent to $n \equiv a(2 - n) \pmod{2n}$.
Given that $n$ is odd, $(2+n:2-n)=1$. Therefore $x \equiv a \pmod{2n}$.
Is this more or less correct?
 A: I don't quite see what you are doing. Let
$$ x-a = tn \; . $$
If $t$ is even, let $t = 2 s,$ so
$$ x-a = s(2n) $$
so $$x \equiv a \pmod {2n}$$
If $t$ is odd, write $t = 2r + 1,$
 so $$  x-a = tn = (2r+1)n = 2rn + n \; ,   $$
$$ x-a = r(2n) + n, $$
$$   x= a + n + r (2n)  $$
$$x \equiv a +n \pmod {2n}$$
A: $\phantom{.} x\equiv a\pmod{\!n} \!\iff\!\! \overbrace{x = a \!+\! n\:\!\color{#0af}k}^{\large\text{for some } \color{#0af}k\ \in\ \Bbb Z}\!\!\! \iff\!\! \left\{\!\!\!\begin{align} &x =\smash[t]{\overbrace{a\!+\!n(\color{#0a0}0\!+\!2j)}^{\large \color{#f00}a\ \ +\ \ 2n\,j}}\ \ {\rm if}\ \ k \  \rm is\ \color{#0a0}{even}\\[.3em] 
&x = \smash[b]{\underbrace{a\!+\!n(\color{#c00}1\!+\!2j)}_{\large \color{#f00}{a+n}\ \ +\ \ 2n\,j}}\ \ {\rm if}\ \ k\ \rm is\ \color{#c00}{odd}\\[-1em] \phantom{.} \end{align}\right.$
We used Division  $\,(k\div 2)\,$ to get  $\ \color{#0af}k = r\! +\! 2j\ $ for $\, r = \color{#0a0}0\,$ or $\,r=\color{#c00}1$

Or $\!\bmod 2n\!:\  nk \overset{\rm\color{#c00}D}\equiv n(k\bmod 2)\equiv \color{#0a0}0, \color{#c00}n,\,$ so $\,a\!+\!nk\equiv a,a\!+\!n\ $ by $\,\small\rm\color{#c00}D=$ mod Distributive law.
