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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 $x \equiv a + n \mod{2n}$. On the other hand, if $n$ is odd, $(n:2)=1$. Hence the Chinese remainder theorem guarantees that the linear congruence system

$$\begin{cases} x \equiv a \mod{n} \\ x \equiv a \mod{2} \end{cases}$$

has a unique solution modulo $2n$. We know the solution is $x \equiv a(2+n) \mod{2n}$, which is congruent to $n \equiv a(2 - n) \mod{2n}$.

Given that $n$ is odd, $(2+n:2-n)=1$. Therefore $x \equiv a \mod{2n}$.

Is this more or less correct?

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    $\begingroup$ Something is fishy here. Take $n=4, x =3, a = 3$. Your proof says that since $n$ is even, 3 is congruent to 7 mod 8, which is not correct. Can you find the error? $\endgroup$ – user113102 May 28 '19 at 19:12
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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 $$ x-a = tn = (2r+1)n = 2rn + n \; , $$ $$ x-a = r(2n) + n, $$ $$ x= a + n + r (2n) $$ $$x \equiv a +n \pmod {2n}$$

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$\phantom{.}\\{\bf Hint}\ \ x\equiv a\pmod{\!n} \iff \overbrace{x = a + n\,k}^{\large\text{for some } k\ \in\ \Bbb Z} \iff \left\{ \begin{align} &x =\smash[t]{\overbrace{a+n(2j)}^{\large \color{#c00}a\ \ +\ \ 2n\,j}}\ \ \ \ \ \ \ \, {\rm if}\ \ k = \,2j\ \ \ \ \ \ \rm is\ even\\ &x = \smash[b]{\underbrace{a+n(1\!+\!2j)}_{\large \color{#c00}{a+n}\ \ +\ \ 2n\,j}}\ \ {\rm if}\ \ k = 1\!+\!2j\ \ \rm is\ odd\\[-1em] \phantom{.} \end{align}\right.$

by $\ k = r\! +\! 2j\ $ for $\, r = 0\,$ or $\,1\,$ by Division with Remainder.

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