0

If $x\equiv a\pmod{n}$, prove that either $x\equiv a\pmod{2n}$ or $x\equiv a+n\pmod{2n}.$

Well I am a bit stuck in this congruence problem. I tried out :-

$n\vert x-a$ so for $2n|x-a$, $x-a$ must be even but I am not able to prove what if $x-a$ is odd, how can I prove that $x-(a+n)$ will be even and divisible by $2n$. Is this the right approach? Though it seems a very vague one, I can't think of any other.

Bill Dubuque
  • 272,048
arnav_de
  • 699
  • As in the proof in the linked dupe, using $ \rm\color{#c00}{mDL}=$ mod Distributive Law we have

    $$\large \begin{align}\bmod 2n!:\ x &\equiv a+kn,, \ {\rm by},\ x\equiv a!!!!\pmod{!n} \ &\equiv a + (kn\bmod 2n)\ &\equiv a + n(\color{#0a0}{k\bmod 2})\ \ {\rm by},\ \rm\color{#c00}{mDL}\ &\equiv a + n\color{#0a0}{{0,1}}\ &\equiv a,: a+n \end{align}\qquad\qquad$$

     – Bill Dubuque Feb 25 '21 at 05:09
  • See this comment for another example. – Bill Dubuque Feb 25 '21 at 08:17

1 Answers1

3

Hint: The hypothesis means that $x = a + kn$ for some integer $k$. What can you conclude about $x$ modulo $2n$ when $k$ is even? How about when $k$ is odd?

Sam Freedman
  • 3,989