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Prove: If n is odd and n|(a-b) and n|(a+b) then n|a and n|b

since n is odd, there is some number k such that

$n=2k+1$

there are also integers p,q such that

$np=(a-b)$ and $nq = (a+b)$

(We want to show there are integers r,s such that)

$nr=a$

$ns=b$


I've only managed to make it this far into the proof, However, I've noted a few observations.

  1. p and q could be either odd or even
  2. a and b must be odd in order for their sum or difference to be divisible by n

Furthermore, my goal is to show that there are some integers r,s that are divisible by n.

I have attempted substitution, but it seems to be leading to dead ends.

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n divides a-b and a+b hence n divides their sum (a-b)+(a+b)=2a. Now, you know that n divides 2a and you want to show that n divides a. Which is easy since... – Did Nov 2 '11 at 18:36
up vote 1 down vote accepted

since $n|a-b$ and $n|a+b$ then $n|2a$ now since $n$ is odd then $n$ and $2$ are coprime, hence by Euclid's Lemma we have that $n|a$.

Similarly $n|-2b$ hence $n|2b$ and again by Euclid's Lemma $n|b$.

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HINT. If $d$ divides $x$ and $y$, then it divides $x+y$ and $x-y$.

HINT 2. If $d$ divides $xz$, and $\gcd(d,z)=1$, then $d$ divides $x$.

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