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Prove or disprove: If $a\mid b + c$ and $a\mid b - c$ and $a$ is odd, then $a\mid b$.

I cannot seem to find a counterexample so I am thinking it might be true, but cannot prove it either.

This is what I have done so far:

Assume $a\mid b + c$ and $a\mid b - c$ and $a$ is odd. Then $b + c = ak_1$ and $b -c = ak_2$ for integers $k_1,k_2$. Then $c = ak_1-b$ and $c=b-ak_2$. So $ak_1-b=b-ak_2$ $\Rightarrow$ $2b = ak_1+ak_2 = (k_1+k_2)a \Rightarrow a\mid 2b$.

Now I am not sure what to do. Any help is appreciated.

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We have $b+c=am_1$ and $b-c=am_2$. Adding both gives us that $2b = a(m_1+m_2)$. This means $a \mid 2b$. However, since $a$ is odd, we have $\gcd(a,2) = 1$. Hence, $a \mid b$.

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  • $\begingroup$ Just as I submitted this for answering I saw this. Nonetheless thank you! $\endgroup$ – Robert May 17 '15 at 19:32
  • $\begingroup$ It can be simplified: $\ 2\nmid a,\,\ 2b = a(m_1\!+m_2)=:am\,\Rightarrow\, 2\mid m\ $ so $\ b = a(m/2),\,$ so $\,a\mid b\ \ $ $\endgroup$ – Gone May 17 '15 at 19:38
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    $\begingroup$ @Robert Perhaps then try to solve the questions yourself before you put a question. Some people try to help you, and they do not deserve to hear "I do not need your answer, but nonetheless thank you". $\endgroup$ – Dietrich Burde May 17 '15 at 19:43

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