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Let $\ m\ge3$, and let $\ a_i$ be the natural numbers less than or equal to $\ m$ that are coprime to $\ m$ put in the following order: $\ a_1

If $\ a_{\frac{\phi(m)}{2}}>\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}\ge\frac{m}{2}$ then $\ a_{\frac{\phi(m)}{2}}+a_{\frac{\phi(m)}{2}+1}>m$ which is wrong.

If $\ a_{\frac{\phi(m)}{2}}\le\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}<\frac{m}{2}$ then $\ a_{\frac{\phi(m)}{2}}+a_{\frac{\phi(m)}{2}+1} which is wrong.

If $\ a_{\frac{\phi(m)}{2}}>\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}<\frac{m}{2}$ then $\ a_{\frac{\phi(m)}{2}+1} which is wrong.

So $\ a_{\frac{\phi(m)}{2}}>\frac{m}{2}$ or $\ a_{\frac{\phi(m)}{2}+1}<\frac{m}{2}$ is wrong, $\ a_{\frac{\phi(m)}{2}}\le\frac{m}{2}$ and $\ a_{\frac{\phi(m)}{2}+1}\ge\frac{m}{2}$ is true, and it gives the result.

Does this proof work?

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    (+1) for all the usual reasons. Hopefully this will set a trend.2012-03-13

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Your proof is correct, but you should clearly indicate where the proof starts and that you are using the result on the sum of two symmetric elements in the proof.