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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?

  • 2
    Notice that $k$ is coprime to $n$ iff $n-k$ is coprime to $n$. What does that tell you about the sequence $a_i$?2011-07-26
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    Your proof is not really a proof. You seem to have anticipated *correctly* that $a_{\frac{\phi(m)}{2}} + a_{\frac{\phi(m)}{2}+1}$ should sum to $m$. But without justifying this, the proof is incomplete.2011-07-26
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    If we require monotonicity from $a_i$, I don't see how $a_{\frac{\phi(m)}{2}}>\frac{m}{2}$ can cohere with anything but $a_{\frac{\phi(m)}{2}+1}>\frac{m}{2}$.2011-07-26
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    I'm using the fact that $\ a_{\frac{phi(m)}{2}}+a_{\frac{phi(m)}{2}+1}=m $ but I'm not seeking to prove it.2011-07-26
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    I can't see the problem: I considered a proposition that leads to a contradiction in all cases so the negation gives the result?2011-07-26
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    @Plane Chon-Ju: Why not prove it? You have not even shown that $\phi(m)/2$ is an integer.2011-07-26
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    @Phane Con-Ju : have you read my comment? You can show the property in a minute without case by case study, and without invoking $a_{\frac{\phi(n)}{2}}+a_{\frac{\phi(n)}{2}+1}=n$2011-07-26
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    (+1) for all the usual reasons. Hopefully this will set a trend.2012-03-13

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