Let $k_1, k_2, k_3,\dots ,k_n$ be positive divisors of $m$ and $n$. Can you prove or disprove the following $$\varphi (k_1) + \varphi (k_2)+\varphi (k_3) +\dots+\varphi (k_n)=\gcd(m, n).$$
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Does $k_1, k_2, k_3,... k_n$ involve all the positive divisors of m and n?? – SchrodingersCat Mar 28 '16 at 14:00
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2In general $\sum_{d|n}\varphi(d)=n$. See, eg, this – lulu Mar 28 '16 at 14:00
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@schrodinger. Yes, – Agus Ahmad Rizqi Mar 28 '16 at 14:02
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@lulu.oh,thanks for the idea, it give a lot for my friends – Agus Ahmad Rizqi Mar 28 '16 at 14:05
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3@lulu An MSE-link is here. Hence $\sum_{d\mid (n,m)}\phi(d)=(n,m)$. – Dietrich Burde Mar 28 '16 at 14:39
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@DietrichBurde Thanks! I searched for such (but obviously not hard enough). – lulu Mar 28 '16 at 14:47
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@dietrich, oh the formula is the key! Thank you, its very useful for my firends presentation. – Agus Ahmad Rizqi Mar 30 '16 at 09:36