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2015-04-20T09:36:54+05:30

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Let a_1,a_2,\ldots,a_{\phi(n)} be the positive integers less than n and relatively prime to n.

Since \gcd(a,n)=\gcd(n-a,n), the numbers  a_1,a_2,\ldots,a_{\phi(n)} are same as the numbers  n-a_1,n-a_2,\ldots,n-a_{\phi(n)} in some order. Thus,
\sum\limits_{i=1}^{\phi(n)}a_i=\sum\limits_{i=1}^{\phi(n)}(n-a_i)=\phi(n)n-\sum\limits_{i=1}^{\phi(n)}a_i

\implies \sum\limits_{i=1}^{\phi(n)}a_i=\frac{n\phi(n)}{2}
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