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Problem: Prove that the sum of all integers $ t \in \{ 1, 2, \cdots, n \} $ and $ (t, n) = 1 $ is $ \dfrac {1}{2} n \phi (n) $, where $ \phi $ is the Euler Totient Function.

My proof:

Define the set $\mathcal{S}$ to be the set of all the elements $t$ such that $ 1 \le t \le n $ and $ (t, n) = 1 $. The cardinality of $\mathcal{S}$ is $\phi(n)$. For $ n = 2 $, the set $\mathcal{S}$ is $ \{ 1 \} $ and so the statement holds, since $ \dfrac {1}{2} \cdot 2 \cdot \phi(1) = 1 $. $ \\ $ $\mathrm{}\\$

Lemma: For natural $ n \ge 3 $, $\phi(n)$ is even.

Proof: If $n$ is a power of $2$, say $2^k$, then $\phi(n)=n \cdot \left( 1 - \dfrac {1}{2} \right) = \dfrac{n}{2}=2^{k-1}$, which is even, since $ n > 2 $. If $n$ is not a power of $2$, as to have at least one odd prime divisor, say $p$, from the Euler Totient Function formula, $$ \phi (n) = n \cdot \displaystyle\prod_{p \text { prime}, \ p \mid n} \left( \dfrac {p-1}{p} \right), $$and so, $(p-1)\mid\phi(n)$. Thus, since $p$ is odd, $2\mid\phi(n)$, which is to say that $\phi(n)$ is even. $ \Box $

$\mathrm{}\\$ From the Lemma, it follows that the cardinality of $\mathcal{S}$ is even, as long as $n\ge3$. We verified the statement for $n=2$, so what remains is $ n \ge 3 $. We will need the even cardinality of $\mathcal{S}$ shortly. Let $\mathcal{S}=\{t_1,t_2,\cdots,a_{\phi(n)}\}$. Let the desired sum be $R$. Now, note the following. If $t_k$ and $n$ are relatively prime, it follows that $n-t_k$ is relatively prime. Note that there is a bijection between $t_k$ and $n-t_k$, from Lemma $1$. That is, for every $t_k$, it follows that there is an $n-t_k$ that corresponds to it. If the cardinality of $\mathcal{S}$ was odd, this would not be the case.

We can now conclude that $ \{ n - t_1, n - t_2, \cdots n - a_{\phi(n)} \} $ is the same set as $ \{ a_1, a_2, \cdots, a_{\phi(n)} \} $, namely $\mathcal{S}$. We add up the two representations of $\mathcal{S}$. We have $$ a_1 + \cdots + a_{\phi(n)} = R = (n - a_1) + \cdots + (n - a_{\phi(n)}). $$Therefore, we solve for $R$ to get $ 2R = n \cdot \phi (n) $, or $ R = \dfrac {1}{2} n \phi (n) $, as desired.

$ \blacksquare $

Are there any other proofs or interesting results?

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You don't need the lemma. For every $1\leq t \leq n$ with $(t, n)=1$, we have $(n-t,n)=1$. The map $t \mapsto n-t$ is a bijection, with itself as inverse: $t=n-(n-t)$. This has nothing to do with the parity of $\varphi(n)$.

The rest of your proof is correct: if $S=\sum_{(t,n)=1} t$, we have:

$$S = \sum_{(t,n)=1} t= \sum_{(t,n)=1} (n-t) = n\varphi(n) - S$$

so $$S = n\varphi(n)/2.$$

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  • $\begingroup$ Will upvote when my vote lock is off. $\endgroup$ – Ahaan S. Rungta Nov 20 '13 at 17:47

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