The Euler’s totient function, denoted $\varphi$ (or $\phi$), is defined at $n$ to be the number of positive integers not exceeding $n$ that are relatively prime to $n$. We can rewrite $\varphi(n)$ in the summation notation as
\[
\varphi(n) = \sum_{\scriptstyle k =1 \atop \scriptstyle (k, n) = 1}^{n} 1.
\] The function $\varphi$ is ubiquitous in number theory, particularly in prime number theory. For instance, the degree of the irreducible polynomial of $e^{2\pi i/n}$ over $\mathbb{Q}$ is $\varphi(n)$ or equivalently, $[\mathbb{Q}(e^{2 \pi i/n}) : \mathbb{Q}] = \varphi(n)$. This can be used to show that the torsion subgroup of the group of units in a number field is finite. More elementarily, the number of generators in a cyclic group of order $n$ is $\varphi(n)$. Next, we find the divisor sum of $\varphi$.
Theorem. If $n \geq 1$, then
\[
\sum_{d | n} \varphi(d) = n.
\]
Proof. We partition the set $\{1, \dots, n\}$ into subsets $A_d = \{1 \leq k \leq n : (k, n) = d\}$, where $d$ is a divisor of $n$, and note that there is a one-to-one bijection between elements of $A_d$ and integers $1 \leq r \leq n/d$ satisfying $(r, n/d) = 1$. This then implies that
\[
n = \sum_{d | n} | A_d| = \sum_{d | n} \varphi(n/d) = \sum_{d \div n} \varphi(d).
\]
There is a relationship between $\varphi$ and $\mu$ which can be deciphered easily from the divisor sum of $\varphi$ once one knows about Möbius inversion formula but we find it directly using what we have at the moment.
Theorem. If $n \geq 1$, then
\[
\varphi(n) = \sum_{d | n} \mu(d) \frac{n}{d}.
\]
Proof. We use the formula for the divisor sum of $\mu$ to obtain
\[
\varphi(n) = \sum_{k = 1}^{n} e((k, n)) = \sum_{k = 1}^{n} \sum_{d | (k, n)} \mu(d) = \sum_{k = 1}^{n} \sum_{\scriptstyle d | n \atop \scriptstyle d | k} \mu(d)
\] Changing the order of summation we find that
\[
\varphi(n) = \sum_{d | n} \sum_{\scriptstyle k = 1 \atop \scriptstyle d | k}^{n} \mu(d)
= \sum_{d | n} \mu(d) \sum_{\scriptstyle k = 1 \atop \scriptstyle d | k}^{n} 1 = \sum_{d | n} \mu(d) \frac{n}{d}.
\]
Next we have a nice product formula for $\varphi$.
Theorem. For $n \geq 1$ we have
\[
\varphi(n) = n \prod_{p | n} \left( 1- \frac{1}{p} \right).
\]
Proof. If $n = 1$, then the product on the right hand side is empty and so the formula trivially holds. Now let $p_1, \dots, p_k$ be the prime divisors of $n$. Then expanding the product, we get
\[\begin{align} \prod_{i = 1}^{k} \left( 1- \frac{1}{p_i} \right) &= \sum_{I \subset [k]} \prod_{i \in I} \left(-\frac{1}{p_i} \right) = \sum_{I \subset [k]} \frac{(-1)^{|I|}}{\prod_{i \in I} p_i} = \sum_{d | n} \frac{\mu(d)}{d} = \frac{\varphi(n)}{n}. \end{align} \]
Corollary. If $m$ and $n$ are positive integers, then
\[
\varphi(mn) = \varphi(m) \varphi(n) \frac{d}{\varphi(d)},
\] where $d = (m, n)$. In particular, if $m$ and $n$ are coprime, then $\varphi(mn) = \varphi(m) \varphi(n)$.
Proof. The assertion follows from the product formula for $\varphi$ as
\[\begin{align} \frac{\varphi(mn)}{mn} &= \prod_{p | mn} \left( 1- \frac{1}{p} \right) \\ &= \prod_{p | m} \left( 1- \frac{1}{p} \right) \prod_{\scriptstyle p | n \atop \scriptstyle p \hspace{1pt} \nmid \hspace{1pt} m} \left( 1- \frac{1}{p} \right) \\ &= \prod_{p | m} \left( 1- \frac{1}{p} \right) \prod_{\scriptstyle p | n} \left( 1- \frac{1}{p} \right)\prod_{\scriptstyle p | n \atop \scriptstyle p | m} \left( 1- \frac{1}{p} \right)^{-1} \\ &= \frac{\varphi(m)}{m} \frac{\varphi(n)}{n} \prod_{p | (n, m)} \left( 1- \frac{1}{p} \right)^{-1} \\ &= \frac{\varphi(m)}{m} \frac{\varphi(n)}{n} \frac{d}{\varphi(d)}. \end{align}\]
From the multiplicativity of $\varphi$ it follows that if $n$ divides $m$, then $\varphi(n)$ divides $\varphi(m)$. This is because if $n$ and $m$ are prime powers, then the statement holds.
For a positive integer $k$ let $A(k)$ denote the number of solutions to the equation $\varphi(n) = k$. It is easy to see that $A(k)$ is finite for every $k$ since $\varphi(n) \to \infty$. In 1999, Kevin Ford (UIUC) proved Sierpinski’s conjecture, which states that all of positive integers greater than $1$ lie in the image of function $A$, using sieve theory. It is not known if $A(k) = 1$ for some $k$. The answers seems to be in the negation and this is known as Carmichael’s conjecture (Robert Carmichael originally stated this as a result instead of a conjecture in his book which he later retracted).
It is easy to see that $\varphi(n) \to \infty$ as $n \to \infty$. To see this let $M > 0$ be fixed and consider all integers $n$ for which $\varphi(n) \leq M$. Take $n = \prod_{i = 1}^{k} p_i^{a_i}$ be such that $\varphi(n) \leq M$. Then we have $\prod_{i = 1}^{n} p_i^{a_i- 1}(p_i- 1) \leq M$. This implies that $p_i \leq M + 1$ and $a_i \leq \log_2(2M)$ for every $1 \leq i \leq k$. Thus there are only finitely many choices of primes $p_i$ and exponents $a_i$ and consequently there are finitely many $n$ for which $\varphi(n) \leq M$. Hence, it follows that $\lim_{n \to \infty} \varphi(n) = \infty$.