\(\textbf{Definition. }\)An Arithmetic function is a function\(f:\mathbb{Z}^+\to\mathbb{C}\).
\(\textbf{Definition. }\)Let \(n\in\mathbb{Z}^+\), we define the tau function \(\tau(n)\) as the number of positive divisors of \(n\). \[ \tau(n) = \#\{d\mid n, d>0\} \]
\(\textbf{Example. }\)Since \(n = 101\) is a prime number, we have \(\tau(101) = 2\).
\(\textbf{Example. }\)Let \(n = 101^7\). We have \[ \tau(n) = \tau(101^7) = \#(1, 101, 101^2, \dots, 101^7) = 7 + 1 = 8. \]
\(\textbf{Note. }\)If \(p\) is a prime number, then \(\sigma(p) = 2\).
\(\textbf{Note. }\)If \(p\) is a prime number, then \(\sigma(p^k) = k + 1\).
\(\textbf{Definition. }\)Let \(n\in\mathbb{Z}^+\), we define the \(\sigma\) function \(\sigma(n)\) as the sum of the positive divisors of \(n\). \[ \sigma(n) = \sum_{d\mid n, d>0}d. \]
\(\textbf{Example. }\)Since \(n = 101\) is a prime number, we have \(\sigma(101) = 101 + 1 = 102\).
\(\textbf{Example. }\)Let \(n = 101^7\). We have \[ \sigma(n) = \sigma(101^7) = \sum_{i = 0}^7 101^i = \frac{101^8 - 1}{101 - 1} = \frac{101^8 - 1}{101 - 1}. \] We use geometric series to calculate the sum here.
\(\textbf{Note. }\)If \(p\) is a prime number, then \(\sigma(p) = p + 1\).
\(\textbf{Note. }\)If \(p\) is a prime number, then \(\sigma(p^k) = \frac{p^k - 1}{p - 1}\).
\(\textbf{Proposition 1. }\)Let \(a, b, n\in\mathbb{Z}^+\) where \(n = ab\). Suppose that \(d\mid n\) where \(d\in\mathbb{Z}^+\). Then \(d = d_1d_2\) where \(d_1\mid a\) and \(d_2\mid b\).
\(\textbf{Proof. }\)Let \(d_1 = \gcd(a, d)\) and \(d_2 = d/\gcd(a, d)\). Since \(d\mid ab\), we can get \(ab = dk) for some \(k\in\mathbb{Z}\). Since \(d = d_1d_2\), we have \[ ab = d_1d_2k. \] Hence, we can get \[ \frac{ab}{d_1} = d_2k. \] We let \(\frac{a}{d_1} = l\). Then \[ bl = d_2k. \] Thus, \(d_2\mid bl).
\(\textbf{Definition. }\)Let \(n\in\mathbb{Z}^+\), we define the Euler function \(\varphi(n)\) as the number of positive integers less than or equal to \(n\) that are relatively prime to \(n\).
\(\textbf{Example. }\)Since \(n = 101\) is a prime number, we have \(\varphi(101) = 101 - 1 = 100\).
\(\textbf{Example. }\)Let \(n = 101^7\). We have \[ \varphi(n) = \varphi(101^7) = 101^7 - 101^6 = 101^6(101 - 1) = 101^6\cdot 100. \]
\(\textbf{Note. }\)If \(p\) is a prime number, then \(\varphi(p) = p - 1\).
\(\textbf{Note. }\)If \(p\) is a prime number, then \(\varphi(p^k) = p^k - p^{k-1}\).
\(\textbf{Definition. }\)Let \(f:\mathbb{Z}^+\to\mathbb{C}\) be an arithmetic function. We say that \(f\) is multiplicative if \(f(1) = 1\) and \(f(ab) = f(a)f(b)\) for all \(a, b\in\mathbb{Z}^+\) where \(\gcd(a, b) = 1\).
\(\textbf{Example. }\)The tau function \(\tau(n)\) is multiplicative.
\(\textbf{Example. }\)The \(\sigma\) function \(\sigma(n)\) is multiplicative.
\(\textbf{Example. }\)The Euler function \(\varphi(n)\) is multiplicative.
\(\textbf{Example. }\)The legendre formula \(\left(\frac{a}{p}\right)\) is multiplicative.
\(\textbf{Proposition 2. }\)Let \(f:\mathbb{Z}^+\to\mathbb{C}\) be a multiplicative function. We define \(g:\mathbb{Z}^+\to\mathbb{C}\) as \[ g(n) = \sum_{d\mid n}f(d). \] Then \(g\) is multiplicative.
\(\textbf{Proof. }\)We need to define a bijective map \(\phi:\mathbb{Z}^+\to\mathbb{Z}^+\to\mathbb{Z}^+\) such that \(g(n) = \sum_{d\mid n}f(d) = \sum_{d\mid n}f(\phi(d)) = h(n)\).