In 18th century, as one of the most outstanding mathematician, Euler first defined the Euler function $ \varphi(n) $ of a positive integer $ n $ to be the number of positive integers not greater than $ n $ but prime to $ n $ [1]. It's well known that as one of the important number theory functions, Euler function was applied widely. Euler function played a key role in RSA public-key cryptosystem since 1970's, and it is also one of the important tools to seek the theoretical basis for the generators of circle groups. There were many interesting open problems on Euler function [2]. For example, Carmichael conjectured that for any positive integer $ n $, there exists a positive integer $ m $ such that $ m\neq n $ and $ \varphi(m) = \varphi(n) $. And then Schinzel conjectured that for any fixed positive integer $ k $, the equation $ \varphi(n+k) = \varphi(n) $ has infinitely many positive integer solutions for $ n $.

On the other hand, in 1938, for any odd prime $ p $, Lehmer [3] established the following important congruence identity

$ \sum\limits_{i = 1}^{\frac{{p - 1}}{2}} {\frac{1}{i}} \equiv - 2{q_2}(p) + pq_2^2(p)({\rm{mod}}{\mkern 1mu} \;{p^2}), $

(1.1)

where $ q_{r}(n) $ denotes the Euler quotient, i.e., $ q_{r}(n) = \frac{r^{\varphi(n)}-1}{n}, n $ and $ r\geq 2 $ are both natural numbers with $ \gcd(n, r) = 1. $

By using (1.1) and the others similar congruences identity, Lehmer obtained many ways to prove the first case of the well-known Fermat's last theorem [4]. Until 2002 and 2007, basing on (1.1) and the other congruence identities given by Lehmer, Cai, etc [5, 6] generalized the modulo from the square of a prime to the square of any positive integer, and defined the generalized Euler function for any positive integer $ n $ to be

$ {\varphi _e}(n) = \sum\limits_{i = 1, \gcd (i, n) = 1}^{[\frac{n}{e}]} 1 , $

i.e., $ \varphi_e(n) $ is the number of positive integers not greater than $ [\frac{n}{e}] $ but prime to $ n $, where $ e $ is a positive integer and $ [x] $ is the greatest integer which is not greater than $ x $. It's easy to verify that $ \varphi_1(n) = \varphi(n) $ is the known Euler function of $ n $, and

$ {\varphi _e}(n) = \sum\limits_{d\mid n} \mu (\frac{n}{d})[\frac{d}{e}], $

(1.2)

where $ \mu(n) $ is the Möbius function, i.e.,

$ \begin{equation} \mu(n) = \left\{ \begin{aligned} 1, \quad &n = 1;\\ {(-1)^k}, \quad &n\geq 2 \quad \mbox{and} \quad {\alpha_1 = \cdots = \alpha_k = 1};\\ 0, \quad &n\geq 2\ \mbox{and there is some}\ {\alpha_i>1(1\leq i\leq k)}, \end{aligned}\right. \end{equation} $

(1.3)

when $ n = \prod\limits_{i = 1}^k p_i^{\alpha_i}(\alpha_i\geq1) $ is a positive integer and $ p_1, \cdots, p_k $ are distinct primes. Easily to see that for any positive integer $ n\geq 2 $,

$ \sum\limits_{d|n} \mu (\frac{n}{d}) = 0. $

(1.4)

On the other hand, for $ e = 1 $, the following computing formula for the generalized Euler function is well-known

$ \varphi (n) = \prod\limits_{i = 1}^k {(p_i^{{\alpha _i}} - p_i^{{\alpha _i} - 1})} . $

Therefore one can naturally to ask the following

Question   For any fixed positive integer $ e $, determine the explicit algorithm formula for the generalized Euler function $ \varphi_e (n) $.

Fixed a positive integer $ n = \prod\limits^k_{i = 1}p^{\alpha_i}_i\geq 2 $, where $ p_1, \cdots, p_k $ are distinct primes and $ \alpha_1, \cdots, \alpha_k $ are positive integers, denote $ \Omega(n) = \sum\limits^k_{i = 1}\alpha_i $ and $ \omega(n) = k $. Especially, $ \Omega(1) = \omega(1) = 0 $.

In recent years, Cai etc [7, 8] obtained the accurate calculation formula for $ \varphi_e(n)\; (e = 2, 3, 4, 6) $, and then, by using properties for Legendre or Jacobi symbols, they also got some necessary and sufficient conditions for that $ \varphi_e(n) $ and $ \varphi_e(n+1)(e = 2, 3, 4) $ are both odd or even numbers.

Proposition 1.1 [7, 8] Let $ p_1, \cdots, p_k $ be distinct primes, $ \alpha_1, \cdots, \alpha_k $ be positive integers, and $ n_1 = \prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}} $.

(1)   If $ \gcd(p_{i}, 3) = 1\; (i = 1, \cdots, k) $ and $ n = 3^{\alpha}n_1>3 $, then

$ \varphi_{3}(n) = \left\{ \begin{array}{ll} \frac{\varphi(n)}{3} +\frac{(-1)^{\Omega(n)}2^{\omega(n)-\alpha-1}}{3} , & \alpha\in\{0, 1\}\ \mbox{and}\ p_{i}\equiv2\; (\mbox{mod}\ 3)\; (i = 1, \cdots, k); \\ \frac{\varphi(n)}{3} , & \mbox{otherwise}. \end{array} \right. $

(2)    If $ \alpha $ is a nonnegative integer and $ n = 2^{\alpha}n_1>4, $ then

$ \varphi_{4}(n) = \left\{ \begin{array}{ll} \frac{\varphi(n)}{4} +\frac{(-1)^{\Omega(n)}2^{\omega(n)-\alpha}}{4} , & \alpha\in\{0, 1\}\ \mbox{and}\ p_{i}\equiv3\; (\mbox{mod}\ 4)\; (i = 1, \cdots, k); \\ \frac{\varphi(n)}{4}, & \mbox{otherwise}. \end{array} \right. $

(3)   If $ \gcd(p_{i}, 6) = 1(i = 1, \cdots, k) $ and $ n = 2^{\alpha}3^{\beta}n_1>6 $, then

$ \varphi_{6}(n) = \left\{ \begin{array}{ll} \frac{1}{6} \varphi(n)+\frac{(-1)^{\Omega(n)}2^{\omega(n)+1-\beta}}{6}, &\alpha = 0, \beta\in\{0, 1\}\ \mbox{and}\ p_{i}\equiv5\; (\mbox{mod}\ 6)\; (i = 1, \cdots, k);\\ \frac{1}{6} \varphi(n)+\frac{(-1)^{\Omega(n)}2^{\omega(n)-1-\beta}}{6}, &\alpha = 1, \beta\in\{0, 1\}\ \mbox{and}\ p_{i}\equiv5\; (\mbox{mod}\ 6)\; (i = 1, \cdots, k);\\ \frac{1}{6} \varphi(n)-\frac{(-1)^{\Omega(n)}2^{\omega(n)-\beta}}{6}, &\alpha\geq2, \beta\in\{0, 1\}\ \mbox{and}\ p_{i}\equiv5\; (\mbox{mod}\ 6)\; (i = 1, \cdots, k);\\ \frac{1}{6} \varphi(n), & \mbox{otherwise}. \end{array} \right. $

Recently, we [9] obtained the formula for $ \varphi_5(n) $ and some sufficient conditions for $ 2|\varphi_5(n) $. The present paper continues the study, based on the elementary methods and techniques, the computing formula for $ \varphi_e(n)\; (e = p, p^2, pq) $ is obtained, where $ p $ and $ q $ are distinct primes (Theorems 1.1–1.5).

For convenience, throughout the paper, we assume that $ p, \; q, \; p_1, \cdots, p_k $ are distinct primes, $ \alpha_1, \cdots, \alpha_k $ are positive integers, $ \alpha $ and $ \beta $ are both nonnegative integers, and

$ n_1 = \prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}}. $

Theorem 1.1   If $ n = p^{\alpha}n_1>p, $ then

$ \varphi_{p}(n) = \left\{ \begin{array}{ll} \frac{\varphi(n)}{p} +\frac{(p-2)(-1)^{\Omega(n)}2^{\omega(n)-\alpha-1}}{p} , & \alpha\in\{0, 1\}\ \mbox{and}\ p_{i}\equiv-1\; (\mbox{mod}\ p)\; (i = 1, \cdots, k); \\ \frac{\varphi(n)}{p}, & \alpha\geq2\mbox{or}\ \alpha\in\{0, 1\}\ \mbox{and}\ p_{i}\equiv1\; (\mbox{mod}\ p)\; (i = 1, \cdots, k). \end{array} \right. $

Theorem 1.2   If $ n = p^{\alpha}n_1>p^2, $ then

$ \varphi_{p^{2}}(n) = \left\{ \begin{array}{ll} \varphi_{p^{2}}(n_{1}), &\alpha = 0;\\ \varphi_{p}(n_{1})-\varphi_{p^{2}}(n_{1}), &\alpha = 1;\\ \varphi(n_{1}) -\varphi_{p}(n_{1}), &\alpha = 2;\\ \frac{1}{p^{2}}\varphi(n), &\alpha\geq3. \end{array} \right. $

Theorem 1.3   For $ n = p^{\alpha}n_1>p^2 $ and $ \alpha\leq 2 $.

(1)    If $ \alpha = 0 $, then

$ \varphi_{p^{2}}(n) = \left\{\begin{array}{ll} \frac{1}{p^{2}}\varphi(n), & p_{i}\equiv 1(\mbox{mod}\ p^{2})(i = 1, \cdots, k).\\ \frac{1}{p^{2}}\varphi(n)+\frac{(p^2-2)(-1)^{\Omega(n)}2^{\omega(n)-1}}{p^2}, & p_{i}\equiv -1\; (\mbox{mod}\ p^{2})\; (i = 1, \cdots, k). \end{array} \right. $

(2)    If $ \alpha = 1 $, then

$ \varphi_{p^2}(n) = \left\{\begin{array}{ll} \frac{\varphi(n)}{p^2}, & p_i\equiv 1(\mbox{mod}\ p^2)(i = 1, \cdots, k);\\ \frac{\varphi(n)}{p^2}+\frac{(p-1)(-1)^{\Omega(n)}2^{\omega(n)-1}}{p^2}, & p_i\equiv -1\; (\mbox{mod}\ p^2)\; (i = 1, \cdots, k). \end{array}\right. $

(3)    If $ \alpha = 2 $, then

$ \varphi_{p^{2}}(n) = \left\{\begin{array}{ll} \frac{\varphi(n)}{p^2}, & p_i\equiv1(\mbox{mod}\ p)(i = 1, \cdots, k).\\ \frac{\varphi(n)}{p^2}+\frac{(p-2)(-1)^{\Omega(n)-1}2^{\omega(n)-2}}{p}, & p_i\equiv -1\; (\mbox{mod}\ p)\; (i = 1, \cdots, k). \end{array}\right. $

Theorem 1.4   If $ n = p^{\alpha}q^{\beta}n_{1}>pq $, then

$ \varphi_{pq}(n) = \left\{ \begin{array}{ll} \varphi_{pq}(n_{1}), & \alpha = \beta = 0; \\ \varphi_{q}(n_{1})- \varphi_{pq}(n_{1}), & \alpha = 1, \beta = 0;\\ \varphi_{p}(n_{1})- \varphi_{pq}(n_{1}), & \alpha = 0, \beta = 1;\\ \varphi(n_{1})+ \varphi_{pq}(n_{1})- \varphi_{p}(n_{1})- \varphi_{q}(n_{1}), & \alpha = \beta = 1;\\ \varphi_{q}(p^{\alpha-1}n_{1}), & \alpha\geq2, \beta = 0;\\ \frac{1}{p(q-1)}\varphi(n)-\varphi_{q}(p^{\alpha-1}n_{1}), & \alpha\geq2, \beta = 1;\\ \frac{1}{pq} \varphi(n) , & \alpha\geq 2, \beta\geq2. \end{array} \right. $

Theorem 1.5   For $ n = p^{\alpha}q^{\beta}n_{1}>pq $.

(1)    If $ \alpha = \beta = 0 $, then

$ \varphi_{pq}(n) = \left\{\begin{array}{ll} \frac{\varphi(n)}{pq}, & p_i\equiv 1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k); \\ \frac{1}{pq}\varphi(n)+\frac{(pq-2)(-1)^{\Omega(n)}2^{\omega(n)-1}}{pq}, & p_i\equiv -1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k). \end{array} \right. $

(2)    If $ \alpha = 1 $ and $ \beta = 0 $, then

$ \varphi_{pq}(n) = \left\{\begin{array}{ll} \frac{\varphi(n)}{pq}, & p_i\equiv 1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k); \\ \frac{\varphi(n)}{pq}+\frac{(p-1)(-1)^{\Omega(n)}2^{\omega(n)-1}}{pq}, & p_i\equiv -1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k). \end{array} \right. $

(3)    If $ \alpha = 0 $ and $ \beta = 1 $, then

$ \varphi_{pq}(n) = \left\{\begin{array}{ll} \frac{\varphi(n)}{pq}, & p_i\equiv 1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k); \\ \frac{\varphi(n)}{pq}+\frac{(q-1)(-1)^{\Omega(n)}2^{\omega(n)-1}}{pq}, & p_i\equiv -1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k). \end{array} \right. $

(4)    If $ \alpha = \beta = 1 $, then

$ \varphi_{pq}(n) = \left\{\begin{array}{ll} \frac{\varphi(n)}{pq}, & p_i\equiv 1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k); \\ \frac{\varphi(n)}{pq}+\frac{(2p+2q-pq-2)}{pq}(-1)^{\Omega(n)}2^{\omega(n)-3}, & p_i\equiv -1\; (\mbox{mod}\ pq)\; (i = 1, \cdots, k). \end{array} \right. $

(5)    If $ \alpha\geq 2 $ and $ \beta = 0 $, then

$ \varphi_{pq}(n) = \left\{\begin{array}{lll} \frac{1}{pq}\varphi(n), & p\equiv p_{i}\equiv1\; (\mbox{mod}\ q)\; (i = 1, \cdots, k); \\ \frac{1}{pq}\varphi(n)-\frac{(q-2)(-1)^{\Omega(n)}\cdot2^{\omega(n)-1}}{q}, & p\equiv p_{i}\equiv-1\; (\mbox{mod}\ q)\; (i = 1, \cdots, k).\\ \end{array} \right. $

(6)    If $ \alpha\geq 2 $ and $ \beta = 1 $, then

$ \varphi_{pq}(n) = \left\{\begin{array}{lll} \frac{1}{pq}\varphi(n), & p\equiv p_{i}\equiv1\; (\mbox{mod}\ q)\; (i = 1, \cdots, k); \\ \frac{1}{pq}\varphi(n)-\frac{(q-2)(-1)^{\Omega(n)}2^{\omega(n)-2}}{q}, & p\equiv p_{i}\equiv-1\; (\mbox{mod}\ q)\; (i = 1, \cdots, k).\\ \end{array} \right. $

(7)    If $ \beta\geq 2 $ and $ \alpha\in\{0, 1\} $, then

$ \varphi_{pq}(n) = \left\{\begin{array}{lll} \frac{1}{pq}\varphi(n), & q\equiv p_{i}\equiv1\; (\mbox{mod}\ p)\; (i = 1, \cdots, k); \\ \frac{1}{pq}\varphi(n)-\frac{(p-2)(-1)^{\Omega(n)}\cdot2^{\omega(n)-\alpha-1}}{p}, & q\equiv p_{i}\equiv-1\; (\mbox{mod}\ p)\; (i = 1, \cdots, k).\\ \end{array} \right. $

Remark   By taking $ p = 3 $ in Theorem 1.1, or $ p = 2 $ in Theorems 1.2–1.3, one can get (1) or (2) of Proposition 1.1, respectively. And by taking $ p = 2 $ and $ q = 3 $ in Theorems 1.4–1.5, one can get (3) of Proposition 1.1. The details is left to interested readers.

Proof for Theorem 1.1   (1)   If $ \alpha = 0 $ and $ p_{i}\equiv-1\; ({\rm mod}\ p)\; (i = 1, \cdots, k) $, i.e., $ n = n_1 $ and for any $ d\mid n_1, d\equiv\pm1({\rm mod}\ p) $. Then by (1.2)–(1.4), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{p}(n) & = \sum\limits_{d\mid n_{1}}\mu\left(\frac{n_{1}}{d}\right)\left[\frac{d}{p}\right]\\ & = \sum\limits_{d\mid n_{1}, d\equiv1({\rm mod} \ p)}\mu\left(\frac{n_{1}}{d}\right)\cdot\left(\frac{d-1}{p}\right)+\sum\limits_{d\mid n_1, d\equiv-1({\rm mod} \ p)}\mu\left(\frac{n_{1}}{d}\right)\cdot\left(\frac{d-(p-1)}{p}\right)\\ & = \frac{1}{p}\left(\sum\limits_{\substack{d\mid n_{1}\\{d\equiv1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)d+\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)d\right)\\ &\quad-\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)+\frac{1}{p}\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)-\frac{1}{p}\sum\limits_{\substack{d\mid n_{1}\\{d\equiv1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)\\ & = \frac{1}{p}\varphi(n_1)-\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)+\frac{1}{p}\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)-\frac{1}{p}\sum\limits_{\substack{d\mid n_{1}\\{d\equiv1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)\\ & = \frac{1}{p}\varphi(n_1)-\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)+\frac{2}{p}\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)-\frac{1}{p}\sum\limits_{\substack{d\mid n_{1}}}\mu\left(\frac{n_{1}}{d}\right)\\ & = \frac{1}{p}\varphi(n_1)-\left(\frac{p-2}{p}\right)\cdot\sum\limits_{\substack{d\mid n_{1}\\{d\equiv-1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)\\ & = \frac{1}{p}\varphi(n_1)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{\substack{d\mid n_{1}\\{d\equiv1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{d}\right)\\ & = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{\substack{(\prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}-1})t\mid n_{1}\\{(\prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}-1})t\equiv1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{n_{1}}{(\prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}-1})t}\right)\\ & = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{\substack{t\mid\prod\limits_{i = 1}^{k}p_{i} \\{(-1)^{\Omega(n)-k}t\equiv1({\rm mod}\thinspace\ p)}}}\mu\left(\frac{\prod\limits_{i = 1}^{k}p_{i}}{t}\right).\end{array} \end{equation*} $

(2.1)

(a)    If $ 2\mid\Omega(n) $ and $ 2\mid k $, then by (2.1) and $ n = n_1 $, we have $ \Omega(n) = \Omega(n_1), \omega(n) = \omega(n_1) $ and

$ \begin{equation*} \begin{array}{ll} \varphi_{p}(n)& = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot \sum\limits_{\substack{t\mid\prod\limits_{i = 1}^{k}p_{i}\\{(-1)^{\Omega(n)-k}t\equiv1({\rm mod}\ p)}}}\mu\left(\frac{\prod\limits_{i = 1}^{k}p_{i}}{t}\right) \\ & = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{\substack{t\mid\prod\limits_{i = 1}^{k}p_{i}\\{t\equiv1({\rm mod}\ p)}}}\mu\left(\frac{\prod\limits_{i = 1}^{k}p_{i}}{t}\right)\\ & = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{j = 0}^{\frac{k}{2}}{{k}\choose {2j}}(-1)^{k-2j} = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot2^{k-1}\\ & = \frac{1}{p}\varphi(n)+\frac{(p-2)(-1)^{\Omega(n_{1})}2^{\omega(n_{1})-1}}{p}\\ & = \frac{1}{p}\varphi(n)+\frac{(p-2)(-1)^{\Omega(n)}2^{\omega(n)-\alpha-1}}{p}. \end{array} \end{equation*} $

(b)    If $ 2\mid\Omega(n) $ and $ 2\nmid k $, then by (2.1) we have

$ \begin{equation*} \begin{array}{ll} \varphi_{p}(n)& = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{\substack{t\mid\prod\limits_{i = 1}^{k}p_{i}\\{(-1)^{\Omega(n)-k}t\equiv1({\rm mod}\ p)}}}\mu\left(\frac{\prod\limits_{i = 1}^{k}p_{i}}{t}\right) \\ & = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{\substack{t\mid\prod\limits_{i = 1}^{k}p_{i}\\{t\equiv-1({\rm mod}\ p)}}}\mu\left(\frac{\prod\limits_{i = 1}^{k}p_{i}}{t}\right)\\ & = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot\sum\limits_{j = 1}^{\frac{k+1}{2}}{{k}\choose {2j-1}}(-1)^{k-2j+1} = \frac{1}{p}\varphi(n)+\left(\frac{p-2}{p}\right)\cdot 2^{k-1}\\ & = \frac{1}{p}\varphi(n)+\frac{(p-2)(-1)^{\Omega(n_{1})}2^{\omega(n_{1})-1}}{p}\\ & = \frac{1}{p}\varphi(n)+\frac{(p-2)(-1)^{\Omega(n)}2^{\omega(n)-\alpha-1}}{p}. \end{array} \end{equation*} $

For the case $ 2\nmid\Omega(n) $ and $ 2\nmid k $ or $ 2\nmid\Omega(n) $ and $ 2\mid k $, in the same proof, we can get

$ \begin{align} \varphi_{p}(n) = \frac{1}{p}\varphi(n)+\frac{(p-2)(-1)^{\Omega(n)}2^{\omega(n)-\alpha-1}}{p}. \end{align} $

(2.2)

(2)    If $ \alpha = 1 $ and for any $ i = 1, \cdots, k, p_{i}\equiv-1({\rm mod}\ p) $, i.e., $ n = pn_1 $ and for any $ d\mid n_1, d\equiv\pm1({\rm mod}\ p) $. Then by (1.2)–(1.3), (2.2) and $ \gcd(p, n_1) = 1 $, we have

$ \varphi(n) = (p-1)\varphi(n_1), \ \Omega(n) = \Omega(n_1)+1, \omega(n) = \omega(n_1)+1, $

and then

$ \begin{equation*} \begin{array}{ll} \varphi_{p}(n) & = \varphi_{p}(pn_{1}) = \sum\limits_{d\mid pn_{1}}\mu\left(\frac{pn_{1}}{d}\right)\left[\frac{d}{p}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p}\right]+\sum\limits_{pd_{1}\mid pn_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{pd_{1}}{p}\right]\\ & = -\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}\\ & = -\varphi_{p}(n_{1})+\varphi(n_{1})\\ & = -\frac{1}{p}\varphi(n_{1})-\frac{(p-2)(-1)^{\Omega(n_{1})}2^{\omega(n_{1})-1}}{p}+\varphi(n_{1})\\ & = \left(\frac{p-1}{p}\right)\cdot\varphi(n_{1})+\frac{(p-2)(-1)^{\Omega(n_{1})+1}2^{\omega(n)-2}}{p}\\ & = \frac{1}{p}\varphi(n)+\frac{(p-2)(-1)^{\Omega(n)}2^{\omega(n)-\alpha-1}}{p}. \end{array} \end{equation*} $

(2.3)

(3)    If $ \alpha\geq2 $, i.e., $ n = p^{\alpha}n_1 $, then by $ \gcd(p, n_1) = 1 $ and (1.2)–(1.3), we have

$ \varphi(n) = \varphi(p^{\alpha}n_1) = (p^{\alpha}-p^{\alpha-1})\varphi(n_{1}), $

and then

$ \begin{equation*} \begin{array}{ll} \varphi_{p}(n) & = \varphi_{p}(p^{\alpha}n_{1}) = \sum\limits_{d\mid p^{\alpha}n_{1}}\mu\left(\frac{p^{\alpha}n_{1}}{d}\right)\left[\frac{d}{p}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{\alpha}n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p}\right]+\sum\limits_{d_{1}\mid n_1, 1\leq\beta\leq\alpha}\mu\left(\frac{p^{\alpha}n_{1}}{p^{\beta}d_{1}}\right)\left[\frac{p^{\beta}d_{1}}{p}\right]\\ & = \sum\limits_{\substack{d_{1}\mid n_{1}\\{1\leq\beta\leq\alpha}}}\mu\left(\frac{p^{\alpha-\beta}n_{1}}{d_{1}}\right)p^{\beta-1}d_{1}\\ & = p^{\alpha-1}\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}+p^{\alpha-2}\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)d_{1}\\ & = p^{\alpha-1}\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}-p^{\alpha-2}\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}\\ & = (p^{\alpha-1}-p^{\alpha-2})\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}\\ & = (p^{\alpha-1}-p^{\alpha-2})\varphi(n_{1}) = \frac{\varphi(n)}{p}. \end{array} \end{equation*} $

(2.4)

(4)    If $ \alpha = 0 $ and $ p_{i}\equiv1({\rm mod} \ p)(i = 1, \cdots, k) $, i.e., $ n = n_1 $ and for any $ d\mid n_1, d\equiv1({\rm mod}\ p) $. Then by (1.2) and (1.4), we have

$ \begin{align} \varphi_{p}(n) = \sum\limits_{d\mid n_{1}}\mu\left(\frac{n}{d}\right)\left[\frac{d}{p}\right] = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\cdot\left(\frac{d-1}{p}\right) = \frac{1}{p}\sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)d-\frac{1}{p}\sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right) = \frac{1}{p}\varphi(n). \end{align} $

(2.5)

(5)    If $ \alpha = 1 $ and $ p_{i}\equiv1({\rm mod}\ p)(i = 1, \cdots, k) $, i.e., $ n = pn_1 $, then $ \varphi(n) = (p-1)\varphi(n_1) $ and for any $ d\mid n_1, d\equiv1({\rm mod}\ p) $. Thus by (1.2)–(1.3), $ \gcd(p, n_1) = 1 $ and (4) we have

$ \begin{equation*} \begin{array}{ll} \varphi_{p}(n) & = \sum\limits_{d\mid pn_{1}}\mu\left(\frac{pn_{1}}{d}\right)\left[\frac{d}{p}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p}\right]+\sum\limits_{pd_{1}\mid pn_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{pd_{1}}{p}\right]\\ & = -\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1} \\ & = -\varphi_{p}(n_{1})+\varphi(n_{1}) = -\frac{1}{p}\varphi(n_{1})+\varphi(n_{1})\\ & = \left(\frac{p-1}{p}\right)\cdot\varphi(n_{1}) = \frac{1}{p}\varphi(n). \end{array} \end{equation*} $

(2.6)

Now from (2.2)–(2.6), Theorem 1.1 is proved.

Proof for Theorem 1.2   (1)   For the case $ \alpha = 0 $, the result is obvious.

(2)    If $ \alpha = 1 $, i.e., $ n = pn_1 $, then by $ \gcd(p, n_1) = 1 $ and (1.2)–(1.3), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{p^{2}}(n) & = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\left[\frac{d}{p^{2}}\right] = \sum\limits_{d\mid pn_{1}}\mu\left(\frac{pn_{1}}{d}\right)\left[\frac{d}{p^{2}}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p^{2}}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_1}\right)\left[\frac{d_{1}}{p}\right]\\ & = -\varphi_{p^{2}}(n_{1})+\varphi_{p}(n_{1}). \end{array} \end{equation*} $

(2.7)

(3)    If $ \alpha = 2 $, i.e., $ n = p^2n_1 $, then by $ \gcd(p, n_1) = 1 $ and (1.2)–(1.3), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{p^{2}}(n) & = \sum\limits_{d\mid p^{2}n_{1}}\mu\left(\frac{p^{2}n_{1}}{d}\right)\left[\frac{d}{p^{2}}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{2}n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p^{2}}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{2}n_{1}}{pd_{1}}\right)\left[\frac{pd_{1}}{p^{2}}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{2}n_{1}}{p^{2}d_{1}}\right)\left[\frac{p^{2}d_{1}}{p^{2}}\right]\\ & = -\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}\\ & = -\varphi_{p}(n_{1})+\varphi(n_{1}). \end{array} \end{equation*} $

(2.8)

(4)    If $ \alpha\geq3 $, i.e., $ n = p^{\alpha}n_1 $, then by $ \gcd(p, n_1) = 1 $ and (1.2)–(1.3), we have $ \varphi(n) = p^2(p^{\alpha-2}-p^{\alpha-3})\varphi(n_1), $ and then

$ \begin{equation*} \begin{array}{ll} \varphi_{p^{2}}(n) & = \sum\limits_{d\mid p^{\alpha}n_{1}}\mu\left(\frac{p^{\alpha}n_{1}}{d}\right)\left[\frac{d}{p^{2}}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{\alpha}n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p^{2}}\right]+\sum\limits_{\substack{d_{1}\mid n_{1}, 1\leq\alpha_{1}\leq\alpha}}\mu\left(\frac{p^{\alpha}n_{1}}{p^{\alpha_1}d_{1}}\right)\left[\frac{p^{\alpha_1}d_{1}}{p^{2}}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d}\right)\left[\frac{p^{\alpha-1}d_{1}}{p^{2}}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha}d_{1}}{p^{2}}\right]\\ & = -p^{\alpha-3}\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}+p^{\alpha-2}\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)d_{1}\\ & = -p^{\alpha-3}\varphi(n_{1})+p^{\alpha-2}\varphi(n_{1})\\ & = p^{\alpha-3}(p-1)\varphi(n_{1}) = \frac{1}{p^{2}}\varphi(n). \end{array} \end{equation*} $

(2.9)

Now from (2.7)–(2.9), we complete the proof of Theorem 1.2.

Proof for Theorem 1.3    (1)   If $ \alpha = 0 $, i.e., $ n = n_1, $ and then $ \gcd(n, p) = 1 $. Suppose that $ p_i\equiv 1\; ({\rm mod}\ p^2)\; (i = 1, \cdots, k) $, then for any $ d\mid n, d\equiv 1\; ({\rm mod}\ p^2) $, thus by (1.2)–(1.4), we have

$ \begin{align} \varphi_{p^2}(n) = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\left[\frac{d}{p^2}\right] = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\left(\frac{d-1}{p^2}\right) = \frac{1}{p^2}\varphi(n)-\frac{1}{p^2}\sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right) = \frac{1}{p^2}\varphi(n). \end{align} $

(2.10)

Suppose that $ p_i\equiv -1\; ({\rm mod}\ p^2)\; (i = 1, \cdots, k) $, i.e, for any $ d\mid n, d\equiv \pm 1\; ({\rm mod}\ p^2) $, and then by (1.2), (1.4) and the proof of Theorem 1.1 (1), we can get

$ \begin{equation*} \begin{array}{ll} \varphi_{p^2}(n) & = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\left[\frac{d}{p^2}\right]\\ & = \sum\limits_{d\mid n, d\equiv 1({\rm mod}\ p^2)}\mu\left(\frac{n}{d}\right)\left(\frac{d-1}{p^2}\right)+\sum\limits_{d\mid n, d\equiv -1({\rm mod}\ p^2)}\mu\left(\frac{n}{d}\right)\left(\frac{d-(p^2-1)}{p^2}\right)\\ & = \frac{1}{p^2}\varphi(n)-\frac{1}{p^2}\sum\limits_{d\mid n, d\equiv 1({\rm mod}\ p^2)}\mu\left(\frac{n}{d}\right)-\left(\frac{p^2-1}{p^2}\right)\sum\limits_{d\mid n, d\equiv -1({\rm mod}\ p^2)}\mu\left(\frac{n}{d}\right)\\ & = \frac{1}{p^2}\varphi(n)+\left(\frac{p^2-2}{p^2}\right)\sum\limits_{d\mid n, d\equiv 1({\rm mod}\ p^2)}\mu\left(\frac{n}{d}\right)\\ & = \frac{1}{p^2}\varphi(n)+\left(\frac{p^2-2}{p^2}\right)(-1)^{\Omega(n)}2^{\omega(n)-1}. \end{array} \end{equation*} $

(2.11)

Now from (2.10)–(2.11), we complete the proof of (1).

(2)    If $ \alpha = 1 $, i.e, $ n = pn_1, $ then by $ \gcd(p, n_1) = 1 $, we have

$ \begin{align} \varphi(n) = (p-1)\varphi(n_1), \Omega(n) = \Omega(n_1)+1, \omega(n) = \omega(n_1)+1. \end{align} $

(2.12)

And so by Theorems 1.1–1.2, (2.12) and (1), we can obtain

$ \begin{equation*} \begin{array}{ll} &\varphi_{p^2}(n) = \varphi_{p}(n_1)-\varphi_{p^2}(n_1)\\ = &\left\{\begin{array}{ll} \frac{\varphi(n_1)}{p}-\frac{\varphi(n_1)}{p^2}, & p_i\equiv 1({\rm mod}\ p^2)(i = 1, \cdots, k);\\ \frac{\varphi(n_1)}{p}-\frac{\varphi(n_1)}{p^2}+\frac{(p-2)(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}}{p}-\frac{(p^2-2)(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}}{p^2}, & p_i\equiv -1({\rm mod}\ p^2)\; (i = 1, \cdots, k);\end{array}\right.\\ = &\left\{\begin{array}{ll} \frac{\varphi(n)}{p^2}, & p_i\equiv 1({\rm mod}\ p^2)\; (i = 1, \cdots, k);\\ \frac{\varphi(n)}{p^2}+\frac{(p-1)(-1)^{\Omega(n_1)+1}2^{\omega(n_1)}}{p^2}, & p_i\equiv -1({\rm mod}\ p^2)\; (i = 1, \cdots, k);\end{array}\right.\\ = &\left\{\begin{array}{ll} \frac{\varphi(n)}{p^2}, & p_i\equiv 1({\rm mod}\ p^2)\; (i = 1, \cdots, k);\\ \frac{\varphi(n)}{p^2}+\frac{(p-1)(-1)^{\Omega(n)}2^{\omega(n)-1}}{p^2}, & p_i\equiv -1({\rm mod}\ p^2)\; (i = 1, \cdots, k). \end{array}\right. \end{array} \end{equation*} $

This completes the proof of (2).

(3)    If $ \alpha = 2 $, i.e, $ n = p^2n_1 $, then by $ \gcd(p, n_1) = 1 $, we have

$ \begin{align} \varphi(n) = p(p-1)\varphi(n_1), \Omega(n) = \Omega(n_1)+2, \omega(n) = \omega(n_1)+1. \end{align} $

(2.13)

Thus by Theorems 1.1–1.2 and (2.13), in the same proof as that of Theorem 1.3 (2), (3) is immediate.

This completes the proof of Theorem 1.3.

Proof for Theorem 1.4   (1)   If $ \alpha = 0, \beta = 0 $, the result is obvious.

(2)    If $ \alpha = 1, \beta = 0 $, i.e., $ n = pn_1 $, then by $ \gcd(pq, n_1) = \gcd(p, q) = 1 $ and (1.2)–(1.3), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n) & = \sum\limits_{d\mid pn_{1}}\mu\left(\frac{pn_{1}}{d}\right)\left[\frac{d}{pq}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{d_{1}}{pq}\right]+\sum\limits_{pd_{1}\mid pn_{1}}\mu\left(\frac{pn_{1}}{pd_{1}}\right)\left[\frac{pd_{1}}{pq}\right] \\ & = -\varphi_{pq}(n_{1})+\varphi_{q}(n_{1}). \end{array} \end{equation*} $

(2.14)

(3)    For the case $ \alpha = 1, \beta = 0 $, in the same proof of (2), the result is obvious.

(4)    If $ \alpha = \beta = 1 $, i.e., $ n = pqn_1 $, then by $ \gcd(pq, n_1) = \gcd(p, q) = 1 $ and (1.2)–(1.3), in the same proof as that of (2), (4) is immediate.

(5)    If $ \alpha\geq2 $ and $ \beta = 0 $, i.e., $ n = p^{\alpha}n_1 $, then by $ \gcd(p, n_1) = 1 $ and (1.2)–(1.3), we can obtain

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n) & = \sum\limits_{d\mid p^{\alpha}n_{1}}\mu\left(\frac{p^{\alpha}n_{1}}{d}\right)\left[\frac{d}{pq}\right] \\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{\alpha}n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{pq}\right]+\sum\limits_{\substack{d_{1}\mid n_{1}, {1\leq \alpha_{1}\leq \alpha}}}\mu\left(\frac{p^{\alpha}n_{1}}{p^{\alpha_{1}}d_{1}}\right)\left[\frac{p^{\alpha_{1}}d_{1}}{pq}\right] \\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}d_{1}}{pq}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha}d_{1}}{pq}\right] \\ & = -\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-2}d_{1}}{q}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}d_{1}}{q}\right]\\ & = -B_{\alpha-2}+B_{\alpha-1}. \end{array} \end{equation*} $

(2.15)

While by $ \alpha\geq 2 $ and (1.2)–(1.4), we know that

$ \begin{equation*} \begin{array}{ll} B_{\alpha-1}& = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}d_{1}}{q}\right] = \sum\limits_{p^{\alpha-1}d_{1}\mid p^{\alpha-1}n_{1}}\mu\left(\frac{p^{\alpha-1}n_{1}}{p^{\alpha-1}d_{1}}\right)\left[\frac{p^{\alpha-1}d_{1}}{q}\right]\\ & = \varphi_q(p^{\alpha-1}n_1)-\sum\limits_{d_{1}\mid n_{1}, 0\leq\beta\leq\alpha-2}\mu\left(\frac{p^{\alpha-\beta-1}n_{1}}{d_{1}}\right)\left[\frac{p^{\beta}d_{1}}{q}\right]\\ & = \varphi_q(p^{\alpha-1}n_1)-\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-2}d_{1}}{q}\right]\\ & = \varphi_q(p^{\alpha-1}n_1)+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-2}d_{1}}{q}\right] = \varphi_q(p^{\alpha-1}n_1)+B_{\alpha-2}, \end{array} \end{equation*} $

(2.16)

thus by (2.15)–(2.16), we have $ \varphi_{pq}(n) = \varphi_q(p^{\alpha-1}n_1) $. Thus (5) is proved.

(6)    If $ \alpha\geq2 $ and $ \beta = 1 $, i.e., $ n = p^{\alpha}qn_1 $, then by $ \gcd(p, n_1) = 1 $, we have

$ \begin{align} \varphi(n) = p^{\alpha-1}(p-1)(q-1)\varphi(n_1). \end{align} $

(2.17)

Thus by (1.2)–(1.4), (2.16) and (2.17), we can get

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n) & = \sum\limits_{d\mid p^{\alpha}qn_{1}}\mu\left(\frac{p^{\alpha}qn_{1}}{d}\right)\left[\frac{d}{pq}\right] \\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{\alpha}qn_{1}}{d_{1}}\right)\left[\frac{d_{1}}{pq}\right]+\sum\limits_{\substack{d_{1}\mid n_{1}, {1\leq\alpha_{1}\leq\alpha}}}\mu\left(\frac{p^{\alpha}qn_{1}}{p^{\alpha_{1}}d_{1}}\right)\left[\frac{p^{\alpha_{1}}d_{1}}{pq}\right] \\ &\quad+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{\alpha}n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{p}\right]+\sum\limits_{\substack{d_{1}\mid n_{1}, {1\leq\alpha_{1}\leq\alpha}}}\mu\left(\frac{p^{\alpha}qn_{1}}{p^{\alpha_{1}}qd_{1}}\right)\left[\frac{p^{\alpha_{1}}qd_{1}}{pq}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pqn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}d_{1}}{pq}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{qn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha}d_{1}}{pq}\right]\\ &\quad+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}d_{1}}{p}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha}d_{1}}{p}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-2}d_{1}}{q}\right]-\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}d_{1}}{q}\right]\\ &\quad-\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left(p^{\alpha-2}d_{1}\right) +\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left(p^{\alpha-1}d_{1}\right)\\ & = B_{\alpha-2}-B_{\alpha-1}-p^{\alpha-2}\varphi(n_1)+p^{\alpha-1}\varphi(n_1) \\ & = -\varphi_q(p^{\alpha-1}n_1)+\frac{1}{p(q-1)}\varphi(n).\end{array} \end{equation*} $

(7)    If $ \alpha\geq2 $ and $ \beta\geq2 $, then by $ \gcd(pq, n_1) = \gcd(p, q) = 1 $, and (1.2)–(1.4), we have

$ \varphi(n) = p^{\alpha-1}q^{\beta-1}(p-1)(q-1)\varphi(n_1), $

and then

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n) & = \sum\limits_{d\mid p^{\alpha}q^{\beta}n_{1}}\mu\left(\frac{p^{\alpha}q^{\beta}n_{1}}{d}\right)\left[\frac{d}{pq}\right] \\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{p^{\alpha}q^{\beta}n_{1}}{d_{1}}\right)\left[\frac{d_{1}}{pq}\right]+\sum\limits_{\substack{d_{1}\mid n_{1}, {1\leq\alpha_{1}\leq\alpha}}}\mu\left(\frac{p^{\alpha}q^{\beta}n_{1}}{p^{\alpha_{1}}d_{1}}\right)\left[\frac{p^{\alpha_{1}}d_{1}}{pq}\right] \\ &\quad+\sum\limits_{\substack{d_{1}\mid n_{1}, 1\leq \beta_{1}\leq\beta }}\mu\left(\frac{p^{\alpha}q^{\beta}n_{1}}{q^{\beta_{1}}d_{1}}\right)\left[\frac{q^{\beta_{1}}d_{1}}{pq}\right]+\sum\limits_{\substack{d_{1}\mid n_{1}, 1\leq\alpha_{1}\leq\alpha, 1\leq\beta_{1}\leq\beta}}\mu\left(\frac{p^{\alpha}q^{\beta}n_{1}}{p^{\alpha_{1}}q^{\beta_{1}}d_{1}}\right)\left[\frac{p^{\alpha_{1}}q^{\beta_{1}}d_{1}}{pq}\right]\\ & = \sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{qn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha}q^{\beta-1}d_{1}}{pq}\right] +\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{n_{1}}{d_{1}}\right)\left[\frac{p^{\alpha}q^{\beta}d_{1}}{pq}\right]\\ &\quad+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}q^{\beta}d_{1}}{pq}\right]+\sum\limits_{d_{1}\mid n_{1}}\mu\left(\frac{pqn_{1}}{d_{1}}\right)\left[\frac{p^{\alpha-1}q^{\beta-1}d_{1}}{pq}\right]\\ & = -p^{\alpha-1}q^{\beta-2}\varphi(n_{1})+p^{\alpha-1}q^{\beta-1}\varphi(n_{1})-p^{\alpha-2}q^{\beta-1}\varphi(n_{1})+p^{\alpha-2}q^{\beta-2}\varphi(n_{1})\\ & = p^{\alpha-2}q^{\beta-2}(-p+pq-q+1)\varphi(n_{1})\\ & = p^{\alpha-2}(p-1)q^{\beta-2}(q-1)\varphi(n_{1}) = \frac{1}{pq}\varphi(n). \end{array} \end{equation*} $

This completes the proof of (7).

Proof for Theorem 1.5   (1)   For the case $ \alpha = \beta = 0 $, i.e., $ n = n_1 $.

(i)    If $ p_i\equiv 1({\rm mod}\ pq)(i = 1, \cdots, k), $ then for any $ d\mid n, d\equiv 1({\rm mod}\ pq). $ Thus by (1.2) and (1.4), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n)& = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\left[\frac{d}{pq}\right] = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\left(\frac{d-1}{pq}\right)\\ & = \frac{1}{pq}\cdot\sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)d-\frac{1}{pq}\cdot\sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right) = \frac{1}{pq}\varphi(n). \end{array} \end{equation*} $

(2.18)

(ii)    If $ p_i\equiv -1({\rm mod}\ pq)(i = 1, \cdots, k), $ i.e., for any $ d\mid n, d\equiv\pm 1({\rm mod}\ pq). $ Then by (1.2)–(1.4) and the proof of Theorem 1.1 (1), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n)& = \sum\limits_{d\mid n, d\equiv 1({\rm mod}\ pq)}\mu\left(\frac{n}{d}\right)\left[\frac{d}{pq}\right]+\sum\limits_{d\mid n, d\equiv -1({\rm mod}\ pq)}\mu\left(\frac{n}{d}\right)\left[\frac{d}{pq}\right] \\ & = \sum\limits_{d\mid n, d\equiv 1({\rm mod}\ pq)}\mu\left(\frac{n}{d}\right)\left(\frac{d-1}{pq}\right)+\sum\limits_{d\mid n, d\equiv-1({\rm mod}\ pq)}\mu\left(\frac{n}{d}\right)\left(\frac{d+1-pq}{pq}\right)\\ & = \sum\limits_{d\mid n}\mu\left(\frac{n}{d}\right)\left(\frac{d-1}{pq}\right)+\sum\limits_{d\mid n_{1}, d\equiv-1({\rm mod}\ pq)}\mu\left(\frac{n}{d}\right)\left(\frac{d-1+2-pq}{pq}\right)\\ & = \frac{1}{pq}\varphi(n)+\left(\frac{2-pq}{pq}\right)\sum\limits_{d\mid n, d\equiv-1({\rm mod}\ pq)}\mu\left(\frac{n}{d}\right) \\ & = \frac{1}{pq}\varphi(n)+\left(\frac{pq-2}{pq}\right)\sum\limits_{d\mid n, d\equiv1({\rm mod}\ pq)}\mu\left(\frac{n}{d}\right)\\ & = \frac{1}{pq}\varphi(n)+\left(\frac{pq-2}{pq}\right)\sum\limits_{\substack{(\prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}-1})t\mid n\\{(\prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}-1})t\equiv1({\rm mod}\thinspace\ pq)}}}\mu\left(\frac{n}{(\prod\limits_{i = 1}^{k}p_{i}^{\alpha_{i}-1})t}\right)\\ & = \frac{1}{pq}\varphi(n)+\left(\frac{pq-2}{pq}\right)\sum\limits_{\substack{t\mid\prod\limits_{i = 1}^{k}p_{i} \\{(-1)^{\Omega(n)-k}t\equiv1({\rm mod}\thinspace\ pq)}}}\mu\left(\frac{\prod\limits_{i = 1}^{k}p_{i}}{t}\right)\\ & = \frac{1}{pq}\varphi(n)+\left(\frac{pq-2}{pq}\right)(-1)^{\Omega(n)}2^{\omega(n)-1}. \end{array} \end{equation*} $

(2.19)

From (2.18)–(2.19), we complete the proof of (1).

(2)    For $ \alpha = 1 $ and $ \beta = 0 $, i.e., $ n = pn_1, $ then by $ \gcd(pq, n_1) = \gcd(p, q) = 1 $, we have

$ \varphi(n) = (p-1)\varphi(n_1), \Omega(n) = \Omega(n_1)+1, \omega(n) = \omega(n_1)+1. $

(i)    If $ p_i\equiv 1({\rm mod}\ pq), $ then $ p_i\equiv 1({\rm mod}\ q)(i = 1, \cdots, k). $ And so by Theorem 1.1, Theorem 1.4 and (2.18), we have

$ \begin{align} \varphi_{pq}(n) = \varphi_q(n_1)-\varphi_{pq}(n_1) = \frac{\varphi(n_1)}{q}-\frac{\varphi(n_1)}{pq} = \frac{\varphi(n)}{q(p-1)}-\frac{\varphi(n)}{pq(p-1)} = \frac{\varphi(n)}{pq}. \end{align} $

(2.20)

(ii)    If $ p_i\equiv -1({\rm mod}\ pq), $ then $ p_i\equiv -1({\rm mod}\ q)(i = 1, \cdots, k), $ thus by Theorem 1.1, Theorem 1.4, and (2.19), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n)& = \varphi_q(n_1)-\varphi_{pq}(n_1)\\ & = \left(\frac{\varphi(n_1)}{q}+\frac{(q-2)(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}}{q}\right)-\left(\frac{1}{pq}\varphi(n_1)+\left(\frac{pq-2}{pq}\right)(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}\right)\\ & = \frac{\varphi(n)}{q(p-1)}-\frac{1}{pq(p-1)}\varphi(n)+\left(\frac{p-1}{pq}\right)(-1)^{\Omega(n_1)+1}2^{\omega(n_1)}\\ & = \frac{1}{pq}\varphi(n)+\left(\frac{p-1}{pq}\right)(-1)^{\Omega(n)}2^{\omega(n)-1}. \end{array} \end{equation*} $

(2.21)

Now from (2.20)–(2.21), we complete the proof of (2).

(3)    For $ \alpha = 0 $ and $ \beta = 1 $, in the same proof as that of (2), the result is immediate.

(4)    For $ \alpha = \beta = 1 $, i.e., $ n = pqn_1, $ then by $ \gcd(pq, n_1) = \gcd(p, q) = 1 $, we have

$ \varphi(n) = (p-1)(q-1)\varphi(n_1), \Omega(n) = \Omega(n_1)+2, \omega(n) = \omega(n_1)+2. $

(i)    If $ p_i\equiv 1({\rm mod}\ pq), $ i.e., $ p_i\equiv 1({\rm mod}\ q) $ and $ p_i\equiv 1({\rm mod}\ q)(i = 1, \cdots, k). $ Then by Theorem 1.1, Theorem 1.4 and (2.18), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n)& = \varphi(n_1)+\varphi_{pq}(n_1)-\varphi_q(n_1)-\varphi_p(n_1)\\ & = \varphi(n_1)+\frac{\varphi(n_1)}{pq}-\frac{\varphi(n_1)}{q}-\frac{\varphi(n_1)}{p}\\ & = \frac{(p-1)(q-1)\varphi(n_1)}{pq}\\ & = \frac{\varphi(n)}{pq}. \end{array} \end{equation*} $

(2.22)

(ii)    If $ p_i\equiv -1({\rm mod}\ pq), $ then by Theorem 1.1, Theorem 1.4 and (2.19), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n)& = \varphi(n_1)+\varphi_{pq}(n_1)-\varphi_q(n_1)-\varphi_p(n_1)\\ & = \varphi(n_1)+\left(\frac{1}{pq}\varphi(n_1)+\left(\frac{pq-2}{pq}\right)(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}\right)\\ &\quad-\left(\frac{\varphi(n_1)}{q}+\frac{(q-2)(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}}{q}\right)-\left(\frac{\varphi(n_1)}{p}+\frac{(p-2)(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}}{p}\right)\\ & = \frac{(p-1)(q-1)\varphi(n_1)}{pq}+\frac{(2p+2q-pq-2)}{pq}(-1)^{\Omega(n_1)}2^{\omega(n_1)-1}\\ & = \frac{\varphi(n)}{pq}+\frac{(2p+2q-pq-2)}{pq}(-1)^{\Omega(n)}2^{\omega(n)-3}. \end{array} \end{equation*} $

(2.23)

Now from (2.22)–(2.23), we complete the proof of (4).

(5)    For $ \alpha\geq 2 $ and $ \beta = 0 $, i.e., $ n = p^{\alpha}n_1, $ then by $ \gcd(n_1, pq) = \gcd(p, q) = 1 $, we have

$ \begin{align} \varphi(n) = p\varphi(p^{\alpha-1}n_1), \Omega(n) = \Omega(p^{\alpha-1}n_1)+1, \omega(n) = \omega(p^{\alpha-1}n_1). \end{align} $

(2.24)

(i)    If $ p\equiv p_i\equiv 1({\rm mod}\ q)(i = 1, \cdots, k), $ then by Theorem 1.1, Theorem 1.4 and (2.24), we have

$ \begin{align} \varphi_{pq}(n) = \varphi_{q}(p^{\alpha-1}n_{1}) = \frac{\varphi(p^{\alpha-1}n_{1})}{q} = \frac{\varphi(n)}{pq}. \end{align} $

(2.25)

(ii)    If $ p\equiv p_i\equiv -1({\rm mod}\ q)(i = 1, \cdots, k), $ then by Theorem 1.1, Theorem 1.4 and (2.24), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n)& = \varphi_{q}(p^{\alpha-1}n_{1})\\ & = \frac{\varphi(p^{\alpha-1}n_{1})}{q}+\frac{(q-2)(-1)^{\Omega(p^{\alpha-1}n_1)}2^{\omega(p^{\alpha-1}n_1)-1}}{q}\\ & = \frac{\varphi(n)}{pq}-\frac{(q-2)(-1)^{\Omega(n)}2^{\omega(n)-1}}{q}. \end{array} \end{equation*} $

(2.26)

Now from (2.25)–(2.26), we complete the proof of (5).

(6)    For $ \alpha\geq 2 $ and $ \beta = 1 $, i.e., $ n = p^{\alpha}qn_1, $ then by $ \gcd(n_1, pq) = \gcd(p, q) = 1 $, we have

$ \begin{align} \varphi(n) = \frac{\varphi(p^{\alpha-1}n_1)}{p(q-1)}, \Omega(n) = \Omega(p^{\alpha-1}n_1)+2, \omega(n) = \omega(p^{\alpha-1}n_1)+1. \end{align} $

(2.27)

(i)    If $ p\equiv p_i\equiv 1({\rm mod}\ q)(i = 1, \cdots, k), $ then by Theorem 1.1, Theorem 1.4 and (2.27), we have

$ \begin{equation*} \begin{array}{ll} \varphi_{pq}(n)& = \frac{\varphi(n)}{p(q-1)}-\varphi_{q}(p^{\alpha-1}n_{1}) = \frac{\varphi(n)}{p(q-1)}-\frac{\varphi(p^{\alpha-1}n_{1})}{q}\\ & = \frac{\varphi(n)}{p(q-1)}-\frac{\varphi(n)}{pq(q-1)} = \frac{\varphi(n)}{pq}.\end{array} \end{equation*} $

(2.28)

(ii)    If $ p\equiv p_i\equiv -1({\rm mod}\ q)(i = 1, \cdots, k), $ then by Theorem 1.1, Theorem 1.4 and (2.27), in the same proof as that of case (5) (ii), one can get

$ \begin{align} \varphi_{pq}(n) = \frac{\varphi(n)}{pq}-\frac{(q-2)(-1)^{\Omega(n)}2^{\omega(n)-2}}{q}. \end{align} $

(2.29)

Now from (2.28)–(2.29), we complete the proof of (6).

(7)    If $ \beta\geq 2 $ and $ \alpha = 0 $ or $ 1 $, in the same proofs as those of (4) and (5), the result is obvious.

From the above, Theorem 1.5 is proved.