1.
Introduction
Let Z and P denote the set of integers and primes, respectively. In order to generalize Lehmer's congruence (see [4] or [7]) for modulo prime squares to be modulo integer squares, Cai et al. [1] defined the following generalized Euler function for a positive integer n related to a given positive integer e:
where [x] is the greatest integer not more than x, i.e., φe(n) is the number of positive integers not greater than [ne] and prime to n. It is clear that φ1(n)=φ(n) is just the Euler function of n, φ2(n)=12φ(n), and
where μ(n) is the Möbius function. There are some good results for the generalized Euler function and its applications, especially those concerning φe(n)(e=2,3,4,6), which can be seen in [3].
In 2013, Cai et al [2] gave the explicit formula for φ3(n) and obtained a criterion regarding the parity for φ2(n) or φ3(n), respectively. In [8], the authors derived the explicit formulae for φ4(n) and φ6(n), and then they obtained some sufficient and necessary conditions for the case that φe(n) or φe(n+1) is odd or even, respectively.
Recently, Wang and Liao [9] gave the formula for φ5(n) in some special cases and then obtained some sufficient conditions for the case that φ5(n) is even. Liao and Luo [5] gave a computing formula for φe(n) (e=p,p2,pq), where p and q are distinct primes, and n satisfies some certain conditions. Liao [6] obtained the explicit formula for a special class of generalized Euler functions. However, the explicit formula for φe(n)(e≠3,4,6) was not obtained in the general case.
In this paper, utilizing the methods and techniques given in [2,5,8], we study the explicit formula and the parity for φe(n)(e=8,12), obtain the corresponding computing formula, and then give a sufficient and necessary condition for the case that φe(n)(e=8,12) is odd or even, respectively.
For convenience, throughout the paper, we denote Ω(n) and ω(n) to be the number of prime factors and distinct prime factors of a positive integer n, respectively. And for k primes p1,…,pk, set Pk={p1,…,pk},
and
We have organized this paper as follows. In Section 2, we obtain the obvious formulas for [m8] and [m12] based on Jacobi symbol, and some important lemmas are given. In Sections 3 and 4, according to (1.1), and by using the property of the Möbius function μ(n), we derive the expressions for φe(8) and φe(12). In Section 5, we give the parities of φ8(n) and φ12(n), respectively. In the last section, we summarize the main advantage of the proposed method, and propose a further problem to be studied.
2.
Preliminaries
In this section, we first present Lemmas 2.1 and 2.2, which are necessary for the derivations of both [m8] and [m12].
Lemma 2.1. For any odd positive integer m, we have
Furthermore, if gcd(m,6)=1, then we have
where (am) is the Jacobi symbol.
Proof. For any odd positive integer m, by properties of the Jacobi symbol, we have
Thus from m≡1(mod8), we can get that 18(m−4+2(−2m)+(−1m))=18(m−1) and [m8]=18(m−1), namely, (2.1) is true. Similarly, if m≡3,5,7(mod8), by direct computation, (2.1) holds.
Furthermore, if gcd(m,6)=1, then by the properties for the Jacobi symbol and the quadratic reciprocity law, we have
Thus by m≡1(mod12), we have that 112(m−6+3(−1m)+2(−3m))=112(m−1)=[m12], i.e., (2.2) is true. Similarly, if m≡5,7,11(mod12), one can get (2.2) also.
This completes the proof of Lemma 2.1.
Now, we give a property for the Möbius function, which unifies the cases of Lemma 1.5 in [2] and Lemmas 1.4 and 1.5 in [8].
Lemma 2.2. Let a be a nonzero integer, p1,…,pk be distinct odd primes, and α1,…,αk be positive integers. Suppose that n=∏ki=1pαii and gcd(pi,a)=1(1≤i≤k); then,
Proof. For a given integer x, set fx(m)=∑d|mμ(md)(xd).
First, if m=pα, where p is an odd prime and α is a positive integer, then, by the definition of the Möbius function, we have
Second, if m=m1pα, where α is a positive integer, p is an odd prime with gcd(m1,p)=1, and m1 is an odd positive integer, then we have
This means that fa(m) is a multiplicative function. Now denote pα‖n to be the case for both pα∣n and pα+1∤n; then, we can get
This completes the proof of Lemma 2.2.
The following lemmas are necessary for proving our main results.
Lemma 2.3. [2] Let p1,…,pk be distinct primes and α,α1,…,αk be non-negative integers. If n=3α∏ki=1pαii>3 and gcd(pi,3)=1(1≤i≤k), then
Lemma 2.4. [8] Let p1,…,pk be distinct odd primes and α,α1,…,αk be non-negative integers. If n=2α∏ki=1pαii>4, then
Lemma 2.5. [8] Let p1,…,pk be distinct primes and α,β,α1,…,αk be non-negative integers. If n=2α3β∏ki=1pαii>6 and gcd(pi,6)=1(1≤i≤k), then
Lemma 2.6. [6] Let p1,…,pk be distinct primes and α1,…,αk be positive integers. If n=∏ki=1pαii and e=∏ki=1pβii with 0≤βi≤αi−1(1≤i≤k), then
3.
The explicit formula for φ8(n)
First, for a fixed positive integer α and n=2α, by Lemma 2.6 we can obtain the following:
Next, we consider the case that n=2αn1, where n1>1 is an odd integer. We have the following theorem.
Theorem 3.1. Suppose that α is a non-negative integer, p1,…,pk are distinct odd primes, and n=2α∏ki=1pαii>8. Then we have the following:
Proof. For n=2α∏ki=1pαii>8, set n1=∏ki=1pαii; then, gcd(n1,2)=1. There are 4 cases as follows.
Case 1. α=0, i.e., n=n1>8. By (1.1), (2.1) and Lemmas 2.1 and 2.2, we have
If 1∈RPk, i.e., there exists an i(1≤i≤k) such that pi≡1(mod8), then (−2pi)=(−1pi)=1. Now by (3.3) we have
If {3,5}⊆RPk, i.e., there exist i≠j such that pi≡3(mod8) and pj≡5(mod8), which means that (−2pi)=(−1pj)=1, then, by (3.3) we also have
If RPk={5,7} or {5}, i.e., for any p∈Pk, we have that p≡5,7(mod8) or p≡5(mod8), respectively. This means that there exists a prime p such that (−2p)=−1 and (−1p)=1. Thus by (3.3) we can obtain
If RPk={3,7} or {3}, i.e., for any p∈Pk,p≡3,7(mod8) or p≡3(mod8), respectively. This implies that for any p∈Pk,(−1p)=−1, and there exists a prime p′∈Pk such that p′≡3(mod8); then, (−2p′)=1. Thus by (3.3) we have
If RPk={7}, i.e., for any p∈Pk,p≡7(mod8), then (−2p)=(−1p)=−1. Thus by (3.3) we have
Now from (3.4)–(3.7) we know that Theorem 3.1 is true.
Case 2. α=1, i.e., n=2n1>8. Then from the definition we have
Now by Lemma 2.4 and the proof for Case 1, we can get the following:
This means that Theorem 3.1 is true in this case.
Case 3. α=2, i.e., n=4n1>8. Then from the definition we have
Now from Lemma 2.4 and the proof for Case 1, we can also get the following:
This means that Theorem 3.1 holds in this case.
Case 4. α≥3. Note that μ(2γ)=0 for any positive integer γ≥2; thus, by (1.1) and Lemma 2.4 we have
If α=3, then
If α≥4, then φ8(n)=−2α−4φ(n1)+2α−3φ(n1)=2α−4φ(n1)=18φ(n), which means that Theorem 3.1 also holds.
From the above, we have completed the proof of Theorem 3.1.
4.
The explicit formula for φ12(n)
In this section, we give the explicit formula for φ12(n). Obviously, φ12(n)=0 when n<12, and φ12(n)=1 when n=12 or 24; then, we consider n>12 and n≠24.
Theorem 4.1. Let α and β be non-negative integers. If n=2α3β>12 and n≠24, then the following holds:
Proof. (1) For the case that α=0, i.e., n=3β>12, and β≥3, then we have
(2) For the case that α=1, i.e., n=2⋅3β>12, and β≥2, by Lemma 2.5,
(3) For the case that α=2, i.e., n=4⋅3β>12, and so β≥2, then we have
(4) For the case that α=3, i.e., n=8⋅3β>12 and n≠24, and β≥2, then
(5) For the case that α≥4, i.e., n=2α⋅3β>12, and so β≥0, if β=0, i.e., n=2α(α≥4), then we have
If β=1, i.e., n=3⋅2α, then we have
If β≥2, we have
This completes the proof of Theorem 4.1.
Now consider the case that n=2α3βn1, where n1>1 and gcd(n1,6)=1. We have the following theorem.
Theorem 4.2. Let α and β be non-negative integers, k,αi(1≤i≤k) be positive integers, and p1,…,pk be distinct primes. Suppose that gcd(pi,6)=1\, (1≤i≤k) and n=2α3β∏ki=1pαii>12; then, we have the following:
Proof. Set n1=∏ki=1pαii; then, gcd(n1,6)=1 and n=2α3βn1.
Case 1. α=0.
(A) If β=0, then n1>1. Thus by (1.1), (2.2) and Lemmas 2.1 and 2.2, we have
If 1∈R′Pk or {5,7}⊆R′Pk, then there exists pi≡1(mod12), or there exist pj and pl such that pj≡5(mod12) and pl≡7(mod12); then, (−1pi)=(−3pi)=1 or (−1pj)=(−3pl)=1, respectively. Thus by (4.3) we can get
If R′Pk={7,11} or {7}, i.e., for any p∈Pk, we have that p≡7,11(mod12) or p≡7(mod12), respectively. This means that (−1p)=−1 and there exists a prime p′≡7(mod12), i.e., (−3p′)=1, in either of the two cases. Thus by (4.3) we can obtain
If R′Pk={5,11} or {5}, i.e., for any p∈Pk, p≡5,11(mod12) or p≡5(mod12), respectively. Then (−3p)=−1, and there exists a prime p′≡5(mod12), i.e., (−1p′)=1 in either case. Thus by (4.3) we can get
If R′Pk={11}, i.e., for any p∈Pk, p≡11(mod12); then, (−1p)=(−3p)=−1. Thus by (4.3) we have
(B) If β≥1, then by (1.1) we have
Now from β=1, Lemma 2.4 and Case 1, we can get the following:
For the case that β≥2, note that μ(3γ)=0 with γ≥2; thus, by Lemma 2.4 we have the following:
From the above (4.3)–(4.9), Theorem 4.2 is proved in this case.
Case 2. α=1.
(A) If β=0, i.e., n=2n1, then by (1.1), Case 1 and Lemma 2.4, we have
(B) If β=1, i.e., n=6n1, then from (1.1) we can get
Now by Lemmas 2.4 and 2.5 and Case 1, we have the following:
(C) If β≥2, then by (1.1) one can easily see that
Now by Lemma 2.4 we can get the following:
From the above (4.10) and (4.12), Theorem 4.2 is true in this case.
Case 3. α=2.
(A) If β=0, i.e., n=4n1, then from Lemmas 2.3 and 2.5, we can obtain
(B) If β=1, i.e., n=12n1, then by the definition we have
Now by Lemmas 2.3 and 2.4 and Case 1, we can get the following:
(C) If β≥2, then from n=4⋅3βn1 and the definition, we know that
From the above (4.13)–(4.15), Theorem 4.2 is proved in this case.
Case 4. α≥3.
(A) If β=0, i.e., n=2αn1, then by Lemma 2.5 we have
(B) If β=1, i.e., n=3⋅2αn1, then by the definition we have
(C) If β≥2, then by Lemma 2.6 we can get
Now from (4.16)–(4.18), Theorem 4.2 is proved in this case.
From the above, we complete the proof for Theorem 4.2.
5.
The parity of the generalized Euler functions φ8(n) and φ12(n)
Based on Theorems 3.1, 4.1 and 4.2, this section gives the parity of φ8(n) and φ12(n), respectively.
Theorem 5.1. If n is a positive integer, then φ8(n) is odd if and only if n=8,16 or n is given by Table 1. In the above table, p,p1,p2 are odd primes with p1≠p2, and α,α1,α2 are positive integers.
Proof. For n=2α, by (3.1) we know that φ8(n) is odd if and only if n=8,16.
Now suppose that n=2α∏ki=1pαii, where α≥0, α1,…,αk are positive integers, and p1,…,pk are distinct odd primes. Set n1=∏ki=1pαii; then, n1>1 is odd. By Theorem 3.1, we have the following four cases.
Case 1. RPk={5,7} or {5}.
(A) If α=0, i.e., n=n1 is odd, then, by (3.2) we have that φ8(n)=18φ(n)+14(−1)Ω(n)2ω(n). Note that there exists a prime factor p of n such that p≡5(mod8); thus, we must have that ω(n)≤2 if φ8(n) is odd. For ω(n)=2, i.e., n=pα11pα22, by (3.2) we have
Therefore φ8(n) is odd if and only if p1≡p2≡5(mod8), which is true. Now for ω(n)=1, i.e., n=pα11 with p1≡5(mod8), similarly, by (3.2) we have
From p1≡5(mod8), we have that p1≡5,13(mod16). If p1≡5(mod16), then
Thus, φ8(n) is odd if and only if 2∣α1. If p1≡13(mod16), then
Thus, φ8(n) is odd if and only if α1 is odd.
(B) If α=1, i.e., ω(n)≥2, by (3.2) we have that φ8(n)=18φ(n)+14(−1)Ω(n)2ω(n)−1. Then we must have that ω(n)≤3 if φ8(n) is odd. For ω(n)=3, namely, n=2pα11pα22, using the same method as (A), φ8(n) is odd if and only if p1≡p2≡5(mod8). Now for ω(n)=2, i.e., n=2pα11 with p1≡5(mod8), similar to (A), φ8(n) is odd if and only if p1≡5(mod16) and α1 is odd, or if p1≡13(mod16) and 2∣α1.
(C) If α=2, i.e., ω(n)≥2, then by (3.2), we have that φ8(n)=18φ(n)=14∏ki=1pαi−1i(pi−1). Thus from the assumption that pi≡5,7(mod8) or pi≡5(mod8), we know that ω(n)=2 if φ8(n) is odd. In this case, n=4pα11 with p1≡5(mod8); then, pα1−11(p1−1)≡4(mod8), namely, φ8(n) is odd.
(D) If α≥3, then by (3.2), φ8(n)=18φ(n)=2α−4∏ki=1pαi−1i(pi−1). Thus we must have that α=3 and k=1 if φ8(n) is odd, namely, n=8pα11 with p1≡5(mod8). In this case, φ8(n)=18φ(n)=12pα1−11(p1−1) is always even.
Case 2. RPk={3,7} or {3}.
(A) If α=0, by (3.2) we have that φ8(n)=18φ(n)+18(−1)Ω(n)2ω(n). Thus we must have that ω(n)≤3 if φ8(n) is odd. For the case that ω(n)=3, i.e, n=pα11pα22pα33, where pi≡3(mod4)(i=1,2,3), it is easy to see that φ8(n) is always even in this case. Therefore we must have that ω(n)=1,2. Consider that ω(n)=2, i.e., n=pα11pα22. Note that RPk={3,7} or {3}; then, by (3.2), φ8(n)=18(pα1−11(p1−1)pα2−12(p2−1)+4⋅(−1)α1+α2) is odd if and only if p1≡p2≡3(mod8). Now, for ω(n)=1, i.e., n=pα11 with p1≡3(mod8), then by (3.2) we have that φ8(n)=18(pα1−11(p1−1)+2(−1)α1). Thus, φ8(n) is odd if and only if p1≡3(mod16) and 2∣α1, or if p1≡11(mod16) and α1 is odd.
(B) If α=1, i.e., ω≥2, by (3.2) we have that φ8(n)=18φ(n)+18(−1)Ω(n)−12ω(n)−1. Thus we must have that ω(n)≤3 if φ8(n) is odd. Using the same method as (A) in case 1, we can get that φ8(n) is odd if and only if n=2pα11pα22 with p1≡p2≡3(mod8), or if n=2pα11 with p1≡3(mod16) and 2∣α1, or if p1≡11(mod16) and α1 is odd.
(C) If α=2, i.e., ω(n)≥2, by (3.2) we have that φ8(n)=18φ(n)+18(−1)Ω(n)−12ω(n). Therefore we must have that ω(n)≤3 if φ8(n) is odd. For the case that ω(n)=3, i.e., n=4pα11pα22, we know that
which is always even. Now for the case that ω(n)=2, i.e., n=4pα11 with p1≡3(mod8), by (3.2) we have that φ8(n)=14(pα1−11(p1−1)+2(−1)α1+1). Since
it follows that φ8(n) is odd.
(D) If α≥3, by (3.2) we have that φ8(n)=18φ(n)=2α−4∏ki=1pαi−1i(pi−1). From RPk={3,7} or {3}, we must have that α=3 and k=1 if φ8(n) is odd, namely, n=8pα11 with p1≡3(mod8). Obviously, φ8(n)=12pα1−11(p1−1) is odd in this case.
Case 3. RPk={7}.
(A) If α=0, by (3.2), φ8(n)=18φ(n)+38(−1)Ω(n)2ω(n). Then we must have that ω(n)≤2 if φ8(n) is odd. For ω(n)=2, i.e., n=pα11pα22, it follows that
Since p1≡p2≡7(mod8), we have that p1−12⋅p2−12≡1(mod4) and
which means that φ8(n) is even. Now for ω(n)=1, i.e., n=pα11, by (3.2) we have
Now from p1≡7(mod8), we have that p1≡7,15(mod16). If p1≡7(mod16), then
namely, φ8(n) is even. Thus, p1≡15(mod16), then
namely, φ8(n) is odd.
(B) If α=1, by (3.2), φ8(n)=18φ(n)+18(−1)Ω(n)2ω(n)−1. Using a similar proof as that for (A) in case 1, φ8(n) is odd if and only if n=2pα11 and p1≡7(mod16).
(C) If α=2, i.e., ω(n)≥2, by (3.2), φ8(n)=18φ(n)+18(−1)Ω(n)−12ω(n). Then we must have that ω(n)≤3 if φ8(n) is odd. For ω(n)=3, i.e., n=4pα11pα22 with p1≡p2≡7(mod8), we know that
is always even. Now for ω(n)=2, i.e., n=4pα11 with p1≡7(mod8), we can verify that
is also even.
(D) If α≥3, by (3.2), φ8(n)=18φ(n)=2α−4∏ki=1pαi−1i(pi−1). Hence, by RPk={7} we know that φ8(n) is odd if and only if α=3 and k=1, i.e., n=8pα11 with p1≡7(mod8).
Case 4. {3,5}⊆RPk or 1∈RPk.
(A) If {3,5}⊆RPk, i.e., k≥2, then by (3.2) we have that φ8(n)=18φ(n)=18φ(2α)∏ki=1pαi−1i (pi−1). Thus we must have that k=2 and α≤1 if φ8(n) is odd, namely, n=pα11pα22 or 2pα11pα22, where p1≡3(mod8) and p2≡5(mod8). Obviously, φ8(n)=18pα1−11(p1−1)pα2−12(p2−1) is always odd in this case.
(B) If 1∈RPk, by (3.2), φ8(n)=18φ(n)=18φ(2α)∏ki=1pαi−1i(pi−1). Thus, we must have that α≤1 and k=1 if φ8(n) is odd. Namely, n=pα11,2pα11 with p1≡1(mod8); then, φ8(n)=18pα1−1(p1−1). Obviously, φ8(n) is odd if and only if p1≡9(mod16).
From the above, we have completed the proof of Theorem 5.1.
Theorem 5.2. If n is a positive integer, then φ12(n) is odd if and only if n=2α(α≥4), 3⋅2α(α≥2), 2⋅3β(β≥2), 4⋅3β(β≥2), or if it satisfies the conditions given in Table 2. Here, p>3 is an odd prime and α≥1.
Proof. Obviously, by the definition of φ12(n) we can get that φ12(n)=0 for n<12 and φ12(n)=1 for n=12,24; then, we consider that n>12 and n≠24. First, we consider the case that n=2α⋅3β.
If α=0, we have that β≥3; then, by (4.1), φ12(n)=12(3β−2−(−1)β) is even.
If α=1, we have that β≥2; then, by (4.1), φ12(n)=12(3β−2−(−1)β+1) is odd.
If α=2, we have that β≥2; then, by (4.1), φ12(n)=3β−2 is odd.
If α=3, we have that β≥2; then, by (4.1), φ12(8⋅3β)=2⋅3β−2 is even.
If α≥4, then that β≥0. For β=0, by (4.1), φ12(n)=13(2α−3+(−1)α) is odd. For β=1, by (4.1), φ12(n)=13(2α−2+(−1)α+1) is odd. For β≥2, by (4.1), φ12(2α⋅3β)=2α−2⋅3β−2, which is always even.
Next, we consider the case that n=2α3βn1, where α≥0, β≥0, n1>1 and gcd(n1,6)=1. For convenience, we set n=2α3β∏ki=1pαii, where αi≥1, pi is an odd prime and pi>3(1≤i≤k). By Theorem 4.2 we have the following four cases.
Case 1. R′Pk={7,11} or {7}.
(A) α=0. If β=0, i.e., ω(n)≥1, from (4.2) we have that φ12(n)=112φ(n)+14(−1)Ω(n)2ω(n). Thus, from the assumption that R′Pk={7,11} or {7}, we must have that ω(n)≤2 if φ12(n) is odd. For ω(n)=2, i.e., n=pα11pα22, note that R′Pk={7,11} or {7}; then,
is always even in this case. Thus, ω(n)=1, i.e., n=pα11; then, by (4.5),
Note that p1≡7(mod12), i.e., p1≡7,19(mod24). If p1≡7(mod24), then pα1−11(p−1)+6⋅(−1)α1≡0(mod24), which means that φ12(n) is even. Thus, p1≡19(mod24); then, pα1−11(p1−1)+6⋅(−1)α1≡12(mod24), namely, φ12(n) is odd.
If β=1, i.e., ω(n)≥2, by (4.2), φ12(n)=112φ(n)=16∏ki=1pαi−1i(pi−1). Similarly, we must have that ω(n)=2 if φ12(n) is odd. In this case, n=3pα1 with p1≡7(mod12); then, φ12(n)=16pα−11(p1−1), easy to see that φ12(n) is always even. If β≥2, i.e., ω(n)≥2, by (4.2), φ12(n)=112φ(n)+14(−1)Ω(n)+12ω(n). Similarly, we must have that ω(n)=2, i.e., n=3βpα11 if φ12(n) is odd. Since p1≡7(mod12), it follows that φ12(n)=3β−2pα1−11⋅p1−12+(−1)β+α1+1 is always even.
(B) α=1. If β=0, i.e., ω(n)≥2, by (4.10), φ12(n)=112φ(n)+14(−1)Ω(n)2ω(n)−1. Similarly, we must have that ω(n)≤3 if φ12(n) is odd. For ω(n)=3, i.e., n=2pα11pα22, note that R′Pk={7,11} or {7}; then, it is easy to see that φ12(n) is always even. Thus, ω(n)=2, i.e., n=2pα11; note that p1≡7(mod12), namely, p1≡7,19(mod24). In this case, φ12(n) is odd if and only if p1≡7(mod24).
If β=1, i.e., ω(n)≥3, by (4.2), φ12(n)=112φ(n)=16∏ki=1pαi−1i(pi−1). Similarly, from R′Pk={7,11} or {7}, we can get that φ12(n) is odd if and only if ω(n)=3, i.e., n=6pα1 with p1≡7(mod12).
If β≥2, i.e., ω(n)≥3, by(4.2), φ12(n)=112φ(n)+14(−1)Ω(n)+12ω(n)−1. Then we must have that ω(n)=3 if φ12(n) is odd, namely, n=2⋅3βpα11 with p1≡7(mod12). Obviously, φ12(n)=3β−2pα−1⋅p−12+(−1)2+β+α is always even in this case.
(C) α=2. If β=0, i.e., ω(n)≥2, by (4.2), φ12(n)=112φ(n)=16∏ki=1pαi−1i(pi−1). Thus, we must have that ω(n)=2 if φ12(n) is odd. In this case, n=4pα11 with p1≡7(mod12); then, pα11(p1−1)≡6(mod12), which means that φ12(n) is always odd in this case.
If β≥1, i.e., ω(n)≥3, by (4.2), φ12(n)=112φ(n)=3β−2∏ki=1pαi−1i(pi−1). Note that R′Pk={7,11} or {7}; then, φ12(n) is always even.
(D) α≥3. By (4.2) and R′Pk={7,11} or {7}, φ12(n)=112φ(n)=13⋅2α−3φ(3βn1) is always even in this case.
Case 2. R′Pk={5,11} or {5}.
(A) α=0. If β=0, i.e., ω(n)≥1, from (4.6), we can get that φ12(n)=112φ(n)+16(−1)Ω(n)2ω(n). Thus, we must have that ω(n)=1 if φ12(n) is odd, namely, n=pα11 with p1≡5(mod12). Hence
Note that p1≡5(mod12), i.e., p1≡5,17(mod24). If p1≡5(mod24), then pα−11⋅p1−14+(−1)α1≡0(mod6), which means that φ12(pα1) is always even. Thus, p1≡17(mod24), in this case pα1−11⋅p1−14+(−1)α1≡3(mod6), namely, φ12(pα1) is odd.
If β=1, i.e., ω(n)≥2, from (4.8) we have that φ12(n)=112φ(n)+16(−1)Ω(n)2ω(n)−1. Thus, we must have that ω(n)=2 if φ12(n) is odd, namely, n=3pα11 with p1≡5(mod12); in this case
is always odd.
If β≥2, i.e., ω(n)≥2, from (4.9) we can get that φ12(n)=112φ(n). We must have that ω(n)=2, i.e., n=3βpα(β≥2), if φ12(n) is odd. From the assumption p1≡5(mod12), φ12(3βpα11)=112φ(3βpα11)=2⋅3β−1pα1−11⋅p1−14 is always even.
(B) α=1, i.e., ω(n)≥2. By (4.10)–(4.12), we must have ω(n)≤3 if φ12(n) is odd. Namely, n=2pα11,2pα11pα22,6pα11, or 2⋅3βpα11(β≥2). Similar to the proof of (A) in case 1, φ12(n) is odd if and only if n=2pα11 with p1≡17(mod24), or if n=6pα11 with p1≡5(mod12).
(C) α=2. If β=0, i.e., ω(n)≥2, by (4.13), φ12(n)=112φ(n)+112(−1)Ω(n)+12ω(n); then, we must have that ω(n)=2 if φ12(n) is odd. In this case, n=4pα11 with p1≡5(mod12). Hence, φ12(n)=16pα1−11(p1−1)+13(−1)α1+3=13(pα1−11p1−12+(−1)α1+3) is always odd.
If β=1, i.e., ω(n)≥3, by (4.14), φ12(n)=112φ(n)+112(−1)Ω(n)+12ω(n)−1; we must have that ω(n)=3 if φ12(n) is odd. In this case, n=12pα11 with p1≡5(mod12); then, φ12(n)=13pα1−11(p1−1)+13(−1)α1+4=13(pα1−11(p1−1)+(−1)α1+4) is odd.
If β≥2, i.e., ω(n)≥3, by (4.15), φ12(n)=112φ(n); we must have that ω(n)=3 if φ12(n) is odd. Namely, n=4⋅3βpα11 with p1≡5(mod12); then,
is always even.
(D) α≥3, i.e., ω(n)≥2. If β=0, then by (4.16) and R′Pk={5,11} or {5}, we konw that φ12(n)=112φ(n)+16(−1)Ω(n)2ω(n) is always even in this case.
If β=1, by (4.17) and R′Pk={5,11} or {5}, we know that φ12(n)=112φ(n)+112(−1)Ω(n)2ω(n) is always even in this case.
If β≥2, by (4.18) and R′Pk={5,11} or {5}, φ12(n)=112φ(n) is always even in this case.
Case 3. R′Pk={11}.
(A) α=0, i.e., ω(n)≥1. From (4.7)–(4.9), we must have that ω(n)≤2 if φ12(n) is odd. Consider that ω(n)=2, i.e., n=pα11pα22, or 3βpα11(β≥1) with p1≡p2≡11(mod12). Thus, by (4.7)–(4.9), φ12(n) is always even. Hence, ω(n)=1, i.e., n=pα11 with p≡11(mod12); then, (4.7) we can get
Note that p1≡11(mod12), i.e., p1≡11,23(mod24). If p1≡11(mod24), then
namely, φ12(n) is even. If p1≡23(mod24), then
namely, φ12(n) is odd.
(B) α=1, i.e., ω(n)≥2. From (4.10)–(4.12), we must have that ω(n)≤3 if φ12(n) is odd. Namely, n=2pα11, 2pα11pα22, 6pα11, or 2⋅3βpα11(β≥2) with p1≡p2≡11(mod12). Using the same method as for (A) in case 1, φ12(n) is odd if and only if n=2pα11 with p1≡11(mod24).
(C) α=2, i.e., ω(n)≥2. If β=0, by (4.13), we must have that ω(n)=2 if φ12(n) is odd, namely, n=4pα11 with p1≡11(mod12). Then by (4.13),
is always even.
If β≥1, i.e., ω(n)≥3, by (4.14)–(4.15), we must have that ω(n)=3 if φ12(n) is odd. Namely, n=4⋅3βpα11(β≥1) with p1≡11(mod12). If β≥2, then by (4.15),
is always even. Thus, β=1; by (4.14), φ12(12pα11)=13(pα1−11(p1−1)+(−1)α1+3) is odd.
(D) α≥3. If β=0, i.e., ω(n)≥2, then by (4.16) and R′Pk={11}, we know that φ12(n)=112φ(n)+16(−1)Ω(n)2ω(n) is always even in this case.
If β=1, i.e., ω(n)≥3, then by (4.17) and R′Pk={11}, we know that φ12(n)=112φ(n)+112(−1)Ω(n)2ω(n) is always even in this case.
If β≥2, i.e., ω(n)≥3, then by (4.18) and R′Pk={11}, we know that φ12(n)=112φ(n)=2α−2⋅3β−1∏ki=1pαi−1i(pi−1) is always even.
Case 4. {5,7}⊆R′Pk or 1∈R′Pk.
(A) If {5,7}⊆R′Pk, by (4.2) we have that φ12(n)=112φ(n) is always even.
(B) If 1∈R′Pk, then by (4.2), φ12(n)=112φ(n); thus, we must have that k=1, α≤1, and β=0 if φ12(n) is odd. Namely, n=pα11 or 2pα11 with p1≡1(mod12). In this case, φ12(n)=112pα1−11(p1−1) is odd if and only if p1≡13(mod24).
From the above, we have completed the proof of Theorem 5.2.
6.
Final remark
In [2,8], Cai, et al. gave the explicit formulae for the generalized Euler functions denoted by φe(n) for e=3,4,6. The key point is that the derivation of [ne] can be obtained by utilizing the corresponding Jacobi symbol for e=3,4,6. In the present paper, by applying Lemmas 2.1 and 2.2, the exact formulae for φ8(n) and φ12(n) have been given and the parity has been determined. Therefore, the obvious expression for [ne] depends on the Jacobi symbol, seems to be the key to finding the exact formulae for φe(n).
We propose the following conjecture.
Conjecture 6.1. Let e>1 be a given integer. For any integer d>2 with gcd(d,e)=1, there exist u∈Q, a1,a2,a3, bi(1≤j≤r)∈Z, and qj(1≤j≤r)∈P, such that
or
where r≥1 and εj∈{1,−1}.
It is easy to see that Conjecture 6.1 is true for e=2,3,4,6,8 and 12. (see [2,8] and (2.1), (2.2)). If the formulas for (6.1) and (6.2) in the above conjecture can be obtained, then, by (1.1), using the properties of Möbius functions, we can find the exact formulae for φe(n).
Use of AI tools declaration
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
Acknowledgments
This work was supported by Natural Science Foundation of China under grant numbers 12361001, 12161001 and 12071321, and Research Projects of ABa Teachers University (AS-XJPT2023-02, AS-KCTD2023-02).
The authors sincerely thanks Professor Cai Tianxin for his guidance and help, as he visited ABa Teachers University in October 2023.
We would like to thank the referee for his/her detailed comments.
Conflict of interest
The authors declare that there are no conflicts of interest regarding the publication of this paper.