1 Introduction

A finite simple group is called a \(K_n\)-group if its order is divisible by exactly n different primes. By a classical \(p^aq^b\) theorem of Burnside, every group of order \(p^a q^b\) is solvable, where p and q are primes, and a and b are positive integers; hence, there is no simple \(K_2\)-group. On the other hand, there are only eight different simple \(K_3\)-groups [1]. In this paper, we concentrate on describing simple \(K_4\)-groups.

In the famous book Unsolved Problems in Group Theory, the following problem is asked: is the number of simple \(K_4\)-groups finite or infinite? See [2]: Problem 13.65. This problem is the first to be posed by the second author Shi [3]. Denote by \({\mathbb {N}}\) and \({\mathbb {P}}\) the set of positive integers and the set of prime numbers, respectively. In [3], the second author claimed that the simple \(K_4\)-group problem can be reduced to the four Diophantine problems:

$$\begin{aligned} p^2-1= & {} 2^a3^bq^c, p, q\in {{\mathbb {P}}}, p>3, q>3,a,b,c\in {\mathbb {N}}, \end{aligned}$$
(1.1)
$$\begin{aligned} 2^m-1= & {} p, 2^m+1=3q^n, p, q\in {\mathbb {P}}, p>3, q>3, m,n\in {\mathbb {N}}, \end{aligned}$$
(1.2)
$$\begin{aligned} 3^m-1= & {} 2p^n, 3^m+1=4q, p, q\in {\mathbb {P}}, p>3, q>3, m,n\in {\mathbb {N}}, \end{aligned}$$
(1.3)
$$\begin{aligned} 3^m-1= & {} 2p, 3^m+1=4q^n, p, q\in {\mathbb {P}}, p>3, q>3, m,n\in {\mathbb {N}}. \end{aligned}$$
(1.4)

In 2001, Bugeaud et al. [4] showed that if \(n>1\), (1.2) and (1.4) have no solution and (1.3) has only the solution \((p,q,m,n)=(11,61,5,2)\).

In Sect. 2, we will prove that if \(c>1\), (1.1) has only the solutions \((p,q,a,b,c)=(97,7,6,1,2)\) and \((p,q,a,b,c)=(577,17,7,2,2)\). Our methods are slightly different from those in [4].

Thus, the infinitude of simple \(K_4\)-groups can be decided by the following three Diophantine problems:

$$\begin{aligned} p^2-1= & {} 2^a3^bq, p, q\in {\mathbb {P}}, p>3, q>3, a,b\in {\mathbb {N}}, \end{aligned}$$
(1.5)
$$\begin{aligned} 2^m-1= & {} p, 2^m+1=3q, p, q\in {\mathbb {P}}, p>3, q>3, m\in {\mathbb {N}}, \end{aligned}$$
(1.6)
$$\begin{aligned} 3^m-1= & {} 2p, 3^m+1=4q, p, q\in {\mathbb {P}}, p>3, q>3, m\in {\mathbb {N}}. \end{aligned}$$
(1.7)

By considering Diophantine equations (1.5), (1.6) and (1.7), one will see that it is very difficult to determine the infinitude of simple \(K_4\)-groups, and this problem goes far beyond what is known about the following Dickson’s conjecture [5] at present. In fact, if the Diophantine equation (1.6) has infinitely many solutions for \(p, q\in {\mathbb {P}}\), \(p>3\), \(q>3\), \(m\in {\mathbb {N}}\), then \(f_1(x)=x\) and \(f_2(x)=3x-2\) represent simultaneously primes for infinitely many integers x. This is the special case of Dickson’s conjecture. Unfortunately, this case is open. The Diophantine equations (1.5) and (1.7) can be similarly discussed. For the details, see Sect. 3.

Dickson Conjecture: Let \(1\le s \in {\mathbb {N}}\), \(f_i(x) = a_i +b_ix\) with \(a_i\) and \(b_i\) integers, \(b_i \ge 1\) (for \(i = 1, \ldots , s \)). If there does not exist any integer \(n > 1\) dividing all the products \(\prod _{i=1}^{i=s}f_i(k)\), for every integer k, then there exist infinitely many natural numbers m such that all numbers \(f_1(m), \ldots , f_s(m)\) are primes.

The case \(s = 1\) is Dirichlet’s theorem. Two special cases are well-known conjectures: there are infinitely many twin primes ( \(f_1(x)=x\) and \(f_2(x)=x+2\) represent simultaneously primes for infinitely many integers x), and there are infinitely many Sophie Germain primes (\(f_1(x)=x\) and \(f_2(x)=2x+1\) represent simultaneously primes for infinitely many integers x). As we know, even these two simple cases, nobody has proved up until now, let alone other special cases.

2 Main Theorem and Its Proof

Theorem 2.1

If \(1<c\in {\mathbb {N}}\), \(p, q\in {\mathbb {P}}\), \(p>3\), \(q>3\), and \(a,b\in {\mathbb {N}}\), then the Diophantine equation \(p^2-1=2^a3^bq^c\) has only the solutions \((p,q,a,b,c)=(97,7,6,1,2)\) and \((p,q,a,b,c)=(577,17,7,2,2)\).

Lemma 2.2

(Zsigmondy’s theorem [6]) If \(a>b>0\), \(\gcd (a,b)=1\) and \(n>1\) are positive integers, then \(a^n+b^n\) has a prime factor that does not divide \(a^k+b^k\) for all positive integers \(k<n\), with the exception \(2^3+1^3\); \(a^n-b^n\) has a prime factor that does not divide \(a^k-b^k\) for all positive integers \(k<n\) unless \(a=2,b=1\) and \(n=6\); or \(a+b\) is a power of 2 and \(n=2\).

Lemma 2.3

[7] The Diophantine equation \(3^m-2y^q=1\) has only the solution \((y,m,q)=(11,5,2)\) satisfying \(2<m\in {\mathbb {N}}\), \(2\le y,q \in {\mathbb {N}}\).

Lemma 2.4

[8] The Diophantine equation \(x^m-y^n=1\) has only the solution \((x,y,m,n)=(3,2,2,3)\) satisfying \(1<m,n\in {\mathbb {N}} \), \(x,y \in {\mathbb {P}}\).

Lemma 2.5

[9] The Diophantine equation \(x^2+1=2y^n\) has only the solution \((x,y,n)=(239,13,4)\) satisfying \(x,y,n\in {\mathbb {N}}\), \(y>1,n>2\).

Lemma 2.6

[10] The Diophantine equation \(3x^2+1=4y^n\) with \(1<n\equiv 1 (\mod 2)\) and \(x,y,n\in {\mathbb {N}}\) has only the solution \(x=y=1\).

Lemma 2.7

[4] The Diophantine equation \(2^m+1=3y^q\) with \(1<m,1<y,1<q\in {\mathbb {N}}\) has no solution.

Proof of theorem 2.1

If \(1<c\in {\mathbb {N}}\), Diophantine equation \(p^2-1=2^a3^bq^c\) has solutions (p, q, a, b, c) such that \(p, q\in {\mathbb {P}}\), \(p>3\), \(q>3\), \(a,b\in {\mathbb {N}}\), then \(a\ge 3\) and \(p^2\equiv 1 (\!\!\!\!\mod 2^a)\). Notice that if \(a\ge 3\), \(x^2\equiv 1 (\!\!\!\!\mod 2^a)\) has only four solutions, say \(x\equiv \pm 1, \pm 1+2^{a-1}(\!\!\!\!\mod 2^a)\). Therefore, we must have \(p=\pm 1 +2^{a-1}+k 2^a\) with \(k \in {\mathbb {N}}\cup \{0\}\). If \(p=1 +2^{a-1}+k 2^a\), then we have \((2k+1)(1 +2^{a-2}+k 2^{a-1})=3^bq^c\) by \(p^2-1=2^a3^bq^c\). But \(\gcd (2k+1, 1 +2^{a-2}+k 2^{a-1})=1=\gcd (3^b, q^c)\). Therefore, \(2k+1=3^b\) or \(2k+1=q^c\). When \(p=-1 +2^{a-1}+k 2^a\), we still have \(2k+1=3^b\) or \(2k+1=q^c\). Thus, when \(k \in {\mathbb {N}}\), \(p^2-1=2^a3^bq^c\) can be reduced to the following Diophantine problems with \(3\le a,1\le b,1<c\in {\mathbb {N}}\), \(p, q\in {\mathbb {P}}\), \(p>3\), \(q>3\):

$$\begin{aligned} 2k+1= & {} 3^b, q^c=1 +2^{a-2}+k 2^{a-1}, p=1 +2^{a-1}+k 2^a, \end{aligned}$$
(2.1)
$$\begin{aligned} 2k+1= & {} q^c, 3^b=1 +2^{a-2}+k 2^{a-1}, p=1 +2^{a-1}+k 2^a, \end{aligned}$$
(2.2)
$$\begin{aligned} 2k+1= & {} 3^b, q^c=-1 +2^{a-2}+k 2^{a-1}, p=-1 +2^{a-1}+k 2^a, \end{aligned}$$
(2.3)
$$\begin{aligned} 2k+1= & {} q^c, 3^b=-1 +2^{a-2}+k 2^{a-1}, p=-1 +2^{a-1}+k 2^a. \end{aligned}$$
(2.4)

\(\square \)

Rewriting these equations, we get (with \(3\le a,1\le b,1<c\in {\mathbb {N}}\), \(p, q\in {\mathbb {P}}\), \(p>3\), \(q>3\)):

$$\begin{aligned} q^c-1= & {} 2^{a-2}3^b, p=-1 +2q^c, \end{aligned}$$
(2.5)
$$\begin{aligned} 3^b-1= & {} 2^{a-2}q^c, p=-1 +2\times 3^b, \end{aligned}$$
(2.6)
$$\begin{aligned} q^c +1= & {} 2^{a-2}3^b, p=1 +2q^c, \end{aligned}$$
(2.7)
$$\begin{aligned} 3^b+1= & {} 2^{a-2}q^c, p=1 +2\times 3^b. \end{aligned}$$
(2.8)

Next, we will prove that if \(1<c\), then (2.6), (2.7), (2.8) have no solution and (2.5) has only two solutions satisfying the conditions.

Clearly, if (2.7) has solutions, then \(c\ne 2\). By Lemma 2.2, \(q^c +1\) has at least one prime factor m that does not divide \(q^r +1\) for all positive integers \(r<c\). However, \(m\ne 2,3\). This leads to a contradiction since \( q^c +1 =2^{a-2}3^b\).

Now, let us consider (2.8). Assume that \( 3^b+1= 2^{a-2}q^c\) has solutions. If b is even, then \(a=3\). By Lemma 2.5, it is impossible. Hence, b is odd. We deduce that \(a=4\) and get that \( 3^b+1= 4q^c\). If c is even, then \( 3^b+1= 4q^c\) has no solution with \(c>1\) [11]. If c is odd, then \( 3^b+1= 4q^c\) has no solution by Lemma 2.6 (since \(c>1\)). So, (2.8) has no solution.

Suppose that (2.6) \( 3^b-1= 2^{a-2}q^c\) has solutions. If b is odd, then \(a=3\) and \( 3^b-1= 2q^c\) has no solution such that \(p=-1 +2\times 3^b\) by Lemma 2.3. Let b be even. Write \(b=2r\). We obtain that \(a\ge 5\) and \(\frac{3^r-1}{2}\frac{3^r+1}{2}=2^{a-4}q^c\). By Lemma 2.4, one can prove that \(\frac{3^r-1}{2}\frac{3^r+1}{2}=2^{a-4}q^c\) has no solution. Thus, (2.6) has no solution.

If \( q^c-1 =2^{a-2}3^b\) has solutions with \(c>1\), then \(c=2\) by Lemma 2.2. By Lemma 2.4, one can obtain that (2.5) has only solutions \((q,a,b)=(7,6,1)\) and \((q,a,b)=(17,7,1)\). It leads that Diophantine equation \(p^2-1=2^a3^bq^c\) has only solutions \((p,q,a,b,c)=(97,7,6,1,2)\) and \((p,q,a,b,c)=(577,17,7,2,2)\) satisfying \(p, q\in {\mathbb {P}}\), \(p>3\), \(q>3\), \(a,b,c\in {\mathbb {N}}\) and \(c>1\).

Finally, we consider the case \(k=0\). Obviously, \(p^2-1=2^a3^bq^c\) can be reduced to the following Diophantine problems with \(3\le a,1\le b,1<c\in {\mathbb {N}}\), \(p, q\in {\mathbb {P}}\), \(p>3\), \(q>3\):

$$\begin{aligned} 3^bq^c= & {} 1 +2^{a-2}, p=1 +2^{a-1}, \end{aligned}$$
(2.9)
$$\begin{aligned} 3^bq^c= & {} -1 +2^{a-2}, p=-1 +2^{a-1}. \end{aligned}$$
(2.10)

By Lemma 2.4, if (2.9) or (2.10) has solutions, then b must be 1. Furthermore, using Lemma 2.7, one can show that (2.9) and (2.10) have no solution satisfying the conditions. This proves Theorem 2.1.

3 Our Conclusion

In this section, we will try to point out that it is not easy to prove the infinitude of simple \(K_4\)-groups. By (1.5), (1.6), (1.7), one will see that the infinitude of simple \(K_4\)-groups can be equivalently decided by the following Diophantine problems:

$$\begin{aligned} 2^m-1= & {} p, 2^m+1=3q, p, q\in {\mathbb {P}}, p>3, q>3, m\in {\mathbb {N}}, \end{aligned}$$
(3.1)
$$\begin{aligned} 3^m-1= & {} 2p, 3^m+1=4q, p, q\in {\mathbb {P}}, p>3, q>3, m\in {\mathbb {N}}, \end{aligned}$$
(3.2)
$$\begin{aligned} q-1= & {} 2^{a-2}3^b, p=-1 +2q, p, q\in {\mathbb {P}}, a\ge 3, a,b\in {\mathbb {N}}, \end{aligned}$$
(3.3)
$$\begin{aligned} 3^b-1= & {} 2^{a-2}q, p=-1 +2\times 3^b, p, q\in {\mathbb {P}}, a\ge 3, a,b\in {\mathbb {N}}, \end{aligned}$$
(3.4)
$$\begin{aligned} q +1= & {} 2^{a-2}3^b, p=1 +2q, p, q\in {\mathbb {P}}, a\ge 3, a,b\in {\mathbb {N}},\end{aligned}$$
(3.5)
$$\begin{aligned} 3^b+1= & {} 2^{a-2}q, p=1 +2\times 3^b, p, q\in {\mathbb {P}}, a\ge 3, a,b\in {\mathbb {N}},\end{aligned}$$
(3.6)
$$\begin{aligned} 3q= & {} 1 +2^{a-2}, p=1 +2^{a-1}, p, q\in {\mathbb {P}}, a\ge 3, a\in {\mathbb {N}},\end{aligned}$$
(3.7)
$$\begin{aligned} 3q= & {} -1 +2^{a-2}, p=-1 +2^{a-1}. p, q\in {\mathbb {P}}, a\ge 3, a\in {\mathbb {N}}. \end{aligned}$$
(3.8)

Hence, if the number of simple \(K_4\)-groups is infinite, then one of the following holds:

$$\begin{aligned} f_1(x)= & {} x \hbox { and} \nonumber \\ f_2(x)= & {} 3x-2 \hbox { represent simultaneously primes for infinitely} \nonumber \\&\hbox {many integers x by } (3.1), \end{aligned}$$
(3.9)
$$\begin{aligned} f_1(x)= & {} x \hbox { and}\nonumber \\ f_2(x)= & {} 2x-1 \hbox { represent simultaneously primes for infinitely} \nonumber \\&\hbox {many integers x by } (3.2)\hbox { or }(20), \end{aligned}$$
(3.10)
$$\begin{aligned} f_1(x)= & {} x \hbox { and}\nonumber \\ f_2(x)= & {} 2x+1 \hbox { represent simultaneously primes for infinitely} \nonumber \\&\hbox {many integers x by } (3.5), \end{aligned}$$
(3.11)
$$\begin{aligned} f_1(x)= & {} x \hbox { and }\nonumber \\ f_2(x)= & {} 4x+1 \hbox { represent simultaneously primes for infinitely many integers x }\nonumber \\&\hbox { by } (3.4) (\hbox { Note that by }(3.4)\hbox { one can deduce that a must be 3.}) , \end{aligned}$$
(3.12)
$$\begin{aligned} f_1(x)= & {} x \hbox { and}\nonumber \\ f_2(x)= & {} 2^{a-1}x-1 \hbox { represent simultaneously primes for infinitely many integers x }\nonumber \\&\hbox {by } (3.6), \hbox { where }a=3 \hbox { or } a=4, \end{aligned}$$
(3.13)
$$\begin{aligned} f_1(x)= & {} x \hbox { and}\nonumber \\ f_2(x)= & {} 6x-1 \hbox { represent simultaneously primes for infinitely} \nonumber \\&\hbox {many integers x by } (3.7),\end{aligned}$$
(3.14)
$$\begin{aligned} f_1(x)= & {} x \hbox { and }\nonumber \\ f_2(x)= & {} 6x+1 \hbox { represent simultaneously primes for infinitely} \nonumber \\&\hbox {many integers x by } (3.8). \end{aligned}$$
(3.15)

Clearly, (3.9), (3.10), (3.11), (3.12), (3.13), (3.14) and (3.15) are all special cases of Dickson’s conjecture. This goes far beyond what is known about Dickson’s conjecture at present. Anyway, due to that fact it is closely tied with many topics in number theory such as Fermat’s primes, Mersenne primes, Dickson’s conjecture and so on, we think that determining the number of simple \(K_4\)-groups is significative. It should be given much attention.