An Abundancy Result for the Two Prime Power Case and Results for an Equations of Goormaghtigh

Abstract Let a represent a positive integer, and σ(a) denote the sum of the (positive) factors of a. The abundancy (or abundancy index) of a, denoted by I(a), is defined by I(a) = σ(a)/a. Suppose that we are given positive integers m0 and n0 and distinct prime numbers p and q. In the current note, new results concerning an equation of Goormaghtigh are established and used to show that the equation I(x) = I(pm0qn0) has, at most, one solution of the form x = pmqn such that pmqn = pm0qn0.


Introduction
We will start with a review of some basic properties that is similar, at times, to the introductory section in [4].If p is a prime number and m is a nonnegative integer, then the sum of the factors of p m is given by Additionally, if a = , where I is multiplicative.It is not difficult to show that, if a is relatively prime to σ(a), then x = a is the unique solution of I(x) = I(a).Suppose we are given a prime number p and positive integer m 0 .Since p does not divide σ(p m 0 ), the only solution of has been studied by the current author [4], [5], [6].When considering this equation, we will assume that p and q represent distinct prime numbers, m 0 and n 0 represent positive integers, and the values of p, q, and m 0 are given.We will assume that the value of n 0 is given too, except in the last paragraph of this article.Also, Z + will denote the set of positive integers.Obviously, if σ(p m 0 q n 0 ) is relatively prime to pq, then x = p m 0 q n 0 is the unique solution of (1).If gcd(σ(p m 0 q n 0 ), pq) > 1, other solutions exist in some instances.When they exist, each additional solution of (1) can be written in the form If a solution of (1) is written in the form of (2), then m = m 0 ; likewise n = n 0 .In (2), m and n represent non-negative integers.
There is some interest in solutions of (1) that can be expressed in the form Since we are assuming that m 0 , n 0 ∈ Z + , it follows that the (integer) values of m and n are positive.(For if (3) is a solution of (1) such that m = 0, then I(q n ) = I(p m 0 q n 0 ), which implies that m 0 = 0, contradicting the assumption that m 0 is positive.Similarly, n = 0.) The investigation concerning solutions in the form of ( 3) is related to the study of another well-known equation.An equation of Goormaghtigh is given by R. Ratat [3] and R. Goormaghtigh [1] noted that solutions to (4) are given by (y, z, u, w) = (2, 5, 5, 3) or (2,90,13,3); it has been conjectured that these are the only solutions.B. He and A. Togbé [2] considered the equation and established the following result.

Theorem 1.1. (He and Togbé) Let z > y > 1 be given integers. Then equation (5) has at most one solution (u, w).
Let us recall another bit of common notation.If c is a nonzero integer, then ν 2 (c) is the nonnegative integer with the property that 2 ν 2 (c) divides c but 2 ν 2 (c)+1 does not divide c.With the aid of theorem 1.1, the following theorem was proved [4], [5].

Theorem 1.2. Suppose that
where p and q represent prime numbers, m 1 , n 1 , m 2 , and n 2 are positive integers, and p m 1 q n 1 = p m 2 q n 2 ; thus p = q.Without loss of generality, we will assume that m 1 < m 2 and p m 2 −m 1 < q n 1 −n 2 throughout this theorem.Let (B) A solution of (4) is given by (C) It turns out that m 1 and m 2 are relatively prime; likewise for n 1 and n 2 ; additionally, The only known solution of ( 6) is {p m 1 q n 1 , p m 2 q n 2 } = {80, 200}.Theorem 1.2 will be useful in the ensuing section.

Main Results
Even though the following theorem is not a major breakthrough in the study of equation ( 5), parts A and B are useful when establishing the primary result in this article.
) is a solution of (5) such that a, b, j 1 , and k 1 are positive integers, a > 1, b > 1, and a is relatively prime to b.
(A) Let j 2 and k 2 represent positive integers such that j 1 u 1 +j 2 j 2 and k 1 (w 1 −1) is a solution of (5).
(B) Let j 3 and k 3 represent positive integers such that j 1 (u 1 −1) is a solution of (5).

(C) Let k 4 and w 4 represent any two positive integers such that w
) is a solution of (5).
(D) Let j 4 and u 4 represent any two positive integers such that u 4 > 1.Then neither (a ) is a solution of (5).
Similarly, neither (a ) is a solution of (5).
Proof.(A) Suppose that (a j 1 , b k 1 , u 1 , w 1 ), and either (a , are solutions of (5).Then and Subtracting 1 from both sides of the equations in ( 8) and ( 9), we observe that and Due to equations ( 8) and ( 11), b k 2 divides a j 1 − 1, and thus b k 2 < a j 1 .However, from equations ( 9) and (10), we see that a j 1 divides b k 2 − 1.Thus a j 1 < b k 2 , and we have a contradiction.

(B)
The proof of part B is similar to the proof of part A of this theorem.
(C) Let us suppose that (a j 1 , b k 1 , u 1 , w 1 ), and either (a , are solutions of (5).Obviously, u 1 > 2 in this case and equation (10) continues to be true.Moreover, In equation ( 10), b k 1 divides a j 1 (u 1 −1) −1 . But, in equation ( 12), b k 1 is relatively prime to the same fraction, and we have a contradiction.If, once again, (a j 1 , b k 1 , u 1 , w 1 ) satisfies ( 5) and we assume that (a j 1 , b k 4 , u 1 + 1, w 4 ) or (b k 4 , a j 1 , w 4 , u 1 +1) satisfies the same equation, then (8) remains true and Thus b k 4 is relatively prime to a j 1 u 1 −1 a j 1 −1 and a factor of the same fraction, which is an obvious contradiction.
(D) The proof of this result is similar to the proof of part C. 2 As you will see, the proof of the next result is easy to attain at this point.We continue to assume that, when considering equation ( 1), the values of the distinct primes p and q, and the positive integers m 0 and n 0 , are given.

Theorem 2.2.
There is, at most, one solution of (1) in the form given by (3).
Proof.Suppose that p m 1 q n 1 , p m 2 q n 2 , and p m 3 q n 3 are distinct solutions of (1).Without loss of generality, we will assume that m 1 < m 2 < m 3 ; thus n 1 > n 2 > n 3 .Furthermore, we will assume that p m 0 q n 0 is equal to one of these solutions.Again, let t p = m 2 +1 m 2 −m 1 and t q = n 1 +1 n 1 −n 2 .Due to theorem 1.2(B), either (7) or is a solution of (4).For the moment, we will assume that (7) is a solution of (4), and thus, a solution of (5).Applying theorem 1.2(B) again, we see that or is a solution of ( 4) and ( 5).Let 14) and (15) can be rewritten as respectively, and we have a contradiction to theorem 2.1(A).We get a similar contradiction, to theorem 2.1(B), if we make the assumption that (13) is a solution of (4).
It was previously established [4] that, if m 0 and n 0 are even and the distinct primes p and q are odd, then there is no solution of (1) in the form of (3).In the same article it was shown that, if m 0 (or n 0 ) is equal to 1, then x = 200 is the only solution of (1) in the form of (3); this solution occurs when pq n 0 (or respectively, p m 0 q) is equal to 80.
In this paragraph, we are assuming that the values of p, q, and m 0 are given.It has been demonstrated [4] that equation ( 1) has (at most) finitely many solutions (x, n 0 ) such that x is of the form specified in (3).It was also verified [4] that, if m 0 + 1 is an odd prime, then there are at most two values of n 0 that will yield a solution of (1) such that x is of the form given by (3).Presently, we can improve on this statement.Due to parts C and D of theorem 2.1, if m 0 + 1 is an odd prime, then there is at most one solution (x, n 0 ) of (1) such that x is of the form given in (3).The proof of this result may be obtained from the author.Lastly, if we apply theorem 2.1, parts C and D, in a similar fashion to example 3.3 from the article cited in this paragraph, we observe that, when m 0 = 11, there are at most four solutions (x, n 0 ) of ( 1) such that x is of the form indicated in (3).

j 1 p j − 1 .
, where p 1 , p 2 , . . ., p k are distinct primes, k is a positive integer, and m 1 , m 2 , . . ., m k are nonnegative integers, then σ(a) = k j=1 p m j +1 j − Consequently, σ is a multiplicative function; that is, if a and b are relatively prime then σ(ab) = σ(a)σ(b).If we continue to express the positive integer a