An Abundancy Result for the Two Prime Power Case ... - HIKARI Ltd

Abundancy result 429
Theorem 1.1. (He and Togbé) Let z > y > 1 be given integers.
equation (5) has at most one solution (u, w).
Then
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 ν2(c) divides c
but 2 ν2(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
I(p m 1
q n 1
)=I(p m 2
q n 2
) (6)
where p and q represent prime numbers, m 1 , n 1 , m 2 , and n 2 are positive integers,
and p m 1
q n1 ≠ 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
t p = m 2+1
m 2 −m 1
and t q = n 1+1
n 1 −n 2
. Then:
(A) The greatest common divisor of m 1 +1 and m 2 +1 is m 2 − m 1 and
gcd(n 1 +1,n 2 +1)=n 1 − n 2 .
(B) A solution of (4) is given by
(y, z, u, w) =(p m 2−m 1
,q n 1−n 2
,t p ,t q ). (7)
(C) It turns out that m 1 and m 2 are relatively prime; likewise for n 1 and n 2 ;
additionally, ν 2 (m 1 +1)≠ ν 2 (m 2 +1)and ν 2 (n 1 +1)≠ ν 2 (n 2 +1).
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.
2 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.
Theorem 2.1. Suppose that (y, z, u, w) =(a j 1
,b k 1
,u 1 ,w 1 ) 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 1u 1 +j 2
are integers greater than 1. Then neither ( a j 2
,b k 2
, j 1u 1 +j 2
(
b
k 2
,a j 2
, k 1(w 1 −1)
k 2
, j 1u 1 +j 2
)
j 2
is a solution of (5).
j 2
and k 1(w 1 −1)
k 2
j 2
, k 1(w 1
)
−1)
k 2
nor
(B) Let j 3 and k 3 represent positive integers such that j 1(u 1 −1)
j 3
and k 1w 1 +k 3
k 3
are integers greater than 1. Then neither ( a j 3
,b k 3
, j 1(u 1
)
−1)
j 3
, k 1w 1 +k 3
k 3
nor
(
b
k 3
,a j 3
, k 1w 1 +k 3
k 3
, j 1(u 1
)
−1)
j 3
is a solution of (5).