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

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430 R. F. Ryan (C) Let k 4 and w 4 represent any two positive integers such that w 4 > 1. Then neither (a j 1 ,b k 4 ,u 1 − 1,w 4 ) nor (b k 4 ,a j 1 ,w 4 ,u 1 − 1) is a solution of (5). Similarly, neither (a j 1 ,b k 4 ,u 1 +1,w 4 ) nor (b k 4 ,a j 1 ,w 4 ,u 1 +1) 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 j 4 ,b k 1 ,u 4 ,w 1 − 1) nor (b k 1 ,a j 4 ,w 1 − 1,u 4 ) is a solution of (5). Similarly, neither (a j 4 ,b k 1 ,u 4 ,w 1 +1) nor (b k 1 ,a j 4 ,w 1 +1,u 4 ) is a solution of (5). Proof. (A) Suppose that (a j 1 ,b k 1 ,u 1 ,w 1 ), and either (a j 2 ,b k 2 , j 1u 1 +j 2 or ( b k 2 ,a j 2 , k 1(w 1 −1) k 2 , j 1u 1 +j 2 ) j 2 , are solutions of (5). Then and j 2 , k 1(w 1 −1) k 2 ) a j 1u 1 − 1 a j 1 − 1 = bk 1w 1 − 1 b k 1 − 1 (8) a j 1u 1 +j 2 − 1 = bk 1(w 1 −1) − 1 . (9) a j 2 − 1 b k 2 − 1 Subtracting 1 from both sides of the equations in (8) and (9), we observe that and a j 1 aj 1(u 1 −1) − 1 a j 1 − 1 = b k 1 bk 1(w 1 −1) − 1 b k 1 − 1 (10) a j 2 aj 1u 1 − 1 a j 2 − 1 = b k 2 bk 1(w 1 −1)−k 2 − 1 b k 2 − 1 . (11) Due to equations (8) and (11), b k 2 divides a j 1 − 1, and thus b k 2
Abundancy result 431 1,w 4 )or(b k 4 ,a j 1 ,w 4 ,u 1 +1) satisfies the same equation, then (8) remains true and a j aj 1u 1 − 1 1 = b k bk 4(w 4 −1) − 1 4 . a j 1 − 1 b k 4 − 1 Thus b k 4 is relatively prime to aj 1 u 1−1 a j 1 −1 which is an obvious contradiction. and a factor of the same fraction, (D) The proof of this result is similar to the proof of part C. ✷ 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 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 (q n 1−n 2 ,p m 2−m 1 ,t q ,t p ) (13) 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 ( p m 3−m 2 ,q n 2−n 3 m 3 +1 , , n ) 2 +1 (14) m 3 − m 2 n 2 − n 3 or ( q n 2−n 3 ,p m 3−m 2 , n ) 2 +1 m 3 +1 , n 2 − n 3 m 3 − m 2 (15) is a solution of (4) and (5). Let j 1 = m 2 −m 1 , k 1 = n 1 −n 2 , j 2 = m 3 −m 2 , and k 2 = n 2 − n 3 . Then (14) and (15) can be rewritten as ( p j 2 ,q k 2 , j 1t p+j 2 j 2 , k 1(t q−1) ) k 2 and ( q k 2 ,p j 2 , k 1(t q−1) k 2 , j 1t p+j 2 ) j 2 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.
Page 1 and 2: International Mathematical Forum, V
Page 3: Abundancy result 429 Theorem 1.1. (

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

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