Research Article | Open Access
Yuanyuan Deng, Wenpeng Zhang, "On the Odd Prime Solutions of the Diophantine Equation ", Abstract and Applied Analysis, vol. 2014, Article ID 186416, 4 pages, 2014. https://doi.org/10.1155/2014/186416
On the Odd Prime Solutions of the Diophantine Equation
Using the elementary method and some properties of the least solution of Pell’s equation, we prove that the equation has no positive integer solutions () with and being odd primes.
Let , be the sets of all integers and positive integers, respectively. In recent years, there are many authors who investigated the various properties of exponential diophantine equation with circulating form (see [1–4]). Recently, Zhang et al.  are interested in the equation Using the -adic lower bound of the log-linear model method, they proved all solutions of (1) satisfying . Meanwhile, they proposed a conjecture as follows.
Conjecture 1. Equation (1) has no positive integer solution .
Using the method in , it seems to be a very difficult problem to improve the upper bound estimate for . In this paper, we use the elementary method and some properties of the least solution of Pell’s equation to solve the conjecture partly. That is, we will prove the following.
Theorem 2. Equation (1) has no positive integer solution with and being odd primes.
2. Several Lemmas
Let be a nonsquare positive integer, and let denote the class number of binary quadratic primitive forms with discriminant . Then we have the following.
Lemma 3. For the equation there is a solution with , and there is a unique positive integer solution satisfying , where pass through all positive integer solutions of (2). We call as the least solution of (2). Every solution of (2) can be expressed as
Proof. See Section 10.9 of .
Lemma 4. Let be an odd prime satisfying and , and then every least solution of (2) satisfies .
Proof. See .
Lemma 5. Consider .
Proof. According to Table II in Chapter 16 of , we know that Lemma 5 holds for , and when , by Theorems and of , we have where is the least solution of (2). If , and , , then by (4), we have a contradiction. This proves Lemma 5.
Lemma 6. Let be an odd prime with . If the equation has the solution , then every solution can be expressed as where are positive integers satisfying and is a solution of (2).
Proof. See Lemma 4 of .
3. Proof of Theorem 2
Let be one of the solutions of (1). Without loss of generality we may assume that , since and are symmetrical in (1). By , we know that are coprime, and . When and are odd primes, must be even. Note that ; by (1) we have ; then , so by the result in , we get
If , by (10), is an odd prime with . We see from (1) that the equation has the solution Since is an odd prime with , applying Lemma 6 to (11) and (12), we have where are positive integers satisfying is a solution of the equation and denotes the class number of binary quadratic primitive forms with discriminant .
Further, , and from (19) we know , since by Lemma 3. Thus, according to Lemma 3 there is such that where is the least solution of (17). For the integer , there exist integers and satisfying Let From (17), (19), and (22), we know that and are integers satisfying And from (18), (20), (21), and (22), we have
Let Then are integers with , and where is the integral part of . From (29), we have Applying (27) to (24), we get From (17) and (26), , by (30), . However, we get from (9) that is an odd prime satisfying and ; then from Lemma 4, we know it is impossible. Thus, the theorem holds for .
Similarly, if , by (10) is an odd prime with . We see from (1) that the equation has the solution Applying Lemmas 5 and 6 to (11) and (12), we have where are positive integers satisfying and is a solution of the equation Applying Lemma 3 to (34) and (35), we have where is the least solution of (36). In addition, the integer can be expressed as Let From (35) and (36), we know that and are integers satisfying And from (34), (37), (38), and (39), we get
If in (38), then, from (41), we have Since is an odd prime, , by (42), we know From (40), we know ; then from (43) we get . Let , and , so By (45), Combining (42) and (46) we may immediately get However, from (42) and (47), we get , but it is impossible. Therefore, we have and
This completes the proof of our theorem.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
This work is supported by the P.S.F. (2013JZ001) and N.S.F. (11371291) of China. The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper.
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