The Diophantine Equation
Let be a fixed odd prime. Using certain results of exponential Diophantine equations, we prove that (i) if , then the equation has no positive integer solutions ; (ii) if , then the equation has only the solutions , where is an odd prime with ; (iii) if and , then the equation has at most two positive integer solutions .
Let , be the sets of all integers and positive integers, respectively. Let be a fixed odd prime. Recently, the solutions of the equation were determined in the following cases:(1)(Sroysang ) if , then (1) has no solutions;(2)(Sroysang ) if , then (1) has no solutions;(3)(Rabago ) if , then (1) has only the solutions , , and .
In this paper, using certain results of exponential Diophantine equations, we prove a general result as follows.
Conjecture 2. If , then (1) has at most one solution .
Lemma 3. If is a prime, where is a positive integer, then must be a prime.
Proof. See Theorem of .
Lemma 4. If is an odd prime with , then the equation has solutions .
Proof. See Section 8.1 of .
Lemma 5. The equation has only the solution .
Proof. See Theorem 8.4 of .
Lemma 6. Let be a fixed odd positive integer. If the equation has solutions , then the equation has at most two solutions , except the following cases:(i), , , , and , where is a positive integer with ;(ii), , , and , where is a positive integer with ;(iii), , , and , where , are positive integers with .
Proof. See .
Lemma 7. If is an odd prime and belongs to the exceptional case (i) of Lemma 6, then .
Proof. We now assume that is an odd prime with . Then we have
If , since , then , and by (7), we have
But, by the second equality of (9), we get , a contradiction.
If , then from (8) we get Further, by the second equality of (10), we have , , and . Thus, the lemma is proved.
Lemma 8. If is an odd prime and belongs to the exceptional case (iii) of Lemma 6, then .
Proof. Using the same method as in the proof of Lemma 7, we can obtain this lemma without any difficulty.
Lemma 9. If belongs to the exceptional case (ii), then (6) has at most one solution with .
Proof. Notice that, for any positive integer , there exists at most one number of 5, , and which is a multiple of 3. Thus, by Lemma 6, the lemma is proved.
Lemma 10. The equation has only the solution .
Proof. See .
3. Proof of Theorem
We now assume that is a solution of (1). Then we have .
Here and below, we consider the remaining cases that . By the above analysis, we have . If , then and (4) has the solution with . But, by Lemma 5, it is impossible. Therefore, we have Substituting (17) into (1), the equation has the solution with . Since , by Lemma 4, (3) has solutions . Therefore, by Lemmas 6–9, (1) has at most two solutions . Thus, the theorem is proved.
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 National Natural Science Foundation of China (11371291).
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