The Diophantine Equation x^n+y^m=c(x^k)(y^l), n,m,k,l,c natural numbers

TL;DR

This paper investigates the solutions of a specific Diophantine equation involving natural numbers, providing new theorems and proofs about the existence and uniqueness of positive integer solutions under various conditions.

Contribution

The paper introduces new theorems and detailed proofs regarding the solutions of the Diophantine equation x^n + y^m = c(x^k)(y^l), expanding understanding of its solution structure.

Findings

No solutions for certain parameter ranges when c ≠ 2

Unique solution (1,1) when c=2 under specific conditions

Exactly two solutions for specific parameter values when c=3

Abstract

The subject matter of this work is the diophantine equation x^n+y^m=c(x^k)(y^l), where n,m,k,l,c are natural numbers.We investigate this equation from the point of view of positive integer solutions.A preliminary examination of sources such as reference[1](L.E.Dickson's History of the Theory of Numbers, Vol.II) and [2](W.Sierpinski's Elementary Theory of Numbers) shows that little or no material can be found regarding this diophantine equation.Note that when c=1, (x,y)=(1,1) is a solution regardless of the values of the exponents n,m,k,and l. In Section3, five results from number theory are listed.The first four are well known and are stated without proof.Result5 is of central importance and it is used in the proofs of most of the nine theorems of this paper.We offer a detailed proof of Result5. The entire paper is organized according to eight cases. Here is a sample of two of the nine theorems. In Theorem2, we prove that if n<or=k and n<m<k+l, then the above has no positive integer solutions if c is not equal to 2;if c=2, then the above equation has the unique solution(x,y)=(1,1). In Theorem5, part(vi), we prove that if c=3, n-k=1,l=1,and m-n=1.Then the above equation has exactly two solutions:the (x,y)=(2,2),(2^k+1,2^k).