On the Diophantine Equation p^x+(p+1)^y=z^2, where p is Prime

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Chantana Wannaphan
Suton Tadee

Abstract

In this paper, we study all non-negative integer solutions of the Diophantine equation  gif.latex?p^{x}+\left&space;(&space;p+1&space;\right&space;)^{y}=z^{2} , where gif.latex?p is a prime number. The research results showed that 1) If gif.latex?y=0, then the equation has only two solutions, namelyy gif.latex?\left&space;(&space;p,x,z&space;\right&space;)\in&space;\left&space;\{&space;\left&space;(&space;2,3,3&space;\right&space;)&space;,\left&space;(&space;3,1,2&space;\right&space;)\right&space;\},  2) if gif.latex?y=1, then the equation has only solutions in the form of gif.latex?\left&space;(&space;p,x,z&space;\right&space;)=\left&space;(&space;2,0,2&space;\right&space;) orgif.latex?\left&space;(&space;p,x,z&space;\right&space;)=\left&space;(&space;4n^{2}+4n-1,0,2n+1&space;\right&space;) , where gif.latex?n is a positive integer, 3) if gif.latex?y=2, then the equation has only two solutions, namely  gif.latex?\left&space;(&space;p,x,z&space;\right&space;)\in&space;\left&space;\{&space;\left&space;(&space;2,4,5&space;\right&space;)&space;,\left&space;(&space;3,2,5&space;\right&space;)\right&space;\}, 4) if gif.latex?p>3 and gif.latex?yis even, then the equation has no solution, and 5) if gif.latex?p=5, then the equation has only one solution. That is gif.latex?\left&space;(&space;x,y,z&space;\right&space;)=\left&space;(&space;4,3,29&space;\right&space;).

Article Details

How to Cite
Wannaphan, C., & Tadee, S. (2024). On the Diophantine Equation p^x+(p+1)^y=z^2, where p is Prime. Journal of Science Ladkrabang, 33(1), 1–7. Retrieved from https://li01.tci-thaijo.org/index.php/science_kmitl/article/view/258509
Section
Research article

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