Geodesic
On the exponential diophantine equation $x^y+y^x=z^z$
Czechoslovak Mathematical Journal, Tome 67 (2017) no. 3, pp. 645-653 Cet article a éte moissonné depuis la source Czech Digital Mathematics Library

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For any positive integer $D$ which is not a square, let $(u_1,v_1)$ be the least positive integer solution of the Pell equation $u^2-Dv^2=1,$ and let $h(4D)$ denote the class number of binary quadratic primitive forms of discriminant $4D$. If $D$ satisfies $2\nmid D$ and $v_1h(4D)\equiv 0 \pmod D$, then $D$ is called a singular number. In this paper, we prove that if $(x,y,z)$ is a positive integer solution of the equation $x^y+y^x=z^z$ with $2\mid z$, then maximum $\max \{x,y,z\}480000$ and both $x$, $y$ are singular numbers. Thus, one can possibly prove that the equation has no positive integer solutions $(x,y,z)$.
For any positive integer $D$ which is not a square, let $(u_1,v_1)$ be the least positive integer solution of the Pell equation $u^2-Dv^2=1,$ and let $h(4D)$ denote the class number of binary quadratic primitive forms of discriminant $4D$. If $D$ satisfies $2\nmid D$ and $v_1h(4D)\equiv 0 \pmod D$, then $D$ is called a singular number. In this paper, we prove that if $(x,y,z)$ is a positive integer solution of the equation $x^y+y^x=z^z$ with $2\mid z$, then maximum $\max \{x,y,z\}480000$ and both $x$, $y$ are singular numbers. Thus, one can possibly prove that the equation has no positive integer solutions $(x,y,z)$.
DOI : 10.21136/CMJ.2017.0645-15
Classification : 11D61
Keywords: exponential diophantine equation; upper bound for solutions; singular number 
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Du, Xiaoying. On the exponential diophantine equation $x^y+y^x=z^z$. Czechoslovak Mathematical Journal, Tome 67 (2017) no. 3, pp. 645-653. doi: 10.21136/CMJ.2017.0645-15

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