Find all pairs $(x, y)$ of integers such that $x \ge 1$ and $y \ge 1$ and $x^{y^2} = y^x$.
work done: if $d = \gcd (x, y)$. then $x = du$ and $y = dv \implies \gcd (u, v) = 1$ and the equation becomes $(du)^dv^2 = (dv)^u$ if $dv^2 = u$ then $u = 1 = v$, $d = 1$ and hence $x = 1 = y \implies (1, 1)$ is one of the solution.
Now my question is, if $dv^2 > u$ and $dv^2 < u$, what are the solutions and how to conclude?
Find all pairs of $(x, y)$ such that $x \ge 1$ and $y \ge 1$ of $x ^{(y^2)} = y^x$.
work done: if $d = \gcd (x, y)$. then $x = du$ and $y = dv\implies \gcd (u, v) = 1$ and the equation becomes $(du)^{(dv^2)} = (dv)^u$ if $dv^2 = u$ then $u = 1 = v$, $d = 1$ and hence $x = 1 = y\implies (1, 1)$ is one of the solutions.
Now my question is, if $dv^2 > u$ and $dv^2 < u$, what are the solutions and how to conclude?