Leningard mathematics league

My solution is similar to the Aditya one-

Examining the sign of both sides, we deduce that ${ y }^{ x } < 19$ . The case $y = 1$ is ruled out by

$\large\ { x }^{ { x }^{ { x }^{ x } } } = (19 - 1).1 - 74 < 0$ .

Hence $y \ge 2$ . But then $x \le 4$ .

For $x = 1$ , the equation becomes $1 = (19 - y)y - 74$ , or ${ y }^{ 2 } - 19y + 75 = 0$ , which has no integral solutions. If $x > 2$ , by requiring that $19 - { y }^{ x }$ be positive, we obtain the case $(x, y) = (2, 2), (2, 3), (3, 2), (4, 2)$ . Performing the computations, we see that only $(x, y) = (2, 3)$ yields a solution.

Since LHS And RHS is positive we must have a inequality,

$y^x$ < 19 for r.h.s to be positive Now only perfect squares are 16,9,4,1 before 19.

Checking for 16 as $4^2$ & $2^4$ yield no result . But, The next attempt for 9 gets us (x,y)=(2,3) and is easy.