Brazil | A Nice Algebra Problem | Math Olympiad 

2^6-5^2= 39 gives x= y= 16: x>0

Please mention in the beginning that x and y are integers. To solve for 2 variables, you need 2 equations, or rather 2 independent pieces of information. The fact that x and y are both integers is that critical missing piece of information that you casually assume in the middle of the solution.

64---25=39 ((2)^3/2)^)^x^1/2=2^6 X=16 (5^)^y^1/2=25=5^2 y=16

(8)^2 ➖ (5)^2={64 ➖ 25}=39 3^13 3^13^1 3^1^1 3^1 ( xy ➖ 3xy+1).

(x，y)＝（ (16/9)(log(2)20)^2，16(log(5)19)^2 )