A awesome mathematics problem | Olympiad Question | can you solve this problem | x=?,y=?

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[@peskarr]

What about x=y=1?

[@grchauvet]

Comparing the given equations, I tried x=y^2, since sqrt(40/10)=2. Then the equations become y^(2(y+y^2))=y^40 and y^(y+y^2)=y^20. Hence y^2+y=20, which can be observed to have positive solution y=4. Hence x=16, y=4 is a solution. x=25, y=-5 would also be a solution if negative integers were permitted.

[@grchauvet]

Don't forget x=1, y=1: If y=1 then raising it to any power will be 1.

[@RealQinnMalloryu4]

y^2^20y^ 2^5^4 y^2^5^1^4 y^2^1^12^2 y^1^1^11^2 1^2 (y ➖ 2y+1) x^2^5 (x ➖ 5x+2)