WakeUp wrote: Solve the system of equations in integers: \[\begin{cases}z^x=y^{2x}\\ 2^z=4^x\\ x+y+z=20.\end{cases}\] Second equation is $z=2x$ First equation is then $(2x)^x=y^{2x}$ and so $x\ne 0$ and either $y^2=2x$, either $y^2=-2x$ and $x$ is even. If $z=y^2=2x$, third equation is $3y^2+2y-40=0$ and so $y=-4$ and the solution $\boxed{(x,y,z)=(8,-4,16)}$ If $z=2x$ and $y^2=-2x$, third equation is $3y^2-2y+40=0$ and so no solution.

case 1--z=2x and z=y^2 case 2--z=2x & z=-[y^2] (x =even) this is enough for solution

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