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I don't know. I wish it doesn't give any solution at all :-D

if x=0, y=an imaginary number

why?

9^0-3^0=1-1=0
and when y=0 then
0+0+0+0=0
same answer, no?

factor out y, you'll end up with \(\ (y^2+2y)(y^2+1)=9^x-3^x \)
when x=0, \(\ y^2=-1, y=\sqrt{-1} \)

but \(\ y^2+2y=0 => y=-2 \)

y(y+2)(y^2+1)=0
so y=0 and y=-2 no?

please write full statements lol, you're getting me lost by ... no? :P

=> my solutions is wrong? not sure... i guess i'll wait for to see the correct one :D

so why wolframalpha does not show y=0?

it isn't as smart as u ;)

Great job everyone, some solution(s) is/are still missing though.

actually for case 2 if \(x>1\)\[|m(3^xm-2)(5+(3^xm^2-4m)3^x)|>|(5+(3^xm^2-4m)3^x)|>5+3^x\]

only solutions\[(x,y)=(0,0)\ ,\ (0,-2)\ , \ (1,1)\]