4^x+16^y+64=8(2^x+4^y)+2^x*4^y ...(*)
Notam a=2^x>0 si b=4^y>0.
Atunci
(*) <=> a^2+b^2+64=8*(a+b)+ab
<=> (1/2)*(a^2-2ab+b^2)+(1/2)*(a^2-16a+64)+(1/2)*(b^2-16b+64)=0
<=> (a^2-2ab+b^2)+(a^2-16a+64)+(b^2-16b+64)=0
<=> (a-b)^2+(a-8)^2+(b-8)^2=0
<=> a=b=8.
a=b=8 <=> 2^x=4^y=8 <=> x=3 si y=3/2. => x=2y.
