a^6+2b^6=4c^6
2 must divide both a and b thus
a=2x, b=2y
64x^6+128y^6=4c^6
16x^6+32y^6=c^6
16(x^6+2y^6)=c^6
2 must divide c as well thus
c=2z
16(x^6+2y^6)=64z^6
x^6+2y^6=z^6
now we are back to the original equation,
thus we can recursively apply the same reasoning
to show that for an integer t>0, 2^t must divide all of a,b, and c.
The only integer which holds this property is 0. Thus the only
solution is a=b=c=0

Posted by Daniel
on 20130901 21:50:26 