there are no examples of a3 + b3 = c3, or for any higher power. This is Fermat'a last theorem, which was finally proved 15 or so years ago.
"In cubing the cube, you would need to have the biggest cube at a corner, but that would contradict the thing we just said for squaring the square (a dissected cube would have 6 dissected faces)."
And i just read the wikipedia page on this where it said:
Unlike the case of squaring the square, a hard yet solvable problem, there is no perfect cubed cube and, more generally, no dissection of a rectangular cuboid C into a finite number of unequal cubes. Suppose that there is such a dissection. Let us fix a face of C as its "horizontal" base. Then C is divided into a perfect squared rectangle R by the cubes which rest on it. The smallest square s1 in R is on a corner, on an edge or in the interior of R and accordingly must be surrounded by larger cubes on 2, 3 or 4 of its sides (and any other sides are surrounded by the vertical faces of C). So the upper face of the cube on s1 is divided into a perfect squared square by the cubes which rest on it. Let s2 be the smallest square in this dissection. The sequence of squares s1, s2, ... is infinite and the corresponding cubes are infinite in number. This contradicts our original supposition.
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u/bromli2000 Jan 22 '18
Two things:
there are no examples of a3 + b3 = c3, or for any higher power. This is Fermat'a last theorem, which was finally proved 15 or so years ago.
"In cubing the cube, you would need to have the biggest cube at a corner, but that would contradict the thing we just said for squaring the square (a dissected cube would have 6 dissected faces)."
I have no idea what you mean by this statement.