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Topic:
Why 2 cubes can never be reassembled into bigger cube
Replies:
1
Last Post:
Jun 28, 2005 12:27 PM



Jacques
Posts:
35
Registered:
12/4/04


Why 2 cubes can never be reassembled into bigger cube
Posted:
Jun 28, 2005 9:10 AM


Imagine a cube made up of lots of small cubes. If the number of cubes added to enlarge the original cube can be rearranged to form another cube then, of course, x^3+y^3=z^3 will hold which is, according the the famous Fermat?s Last Theorem, impossible.
The number of cubes needed to enlarge the original n*n*n cube by one layer is 3*n^2+3*n+1, ie a layer on each of three faces, a row between each of the new faces and a single cube in the corner. Or, maybe easier to see, (n+1)^3n^3.
The number of cubes to enlarge the cube by k layers is then 3*k*n^2+3*k^2*n+k^3, ie (n+k)^3n^3.
The series generated by taking a single cube (1*1*1) and adding a single layer, then a double layer, then three layers etc (ie n=1 and k=1, 2, 3, ?) is 7, 26, 63, 124, ?. It is clear, sommer by inspection, that this can be rewritten in the format (n+1)^31. Any cubic number minus one can never be a cube (integers, of course). In the row of positive integers there are no consecutive cubes.
The series generated by taking the next sized cube (2*2*2) and adding a single layer, then a double layer, then three layers etc (ie n=1 and k=1, 2, 3, ?) is 19, 56, 117, 208, ..etc which can, of course, be rewritten as a cubic minus eight, ie (n+2)^38. The differences between these consecutive numbers are the same as the differences between the series of cubic numbers and as the ?starting point?, namely 19, is not a cubic number and, given the geometric nature of the differences, another cubic number will never be ?hit?. In other words, adding any number of layers to a 2*2*2 cube a number of eight less than a cubic number will always be obtained.
The same goes for adding layers to a 3*3*3, 4*4*4, ?q*q*q cubes. The differences between the consecutive numbers of any series so created are the same as the differences between the cubic numbers (you can check that with some simple algebra) and the ?starting points? are always of such magnitude that another cubic will never be hit.
Thus, as long as no starting point is a cubic number FLT is OK. If one looks at the series of starting points, ie where k=1 and n=1, 2, 3, .. etc it is of the format (n+k)^3k^3 and, as k=1, it is a cubic number less one which can never be a cubic number.
One can therefore safely say that x^3+y^3 < > z^3 for any integer x,y or z. But can one?
Incidentally, the series generated by the differences between the square numbers is linear and not geometric as the differences above. The same ?method? as employed above shows that many squares can be hit, no matter what number the starting point is (as we all know, given the multitude of Pythagorean triplets).



