Tuesday, April 30, 2019

Fairing Fermat's Last Theorem


Fermat was a Superior Judge in France, who presided over several trials and had ordered execution of many a criminal, like fake gurus of that era, on infamous stakes. As a judge, he had to keep his distance and in his splendid isolation, mathematics was his chief love and companion. Nothing else comes between a proposition and its proof that is either clearly there in one’s mind or it simply does not exist. He enjoyed teasing and taunting his peers making outlandish mathematical claims sans proofs, challenging them to come with one. He wrote it is impossible to separate a cube into two cubes, and more generally, no three positive integers a, b, c satisfy the equation a^n + b^n = c^n for any integer value of n greater than 2. He remarked that the proof was too long to fit in the narrow margin of the book he made the note.
A cube to be separated into a pair of cubes, must first be capable of being separated into squares. The smallest Pythagorean triple is 3, 4 and 5. When squared, the numbers become 9, 16 and add to a perfect square of 25. If they are cubed, we get, 27, 64 and 125, but the cubes of 3 and 4 add up only to 91. If the same triple is scaled by two, the triple 6, 8 and 10 yield perfect squares that add up, i.e. 34 + 64 = 100, but when cubed, yield, 216 and 512 that add to 728, woefully short of 10^3. The woeful shortage observed while cubing Pythagorean triples is because while the diagonal of a square is square root of two, the diagonal of the inner space of a cube is square root of three, and the ratio between the two is 0.816. It is no coincidence that 91/125 = 728/1000 = 0.728. The ratio of the shortage is larger in the case of the other Pythagorean triples. When n > 3, it is only a reflection of the cube, n times in nDspace, and therefore there is never any let in the shortage.
However, 728 is interestingly very close to 9^3 – just one cube short of 729. Most obviously then
6^3 + 8^3 + 1^3 = 9^3
Curiously, 729 is both a cube and a square: 729 = 9^3 = 27^2
It is clearly possible to separate some cubes into three cubes, in the best case with the smallest Pythagorean triple 3, 4, 5 scaled by 2.
This could be scaled to infinity.
For example,
12^3 + 16^3 + 2^3 = 18^3
Numbers like 729 and 4096, that are both cubes and squares, must be called the Pythagorean lame doubles – lame right triangles with a hypotenuse and arm coinciding but with a missing arm, and are cubes that can be separated into three cubes, more or less.
4096 = 64^2 = 16^3 = 8^4 = 4^6
Again, 4096 = 15^3 + 9^3 – 2^3
The best it can be expressed in terms of cubes.
Only lame doubles come anywhere close to aiding the separation a cube into three smaller cubes, and separating a cube into just two cubes is simply impossible.
The judge would hardly have asked for more proof.