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.