The case of the cubic triples:
Fermat’s Last Theorem states that there are no three positive integers a, b and c such thata^3 + b^3 = c^3 and more generally, there are no three positive integers a, b and c such that a^n + b^n = c^n where n > 2.
The simple proof for that is as follows.
Pythagoras theorem states:
c^2 = a^2 + b^2
If we multiply both sides by c,
c^3 = ca^2 + cb^2
Since the hypotenuse is greater than the sides, c > a and c > b
∴ ca^2 > a^3 and cb^2 > b^3
∴ ca^n > a^n and cb^n > b^n where n>2
Fermat’s Last Theorem. QED.
The case of the cubic quadruples:
In the case of the cubic quadruples, a^3 + b^3 + c^3 = d^3Take a cuboid with length l, breadth b, (diagonal of l, b is c) height h and diagonal d. It may be taken to be a cube where l = b = h. Then,
d^2 = l^2 + b^2 + h^2
Multiplying both sides by d,
d^3 = dl^2 + db^2 + dh^2
Since d > l, d > b and d > h, it follows
dl^2 > l^3, db^2 > b^3 and dh^2 > h^3
However, we can’t conclude d^3 > l^3 + b^3 + h^3 while its absolutely correct to say c^3 > a^3 + b^3. Why?
Cubic triples vs. Cubic quadruples:
The following may be noted:
[1] Non-Pythagorean integer triples may correspond to non-right triangles whose base may correspond to a cube of integer length that may be the sum of three cubes, with two cubes corresponding to the integer lengths of the other sides of such a triangle.
Non-Pythagorean integer triples d, e, f could correspond to non-right (acute or obtuse) triangles on a plane with d^2+e^2>f^2 or d^2+e^2For eg. 6^2 + 8^2 = 10^2 = 36 + 64 = 100
Also, 6^3 + 8^3 = 10^3 = 1000. 1000 does not divide into two cubes of integer lengths, but apparently, 6^2 + 8^2 > 9^2, and, 9^3 = 6^3 + 8^3 + 1^3 = 729.
In cases where d^2+e^2
The same could be scaled up, for eg. 12^3 + 16^3 + 2^3 = 18^3
It is seen that while it is impossible for a right triangle to scale up to a cube, acute or obtuse triangles may scale up to form cubes with positive integer lengths and can be divided to three cubes corresponding to the cubes of the sides of the said triangles.
[2] In the case of c^3 > a^3 + b^3, always a, b, c > 1 as otherwise one of the sides of the right triangle would be irrational vide Spiral of Theodorus and their cubes would also be correspondingly irrational. Only the Pythogorean Triples need to be considered at all as they are the only positive integers such that a^2 + b^2 = c^2. Since a * a^2 = a^3, b * b^2 = b^3 and c * c^2 = c^3, if an integer cannot even be expressed as the sum of two squares, it could not be expressed as the sum of two cubes. The smallest Pythogorean Triple is 3, 4, 5 and therefore, a, b are always greater than 2. It has already been seen that since the hypotenuse is greater than the sides, c > a and c > b, therefore ca^2 > a^3 and cb^2 > b^3.
[3] d^3 could be equal to l^3 + b^3 + h^3, as one of the three cubes could be 1, which is its own cube, and could be summed with two other cubes. For example eg. 9^3 = 6^3 + 8^3 + 1^3. It may be noted that the irrational diagonal of the square in the face of the cube gets omitted. We start with c^2 = l^2 + b^2 and since d^2 = h^2 + c^2, d^2 = l^2 + b^2 + h^2, any irrationality in the diagonal of the square face of a cube is eliminated as the internal diagonal of a cube can be expressed in terms of the length, breadth and height. While the cubic triples would never have 1 as one of the integers, cubic quadruples could have it and make a^3 + b^3 + c^3 = d^3 possible. In the case of the example, the cubes of the Pythogorean Triple 6 and 8 by themselves won’t make a cube but along with 1 cube, it adds up to 9 cube.
The cubic triples when compared with the cubic quadruples show that Fermat’s Last Theorem is accurate beyond any shadow of doubt.
Ref:
[1] https://www.calculator.net/triangle-calculator.html used to draw the figures on this page.
[2] https://plus.maths.org/content/triples-and-quadruples Triples and quadruples: from Pythagoras to Fermat by Chandrahas Halai
[3] Private email from Ravi Sundaram pointing to the Non-Pythogorean integer triples corresponding to non-right triangles.
