Fermat

Every cube is made up of the complete sequence of the difference between the cubes plus 1.

Thus the complete sequence is 7, 19, 37, 61….infinity.

7+1 =8. 7+19+1 =27 etc.

The sequence can also run in reverse,

Infinity…..61,37,19,7. and again adding 1 will always produce a cube.

This allows us to find a proof that any cube between two cubes cannot add to a third cube.

The difference between every cube can be calculated in many ways.

A cube is the sum of all the integers in pyramidal or triangular sequence starting from 0 for the first cube

plus the root of the cube hence

1=1 .. 6×0 +1, 2 =8.. 6×1+2, 3= 27.. 6×4 plus 3, 4=64 .. 6×10 +4, etc

The triangular sequence sum being 0, 1 4, 19, 20, 35, etc

from the triangular sequence 0, 1, 3, 6, 10, 15, 21 etc.

So given any two cubes we know what the difference is

Which is 7 between the first two cubes or 6+1, 19 between the second and third or 6×3 plus 1

i.e the triangular sequence number x6 plus 1 where the triangular sequence starts at 0 for the first cube and 1 for the second cube etc.

Thus the difference between any two cubes is the sum of the 6n+1 numbers between the two cubes.

Any sum starting with 7 will form a complete sequence upwards [CS] which as previously described will always equal a cube.

The problem is that as the starting cubes get larger and the difference between the cubes gets larger

it is very difficult to prove definitively that a smaller cube might not somehow join to a larger cube to form a third cube.

Part of the problem stems from the fact that though a cube nominally needs a 6 plus 2 [8] for the second cube

and every cube thereafter needs a set of 6 plus 1 combinations this cannot be provable by simple maths.

The problem being a cube like 7 cubed, 343, also forms a 6 plus 1 number which could conceivably be an intervening partial sequence.

Other larger cubes could form a larger combination of PS seemingly doing away with the requirement for a 6 plus 1 shape at all.

The problem can be easily solved by tackling the sequence from the other direction, in reverse.

Every partial sequence can be seen to make a cube either by attaching to the cube made by the complete sequence below it and adding 1.

cube or by attaching to the partial sequence 19, + 7+1

Or by making it 1 bigger and adding it to the Complete sequence below it 19 +1, +7.

So we know that every PS or group of PS plus 1 can find a complete sequence below them to make a cube.

How does this rule out the fact that 19, or 19+37 [56], or a larger PS might be a cube?

Might be able to find a cube rather than the complete sequence to joint to form a cube?

By taking the sequence as a complete sequence in reverse we can say that whatever large PS we start with we can make a cube by making it 1 bigger, then adding the PS ending in 7.

The difference is that the PS “cube” can only form a cube with a CS when it, not the CS, has 1 added to it.

Now the “new” PS is making the cube.

But it is also a CS [in reverse]

So it is making a cube with a cube plus 1

If we add 1 to the CS we get our smaller cube but we also make the complete cube larger than a cube by 1. to make a cube plus 1

But it is a cube minus 1 so to be a cube it has to have 1 added to it.

When it is a cube it adds to a cube plus 1

If every part of any sequence reduces to zero it is part of a CS and cannot be a cube

37 reduces to zero hence cannot be a cube, ditto 56 or 63 What is left is the CS missing the parts reducing to zero which obviously cannot be a cube.

If the known bit is a PS adding 1to it enables it to form a cube with the missing CS but if the PS is a cube it must also be a CS minus 1

Now we have a complete sequence trying to form a cube with a cube.

It can only form a cube with 1 as a CS

not allowed per definition

or a PS plus 1 since the PS is also the other cube it can only equal the other cube plus 1

since it must go down to 1, not zero.

no PS being a cube can be anything other than a complete sequence plus 1

Taking 1 off it always reduces it to a CS

adding 1 to it makes it a cube

but if it is a complete sequence then the other cube must be a CS plus 1 as well

but a complete sequence can only join with a partial sequence plus 1

so a CS would have to equal a PS plus 1