Fermat’s theorem solved correctly and simply.

Paper written on 28/12/2021.

By Dr H W. Lee. Mb, Bs

Fermat made a conjecture that no two positive numbers greater than 2

could be raised to a power greater than 2 and yet add together to form a third positive number to the same power.

This is trivially true due to the fact that when any two such numbers are added it creates an extra factor of 2 in the resultant third number.

which means that any reduction in the number by the power cannot be a positive complete number.

The easiest way to show this is to take two simple cubes such as 8 and 27 or a and b.

There is always a gap between the first and second cubes in any sum

The resultant addition is always the first cube added to itself plus the gap between the first and second cubes.

Therefore any cube has to equal the first cube doubled plus the gap between the first and second cubes.

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Since both the gap b-a and and the first cube a are cubes

this can be written in 2 ways

2a^3 +b ^3

or 2b^3 +a^3

When a and b are two different numbers they add up to 2 different sums.

Thus the one simple sum of addition gives 2 different answers

If the numbers do add up to a third number it would have to be the same either way.

Since it is not, neither answer can be correct

so two cubes cannot equal a third numerical cube.

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Further every a and b as a gap does form a cube with two cubes.

But only when those cubes are half of a real cube.

Hence 2[1/2 a^3] +b^3 = c^3 but only when the 1/2 cubes add to form a real cube.

Similarly 2[1/ b^3] + a^3 = c^3 but only when the gap is a half gap or cube

In other words for every gap b-a there is a cube c^3 which = 1/2 a^3

where two such cubes can form a cube but only if they themselves first add to a cube.

So when the cube is doubled it always has a factor of the square root of two as a cube and

therefore cannot form a cube with a numerical cube root when added to a cube with a numerical cube root.

When swapped around the same logic applies to the gap as a cube.

If it is a cube b^3 [ eg 19] then it will form a doubled cube 2b^3 that has to add to a cube a [8] to form a numeric cube and again,

as a doubled cube it cannot make a rational numeric cube root capable of adding to the gpa which is now the cube 8.