So first of all the equation, for which I thick i can prove there is no whole number solution for it. At the end you can know why I can do it only for 3 (because of the method I used).
Proof by contradiction.
So now for to be an whole number, the term must be a cube
Now let us think about the term and what will be its form so that its subtraction will lead to a cube.
- Imagine a (4 by 4) cube to turn it into (3 by 3)
- there are 64 small cubes in 4 by 4
- there are 27 small cubes in 3 by 3
- to convert it will be
- here the has form
- Now similarly we convert into any other cube by subtracting
- With above logic we note that must be of form,
- , where is a natural number [IMP step]
- so now we prove that this equation is cannot be true(and thus contradiction)
- by inspecting we conclude that is multiple of 3 so let us assume that
we conclude that and thus ,where is natural number.
now note that is more than 3 for natural (not proving that)
so in everything is positive and in there is zero
so we conclude that there exist no natural for which which is necessary condition for whole number solutions of .
so it is a
The proof is incomplete, and it is in progress when doesn't divides