So first of all the equation, \((a^{3}+b^{3}=c^{3})\) 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, as \((a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2}\)

\(a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)\)

\(c^{3}=(a+b)^{3}-3ab(a+b)\)

\(c=[(a+b)^{3}-3ab(a+b)]^{\frac{1}{3}}\)

So now for \("c"\) to be an whole number, the term \((a+b)^{3}-3ab(a+b)\) must be a cube

Now let us think about the term \(3ab(a+b)\) 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 \(4^{3}-x=3^{3}\)
- here the \(x\) has form \( {\color{Red} [3\cdot 1\cdot (4)^{2}]}-{\color{Blue} [3\cdot 1^{2}\cdot 4]}+{\color{Green} [1]^{3}}\)

- Now similarly we convert \((a+b)^{3}\) into any other cube by subtracting \(3ab(a+b)\)
- With above logic we note that \(3ab(a+b)\)
**must**be of form, - \(3ab(a+b)=3k(a+b)^{2}-3k^{2}(a+b)+k^{3}\), where \(k\) is a natural number [IMP step]
- so now we prove that this equation is cannot be true(and thus contradiction)
- \(3k(a+b)[a+b-k]+k^{3}=3ab(a+b)\)
- by inspecting we conclude that \(k\) is multiple of 3 so let us assume that\(k=3m\)

\(\Rightarrow 9m(a+b)[a+b-3m]+27m^{3}=3ab(a+b)\)

\(\Rightarrow 3m(a+b)[a+b-3m]+9m^{3}=ab(a+b)\)

\(\Rightarrow 3m(a+b)^{2}-9m^{2}(a+b)+9m^{3}=ab(a+b)\)

\(\Rightarrow 3m(a+b)-9m^{2}+\frac{9m^{3}}{a+b}=ab\)

\(\Rightarrow\) we conclude that \(m=(a+b)n\) and thus \(k=3(a+b)n\),where \(n\) is natural number.

\(\Rightarrow 3(a+b)^{2}n-9n^{2}(a+b)^{2}+9n^{3}(a+b)^{2}=ab\)

\(\Rightarrow (a+b)^{2}[3n+9n^{3}-9n^{2}]=ab\)

\(\Rightarrow\) Let \([3n+9n^{3}-9n^{2}]\) be \(S\)

\(\Rightarrow (a+b)^{2}S=ab\)

\(\Rightarrow Sa^{2}+Sb^{2}+2Sab=ab\)

\(\Rightarrow Sa^{2}+Sb^{2}+(2S-1)ab=0\)

\(\Rightarrow\) now note that \(S\) is more than 3 for natural \(n\) (not proving that)

\(\Rightarrow\) so in \(L.H.S.\) everything is positive and in \(R.H.S.\) there is zero

\(\Rightarrow\) so we conclude that there exist no natural \(k\) for which \(3ab(a+b)=3k(a+b)^{2}-3k^{2}(a+b)+k^{3}\) which is necessary condition for whole number solutions of \((a^{3}+b^{3}=c^{3})\).

\(\Rightarrow\) so it is a \(\boxed{\mathbf{CONTRADICTION}}\)

The proof is incomplete, and it is in progress when \((a+b)\) doesn't divides \(m\)

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The beginning is confusing and a little more complicated than it needs to be. You're writing \(a^3+b^3=c^3,\) and letting \(c=a+b-k.\) Expanding this leads to the "IMP step" \[ 3ab(a+b) = 3k(a+b)^2-3k^2(a+b)+k^3. \] I agree that \(k=3m\) for some \(m,\) and that leads to \[ 3m(a+b)-9m^2+\frac{9m^3}{a+b} = ab. \] (There's a typo in your version of this equation--you temporarily forgot to cancel \(a+b\) from the right side, but that's corrected on the next line.)

The argument fails here: you claim that \(a+b\) divides \(m.\) It divides \(9m^3,\) yes, but not necessarily \(m.\) Your argument works when \(a+b\) is a prime not equal to \(3,\) but that case is easy from the original equation: \((a+b)(a^2-ab+b^2) = c^3,\) so \(p=a+b\) divides \(c^3,\) hence \(c,\) so \(p|(a^2-ab+b^2),\) but it also divides \(p^2 = a^2+2ab+b^2,\) so \(p|3ab,\) but it can't divide \(3,a,b,\) done.

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Hi, I appreciate for your point of view but I don't get what you are saying actually ( why c=a+b-k) . And thanks for the typo

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Don't worry about that part, I'm just re-deriving your formula more simply and quickly. The important part of the post is where I point out that the argument fails (the "we conclude that..." statement is not necessarily true).

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e.g \((a+b)=7\) and if \(n=1\) then \(m=7\) and \(k=21\)

if \(n=2\) then \(m=14\) and \(k=42\) and so on.....

e.g \((a+b)=6\) and if \(n=1\) then \(m=6\) and \(k=18\)

if \(n=2\) then \(m=12\) and \(k=36\) and so on.....

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\(a+b=3\)

\(a+b=9\)

or \(m=(a+b)n\) correct me I am wrong

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also you can do similar thing for n=2 (square) but there you will end up with condition and not contradiction

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don't judge grammar and typos

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