Judge this proof: Fermat's last theorem for n=3

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 \(m\)

Note by Yash Ghaghada
3 months, 3 weeks ago

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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.

Patrick Corn - 3 months, 3 weeks ago

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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

Yash Ghaghada - 3 months, 3 weeks ago

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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).

Patrick Corn - 3 months, 3 weeks ago

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@Patrick Corn but why because \(m\) is some natural number \(n\) is some natural number

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.....

Yash Ghaghada - 3 months, 3 weeks ago

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@Yash Ghaghada The problem is that \(n\) is not necessarily a natural number: \(\frac{9m^3}{a+b}\) is an integer, but it's not necessarily true that \(\frac{m}{a+b}\) is an integer. For instance \(a+b=12, m =2.\)

Patrick Corn - 3 months, 3 weeks ago

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@Patrick Corn ok ok I get it. but there are only 3 cases i.e

\(a+b=3\)

\(a+b=9\)

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

Yash Ghaghada - 3 months, 3 weeks ago

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@Yash Ghaghada No, this is still wrong. Again, look at my example, where \(a+b=12.\) There are infinitely many counterexamples. It’s a serious hole.

Patrick Corn - 3 months, 3 weeks ago

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@Patrick Corn oh I get it, but my idea is correct that the equation is not true. I will update the note as I figure out to solve that problem and thank you

Yash Ghaghada - 3 months, 3 weeks ago

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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

Yash Ghaghada - 3 months, 3 weeks ago

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don't judge grammar and typos

Yash Ghaghada - 3 months, 3 weeks ago

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