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

So first of all the equation, (a3+b3=c3)(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=a3+b3+3a2b+3ab2(a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2}

a3+b3=(a+b)33ab(a+b)a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)

c3=(a+b)33ab(a+b)c^{3}=(a+b)^{3}-3ab(a+b)

c=[(a+b)33ab(a+b)]13c=[(a+b)^{3}-3ab(a+b)]^{\frac{1}{3}}

So now for "c""c" to be an whole number, the term (a+b)33ab(a+b)(a+b)^{3}-3ab(a+b) must be a cube


Now let us think about the term 3ab(a+b)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 43x=334^{3}-x=3^{3}
  • here the xx has form [31(4)2][3124]+[1]3 {\color{#D61F06} [3\cdot 1\cdot (4)^{2}]}-{\color{#3D99F6} [3\cdot 1^{2}\cdot 4]}+{\color{#20A900} [1]^{3}}

  • Now similarly we convert (a+b)3(a+b)^{3} into any other cube by subtracting 3ab(a+b)3ab(a+b)
  • With above logic we note that 3ab(a+b)3ab(a+b) must be of form,
  • 3ab(a+b)=3k(a+b)23k2(a+b)+k33ab(a+b)=3k(a+b)^{2}-3k^{2}(a+b)+k^{3}, where kk is a natural number [IMP step]
  • so now we prove that this equation is cannot be true(and thus contradiction)
  • 3k(a+b)[a+bk]+k3=3ab(a+b)3k(a+b)[a+b-k]+k^{3}=3ab(a+b)
  • by inspecting we conclude that kk is multiple of 3 so let us assume thatk=3mk=3m

9m(a+b)[a+b3m]+27m3=3ab(a+b)\Rightarrow 9m(a+b)[a+b-3m]+27m^{3}=3ab(a+b)

3m(a+b)[a+b3m]+9m3=ab(a+b)\Rightarrow 3m(a+b)[a+b-3m]+9m^{3}=ab(a+b)

3m(a+b)29m2(a+b)+9m3=ab(a+b)\Rightarrow 3m(a+b)^{2}-9m^{2}(a+b)+9m^{3}=ab(a+b)

3m(a+b)9m2+9m3a+b=ab\Rightarrow 3m(a+b)-9m^{2}+\frac{9m^{3}}{a+b}=ab

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

3(a+b)2n9n2(a+b)2+9n3(a+b)2=ab\Rightarrow 3(a+b)^{2}n-9n^{2}(a+b)^{2}+9n^{3}(a+b)^{2}=ab

(a+b)2[3n+9n39n2]=ab\Rightarrow (a+b)^{2}[3n+9n^{3}-9n^{2}]=ab

\Rightarrow Let [3n+9n39n2][3n+9n^{3}-9n^{2}] be SS

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

Sa2+Sb2+2Sab=ab\Rightarrow Sa^{2}+Sb^{2}+2Sab=ab

Sa2+Sb2+(2S1)ab=0\Rightarrow Sa^{2}+Sb^{2}+(2S-1)ab=0

\Rightarrow now note that SS is more than 3 for natural nn (not proving that)

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

\Rightarrow so we conclude that there exist no natural kk for which 3ab(a+b)=3k(a+b)23k2(a+b)+k33ab(a+b)=3k(a+b)^{2}-3k^{2}(a+b)+k^{3} which is necessary condition for whole number solutions of (a3+b3=c3)(a^{3}+b^{3}=c^{3}).

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

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

Note by Yash Ghaghada
1 year, 8 months ago

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don't judge grammar and typos

Yash Ghaghada - 1 year, 8 months 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 - 1 year, 8 months ago

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The beginning is confusing and a little more complicated than it needs to be. You're writing a3+b3=c3,a^3+b^3=c^3, and letting c=a+bk.c=a+b-k. Expanding this leads to the "IMP step" 3ab(a+b)=3k(a+b)23k2(a+b)+k3. 3ab(a+b) = 3k(a+b)^2-3k^2(a+b)+k^3. I agree that k=3mk=3m for some m,m, and that leads to 3m(a+b)9m2+9m3a+b=ab. 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+ba+b from the right side, but that's corrected on the next line.)

The argument fails here: you claim that a+ba+b divides m.m. It divides 9m3,9m^3, yes, but not necessarily m.m. Your argument works when a+ba+b is a prime not equal to 3,3, but that case is easy from the original equation: (a+b)(a2ab+b2)=c3,(a+b)(a^2-ab+b^2) = c^3, so p=a+bp=a+b divides c3,c^3, hence c,c, so p(a2ab+b2),p|(a^2-ab+b^2), but it also divides p2=a2+2ab+b2,p^2 = a^2+2ab+b^2, so p3ab,p|3ab, but it can't divide 3,a,b,3,a,b, done.

Patrick Corn - 1 year, 8 months 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 - 1 year, 8 months 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 - 1 year, 8 months ago

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@Patrick Corn but why because mm is some natural number nn is some natural number

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

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

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

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

Yash Ghaghada - 1 year, 8 months ago

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@Yash Ghaghada The problem is that nn is not necessarily a natural number: 9m3a+b\frac{9m^3}{a+b} is an integer, but it's not necessarily true that ma+b\frac{m}{a+b} is an integer. For instance a+b=12,m=2.a+b=12, m =2.

Patrick Corn - 1 year, 8 months ago

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

a+b=3a+b=3

a+b=9a+b=9

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

Yash Ghaghada - 1 year, 8 months ago

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

Patrick Corn - 1 year, 8 months 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 - 1 year, 8 months ago

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Watch Mathologer's proof of Fermat's Last Theorem.

A Former Brilliant Member - 1 year, 2 months ago

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