# 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).

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 $$m$$

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.

- 3 months, 3 weeks ago

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

- 3 months, 3 weeks ago

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).

- 3 months, 3 weeks ago

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.....

- 3 months, 3 weeks ago

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 $$a+b=12, m =2.$$

- 3 months, 3 weeks ago

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

- 3 months, 3 weeks ago

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

- 3 months, 3 weeks ago

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

- 3 months, 3 weeks ago

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

- 3 months, 3 weeks ago

don't judge grammar and typos

- 3 months, 3 weeks ago

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