{ x }^{ 3 }+{ y }^{ 3 }={ z }^{ 3 }

According to Fermat's Last Theorem, ${ a }^{ n }+{ b }^{ n }={ c }^{ n }$ have no solutions in positive integers, if n is an integer greater than 2.

But I can (hopefully) only prove ${ x }^{ 3 }+{ y }^{ 3 }={ z }^{ 3 }$ .

Expand ${ x }^{ 3 }+{ y }^{ 3 }={ z }^{ 3 }$ and you will get $(x+y)({ x }^{ 2 }-xy+{ y }^{ 2 })$

So, ${ x }^{ 3 }+{ y }^{ 3 }$ = $(x+y)({ x }^{ 2 }-xy+{ y }^{ 2 })$

Any integer can be expressed as $x+y$ .

Hence, let $z$ be $x+y$

$(x+y)({ x }^{ 2 }-xy+{ y }^{ 2 })$ = ${ (x+y) }^{ 3 }$

Dividing both sides by $(x+y)$

${ x }^{ 2 }-xy+{ y }^{ 2 }$ = ${ (x+y) }^{ 2 }$

Expanding ${ (x+y) }^{ 2 }$ , you will get ${ x }^{ 2 }+2xy+{ y }^{ 2 }$

Simplifying the equation, $-xy=2xy$

$0=3xy$

Hence, $x=0$ and $y=$ any integer.

$x=$ any integer and $y=0$ .

But anybody know that $0$ can be a solution.

Note that $0$ is a neutral number

Now consider the case, $z\neq x+y$

Assume $z=k(x+y)$

Where $k$ is any integer.

Using graph theory it is easy to realise that there are no rational solutions.

If interested, view Leonhard Euler's proof of this.

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
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`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
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`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Calvin Lin I think you would like this.

Seems legid but it means that $-1=2$ . I can't find the error and how to resolve it.

Julian Poon - 6 years, 4 months ago

Only hitch in the solution is that $3xy = 0$ does not imply that BOTH $x = 0$ and $y = 0$ . Only one the statements has to be true.

@Julian Poon . I don't see the problem you're having. Where does $-1 = 2$ ? Or has the note been edited?

Siddhartha Srivastava - 6 years, 4 months ago

I dont really think theres a problem with this but @Luke Zhang says that this step isnt valid even though this step is valid to me:

$-xy=2xy$

However, this step (to me) isn't valid:

$-1=2$

Since this proof does not have the division by $0$ , I don't think there is a problem.

For that part you can't prove -1=2 as you will be dividing both sides by 0 which is invalid. No problem with that prove. Just lacking 3 more cases.

Luke Zhang - 6 years, 4 months ago

Please comment about any loop holes (I'm only 14)

The Boh is right.

The BOH?

I still dont think that there is any error in this cause there is no division by $0$ . It is just proving $x$ and/or $y$ is $0$ .

Yes Julian But I said both x and y bust be 0

@Luke Zhang – THats why there isnt an error.

@Julian Poon – But there is! As I claimed both x and y must be 0 Why not ask Calvin to vet it?

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`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

x3+y3=z3{ x }^{ 3 }+{ y }^{ 3 }={ z }^{ 3 }x3+y3=z3

Comments

${ x }^{ 3 }+{ y }^{ 3 }={ z }^{ 3 }$