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.

OR

\(x=\)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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewest@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.

Log in to reply

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?

Log in to reply

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.

Log in to reply

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.

Log in to reply

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

Log in to reply

The Boh is right.

Log in to reply

The BOH?

Log in to reply

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 \(0\).

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply