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:

TopNewestOnly 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

@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

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

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

The Boh is right.

Log in to reply

The BOH?

Log in to reply