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

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.

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.

Note by Luke Zhang
3 years, 1 month ago

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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?

- 3 years, 1 month 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.

- 3 years, 1 month ago

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.

- 3 years, 1 month ago

@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.

- 3 years, 1 month ago

- 3 years, 1 month ago

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 $$0$$.

- 3 years, 1 month ago

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

- 3 years, 1 month ago

THats why there isnt an error.

- 3 years, 1 month ago

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

- 3 years, 1 month ago

The Boh is right.

- 3 years, 1 month ago

The BOH?

- 3 years, 1 month ago