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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
2 years, 10 months 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?

Siddhartha Srivastava - 2 years, 10 months ago

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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.

Julian Poon - 2 years, 10 months ago

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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 - 2 years, 10 months ago

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@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 - 2 years, 10 months ago

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

Luke Zhang - 2 years, 10 months ago

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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 \(0\).

Julian Poon - 2 years, 10 months ago

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

Luke Zhang - 2 years, 10 months ago

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@Luke Zhang THats why there isnt an error.

Julian Poon - 2 years, 10 months ago

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

Luke Zhang - 2 years, 10 months ago

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The Boh is right.

Julian Poon - 2 years, 10 months ago

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The BOH?

Luke Zhang - 2 years, 10 months ago

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