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

According to Fermat's Last Theorem, an+bn=cn{ 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 x3+y3=z3{ x }^{ 3 }+{ y }^{ 3 }={ z }^{ 3 }.

Expand x3+y3=z3{ x }^{ 3 }+{ y }^{ 3 }={ z }^{ 3 } and you will get (x+y)(x2xy+y2)(x+y)({ x }^{ 2 }-xy+{ y }^{ 2 })

So, x3+y3{ x }^{ 3 }+{ y }^{ 3 }= (x+y)(x2xy+y2)(x+y)({ x }^{ 2 }-xy+{ y }^{ 2 })

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

Hence, let zz be x+yx+y

(x+y)(x2xy+y2)(x+y)({ x }^{ 2 }-xy+{ y }^{ 2 })=(x+y)3{ (x+y) }^{ 3 }

Dividing both sides by (x+y)(x+y)

x2xy+y2{ x }^{ 2 }-xy+{ y }^{ 2 }=(x+y)2{ (x+y) }^{ 2 }

Expanding (x+y)2{ (x+y) }^{ 2 }, you will get x2+2xy+y2{ x }^{ 2 }+2xy+{ y }^{ 2 }

Simplifying the equation, xy=2xy-xy=2xy

0=3xy0=3xy

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

OR

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

But anybody know that 00 can be a solution.

Note that 00 is a neutral number

Now consider the case, zx+yz\neq x+y

Assume z=k(x+y)z=k(x+y)

Where kk 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
4 years, 8 months ago

No vote yet
1 vote

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@Calvin Lin I think you would like this.

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

Julian Poon - 4 years, 8 months ago

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Only hitch in the solution is that 3xy=0 3xy = 0 does not imply that BOTH x=0 x = 0 and y=0 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 -1 = 2 ? Or has the note been edited?

Siddhartha Srivastava - 4 years, 8 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-xy=2xy

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

1=2-1=2

Since this proof does not have the division by 00, I don't think there is a problem.

Julian Poon - 4 years, 8 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 - 4 years, 8 months ago

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

Luke Zhang - 4 years, 8 months ago

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The Boh is right.

Julian Poon - 4 years, 8 months ago

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The BOH?

Luke Zhang - 4 years, 8 months ago

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I still dont think that there is any error in this cause there is no division by 00. It is just proving xx and/or yy is 00.

Julian Poon - 4 years, 8 months ago

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

Luke Zhang - 4 years, 8 months ago

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@Luke Zhang THats why there isnt an error.

Julian Poon - 4 years, 8 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 - 4 years, 8 months ago

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