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I'm aware that one can use a method of infinite descent, or even just refer to the more general case which has been proven by Andrew Wiles, but I was thinking about it the other day, and I remember seeing a proof which used the equation

$\left(x+y-z\right)^3=x^3+y^3-z^3+3\left(x+y\right)\left(x-z\right)\left(y-z\right)$

And assuming you had positive integer solutions so that

$x^3+y^3-z^3=0$

You eventually get to a contradiction which shows that your prior assumption can never be true.

I had a go at it, and as we know that z must be in the form of

$x+y-3a$

With a being some positive integer, you can form the equation.

$\left(3a\right)^3=3\left(x+y\right)\left(3a-x\right)\left(3a-y\right)$

And then I believe you should be able to find some contradiction.

I don't think the original proof that I saw used this technique, but I have scoured the internet and found nothing on this method of forming the equation

$\left(x+y-z\right)^3=x^3+y^3-z^3+3\left(x+y\right)\left(x-z\right)\left(y-z\right)$

And then finding a contradiciton. I believe (if my memory has served me well) the original proof looked at the fact that x,y and z must all be coprime and then there is a contradiction, however I could just be imagining that.

If anyone has the proof or knows how to do it that would be very useful as this has been bugging me recently.

Thanks in advance

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  • $\begingroup$ Here you find the same question with a proof using a different technique. $\endgroup$ – Rolandb Jun 3 at 21:16

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