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My friend pass me a simple proof that there is no positive integer solution for equation $a^3+b^3=c^3$.

I'm not sure whether the proof is right or not.

The proof:

suppose $a,b,c$ are positive integers , coprime pairwisely and satisfy the equation $a^3+b^3=c^3$

then we get

$(a+b-c)^3=3(a+b)(c-a)(c-b)~~~~~~(1)$

let $x=a+b$, so $x$ divides $c^3$ ,since $c^3 = a^3+b^3=(a+b)(a^2+ab+c^2)$

so, let $y=c-b,z=c-a$,then $y$ divides $a^3$, then $z$ divides $b^3$

we have $x,y,z$ are pairwisely coprime since $a^3,b^3,c^3$ are pairwisely coprime

by $(1)$, we can see

$(x-y-z)^3=8(a+b-c)^3=24xyz~~~~~(2)$

then $24xyz$ should have factor $x^3$, so $x^2$ divides $24$

Then $x=1$ or $2$

It's imposible.

Leitingok
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1 Answers1

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Your main result says that:

$$(x-y-z)^3=24xyz$$

...whcih basically means that $P=24xyz$ must be a perfect cube. Fair enough, but why does it implicate that $x^3$ divides $P$?

For example, $P$ is a perfect cube for $x=9$, $y=10^3$, $z=11^3$. But $9^3\nmid24\times9\times10^3\times11^3$.

Oldboy
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    this result in this post is by the assumption $a^3+b^3=c^3$ – Leitingok Nov 18 '18 at 09:40
  • It's easy to see that $x$ divides $c^3$, $y$ devides $a^3$,$z$ dvides $b^3$. $a^3,b^3,c^3$ are pairwise coprime,if $x,y,z$ are not pairwise coprime, for example $(x,y)=d$, then $d$ is common factor of $a^3$ and $c^3$, contradction. – Leitingok Nov 18 '18 at 09:48