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According to Fermat´s last theorem, equality in $ x^n+y^n= z^n $ is not possible for positive integers x, y, z, n if n≥3 .

Following proof for n=3 is based upon the assumption that there is equality and that this assumption leads to contradiction. The assumption that there is equality in (I) $ x^3+y^3= z^3 $ for some positive integers x, y, z is combined with (II) ${zover{x+y}}=sqrt[3]{z^3over(x+y)^3}$. The combination leads to contradiction and shows that $ {zover {x+y}}$ is irrational, then at least one of x,y,z is irrational and $x^3+y^3neq z^3$.

Assume that $ x^3+y^3= z^3 $ is in its reduced form, where x, y, z are pairwise coprime.

Then, assume that there is equality in (I) $ x^3+y^3= z^3 $ for some positive integers x, y, z . The expression $zover{x+y}$ is a rational number. Whether $zover{x+y}$ is reduced or not is unknown.

A. In the case that $zover{x+y}$ is reduced $Rightarrow$ z and x+y are coprime. Let ${zover{x+y}}={rover s}$, where $rover s$ is reduced $Rightarrow$ r and s are coprime $Rightarrow$ $sqrt[3]{z^3over(x+y)^3}={rover s}$ ; ${z^3over(x+y)^3}={r^3over s^3}$

Since both sides of ${zover{x+y}}={rover s}$ are reduced $Rightarrow$
$s^3=(x+y)^3 $ and $r^3=z^3$. Insert (I) $Rightarrow$ $r^3= x^3+y^3=(x+y)(x^2-xy+y^2)$. But then $r^3$ and $s^3$ have a common factor $x+y$ $Rightarrow$ $r$ and $s$ have a common factor $Rightarrow$ contradiction $Rightarrow$ $zover{x+y}$ is
irrational $Rightarrow$ at least one of x,y,z is irrational $Rightarrow$
$x^3+y^3neq z^3$.

B. In the case that $zover{x+y}$ is not reduced $Rightarrow$ z and x+y have
a common factor k $Rightarrow$
$ {rover s}={{z/k}over{(x+y)/k}}$ . Let ${rover s}=sqrt[3] {{(z/k)^3}over{((x+y)/k)^3}}$;
${r^3over s^3}={{(z/k)^3}over{((x+y)/k)^3}}$ ; both sides of ${rover s}={{z/k}over{(x+y)/k}}$ are reduced $Rightarrow$
$s^3={{x+y}over k^3} $ and and $r^3={z^3over k^3}$. Insert (I) $Rightarrow$ $r^3={{x^3+y^3}over k^3}$ = ${{(x+y)(x^2-xy+y^2)}over k^3}$.

B1. $k$ but not $k^2$or $k^3$ divides $x+y$.

What seems to be a shared common factor between $s^3$ and $r^3$ is ${x+y}over k$, which is an integer, so there seems to be a contradiction. Since $r^3$ is an integer,${{x^2-xy+y^2}over k^2}$ should be an integer. But
$x^2-xy+y^2$ is coprime to $x+y$ (*See below) and is not divisible by $k^2$ (**See below) $Rightarrow$ ${{x^2-xy+y^2}over k^2}$ is not an integer $Rightarrow$ $r^3={z^3over k^3}$ is an integer, but $r^3={{x^3+y^3}over k^3}$ is not an integer. That is a contradiction $Rightarrow$
$x^3+y^3neq z^3$

B2. $k^3$ divides $x+y$ $Rightarrow$ $x^2-xy+y^2$ is an integer.

$r^3$ $={{x+y}over k^3}cdot {(x^2-xy+y^2)}$and ${s^3={{x+y}over k^3}}{cdot(x+y)^2}$ $Rightarrow$ $r^3$ and $s^3$ have a common factor ${{x+y}over k^3}$ $Rightarrow$ r and s have a common factor. That is a contradiction $Rightarrow$ $zover{x+y}$ is irrational $Rightarrow$ at least one of x,y,z is irrational $Rightarrow$$x^3+y^3neq z^3$.

B3. $k^2$ but not $k^3$ divides $x+y$. $Rightarrow$ $r^3$ $={{x+y}over k^2}cdot {{x^2-xy+y^2}over k}$ and $s^3={(x+y)over k^2}{cdot(x+y)^2over k}$. If $k=3$, $x^2-xy+y^2$ can be divided by 3 and the common factor is ${{x+y}over k^2}$ $Rightarrow$ contradiction like in B2.

If $kneq 3$ $Rightarrow$ $x^2-xy+y^2$ is not divided by $k$. $x^2-xy+y^2$ is not an integer.There is a contradiction the same way as in B1$Rightarrow$ $x^3+y^3neq z^3$.

Fermat´s last theorem is true for n=3.

*According to the conditions, x is coprime to y $Rightarrow$ x and y both are coprime to x+y. Also, $x+y$ and $x^2-xy+y^2=(x+y)^2-3xy$ have no common factors (if 3 is not the common factor, see **). If the differens ${(x+y)^2-3xy}$ had a factor in common with one of the terms $Rightarrow$ the other term would have the same factor. There is also a theorem which could be used. see this link:
If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $frac{a^n+b^n}{a+b}$ is co-prime to $a+b$.

**If 3 divides x+y, then 3 does not divide x or y $Rightarrow$ $x^2-xy+y^2=(x+y)^2-3xy$ has at most one factor =3, since the expression is limited by the term $3xy$ which has only one factor 3 $Rightarrow$ $x+y$ and $x^2-xy+y^2$ have no common factors , if 3 is not the common factor $Rightarrow$ $x^2-xy+y^2$can not be divided by k, if $kneq3$. If k=3, $x^2-xy+y^2$ can be divided by 3 but not by $k^2=9$ $Rightarrow$ ${x^2-xy+y^2}over k^2$ is not an integer.