Another proof of Fermat's Last Theorem

Let $n$, $x$, $y$ and $z$ be positive integers. Fermat’s Last Theorem states that there are no solutions of the following equations for $n≧3$.
$$x^n+y^n=z^n …(1)$$
If two variables of $x$, $y$ and $z$ have a same prime as factor, a rest of those must include the prime. Therefore, we suppose these variables do not have a same prime factor and there is no loss of generality in this way. We will consider below by dividing the cases depending on whether $n$ is an odd prime or not.
Ⅰ. When $n$ is an odd prime
Let $a$ be a positive integer. We suppose the following equation holds.
$$z^n=(x+y)a$$
We will consider as follows by dividing the cases according to whether $z^n/a$ is divisible by $z$ or not.
ⅰ. When $z^n/a$ is a multiple of $z$
Let $b$ be a positive integer. We suppose the following equation holds.
$$z^n/a=x+y=bz …(2)$$
We consider the case where $n≡0\ (mod\ x+y)$ holds at first. At this time, $n=x+y$ holds since $n$ is a prime. In this case, $b=1$ holds since $x+y$ is a prime and $z≠1$ holds. When $b=1$ holds, $x^n+y^n=(x+y)^n$ holds. However, this equation does not hold.
In the second place, we consider the case where $n≢0\ (mod\ x+y)$ holds. Since $a= \sum_{i=0}^{n}x^{n-i}(-y)^{i-1}$ holds,
$$a≡nx^{n-1}≢0\ (mod\ x+y) …(3)$$
holds. By the equation (2),
$$(x+y)a=((x+y)/b)^n$$
$$ab^n=(x+y)^{n-1}$$
holds. Let $c$ be a positive integer. By the expression (3),
$$b^n=(x+y)^{n-1}c$$
holds. $a=1$ holds since $ac=1$ holds. By the equations (1) and (2), $x^n+y^n=x+y$ holds. $x=1$ and $y=1$ are the positive solutions when $n≧3$ holds. Then $z$ is not an integer since $z^n=2$ holds.

ⅱ. When $z^n/a$ is not a multiple of $z$
Let $d$, $e$ and $z'$ be positive integers and $z'$ is a prime factor of $z$. We suppose $z^n/a=dz'+e$ and $e≢0\ (mod\ z')$ hold.
$$z^n=(dz'+e)a$$
$$ae≡0\ (mod\ z')$$
Hence, $a$ is a multiple of $z'$. Let $f$ and $g$ be positive integers and we suppose $a=fz'^g$, $f≢0\ (mod\ z')$ and $1≦g≦n-1$ hold.
$$z'^{n-g}=(dz'+e)f$$
$$ef≡0\ (mod\ z')$$
It becomes a contradiction since this expression does not hold.
From the above, there are no positive solutions of the equation (1) in the case Ⅰ.

Ⅱ. When $n$ is not an odd prime and $n≧6$ holds.
We will take account of two cases depending on the order as follows.
ⅰ. When the order includes an odd prime
Let $p_i$ be an odd prime and $q_i$ and $r$ be positive integers. Let $h$, $j$, $x'$, $y'$ and $z'$ be positive integers and $j= \prod_{i=0}^{r}{p_i}^{q_i}×2^h$ holds. We suppose the following equation holds when the order is $j$.
$$(x'^{j/p_i})^{p_i}+(y'^{j/p_i})^{p_i}=(z'^{j/p_i})^{p_i}$$
However, there are no positive solutions to this equation since this is the equation (1) when $x=x'^{j/p_i}$, $y=y'^{j/p_i}$, $z=z'^{j/p_i}$ and $n=p_i$ hold.

ⅱ. When the order does not include an odd prime
In this instance, we use the fact that there is no solution when $n=4$ holds, which was proven by Fermat. Let $k$ be an integer and $k≧3$ holds. Let $x''$, $y''$ and $z''$ be positive integers. We suppose the following equation holds when the order is $2^l$.
$$(x''^{2^{k-2}})^4+(y''^{2^{k-2}})^4=(z''^{2^{k-2}})^4$$
Though, no positive solutions exist to this equation since this is the equation (1) when $x=x''^{2^{k-2}}$, $y=y''^{2^{k-2}}$, $z=z''^{2^{k-2}}$ and $n=4$ hold.
Therefore, there are no positive solutions to the equation (1) in the case Ⅱ. From the above, it is proved that Fermat's Last Theorem is true. (Q.E.D.)