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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 positibe integer solutions to the following equation when $n≧3$ holds.
$$x^n+y^n=z^n\ …(1)$$
If two variables of $x$, $y$ and $z$ have a same prime as factor, then a rest of those must include the prime. We suppose that these variables do not have a same prime factor. There is no loss of generality in making this supposition. We will consider below by dividing the cases depending on whether $n$ is an odd prime, $4$ or some other value.

Ⅰ. When $n$ is an odd prime

$ $Let $a$, $b$ and $z'$ be positive integers and $z'$ be a prime factor of $z$. We suppose that the following expressions hold.
$$z^n/a=x+y=b$$
$$b≡0\pmod{z'}$$
In this case,
$$a=\sum_{i=0}^{n-1}(x-b)^ix^{n-1-i}$$
$$a≡nx^{n-1}\pmod{z'}$$
holds. If $b≢0\pmod{n}$ holds, then the value on the right-hand side is not $0$ and there are no common prime factors of $a$ and $b$ since $x$ and $z$ are relatively prime. Conversely, if $b≡0\pmod{n}$ holds, then the value on the right-hand side is $0$ for $z'=n$, the one is not $0$ for $z'≠n$ and there is a common prime factor of $a$ and $b$ $n$.
$ $To start with, we consider the case where $b≢0\pmod{n}$ holds. Let $a_i$, $c$, $d$, $e$, $f$ and $g$ be positive integers. We suppose that $z^n/a=z+c$ holds, $d$ is the greatest common divisor of $a$ and $c$, $e=a/d$ holds, $f$ is the greatest common divisor of $b$ and $c$ and $g=b/f$ holds. At this time, $z=\sqrt[n]{defg}$ holds, $c$ is divisible by $df$ and $c$, $e$ and $g$ are prime to one another. We consider the following equation.
$$z^n=(z+c)a$$
Since $ac≡0\pmod{\sqrt[n]{defg}}$ holds, $a=a_1\sqrt[n]{eg}$ holds.
$$z^n/\sqrt[n]{eg}=(z+c)a_1$$
Since $a_1c≡0\pmod{\sqrt[n]{defg}}$ holds, $a_1=a_2\sqrt[n]{eg}$ holds. Repeating this operation, we obtain the following congruent expression.
$$a≡0\pmod{eg}$$
Because $a$ and $b$ are coprime, $g$ must be $1$ and $b=f$ holds. However, this is a contradiction since $c$ must be a multiple of $b$ in this case.
$ $Secondly, we consider the case where $b≡0\pmod{n}$ holds. Let $h$ be a positive integer and the greatest common divisor of $a$ and $b$. We obtain a similar proof by replacing $a$ with the value of $a$ divided by $h$ and supposing that $z=\sqrt[n]{defgh}$ holds in the proof described above.
$ $From the above, there are no positive integer solutions to the equation (1) when $n$ is an odd prime.

Ⅱ. When $n$ is $4$

$ $If $x$ and $y$ are odd, then the equation (1) has no positive integer solutions when $n$ is $4$. Therefore, we suppose that $x$ and $z$ are odd and $y$ is even. Since $y$ is even and $z$ is odd, $y+z$ is odd. Let $a$, $b$ and $x'$ be positive integers and $x'$ be a prime factor of $x$. We suppose that the following expressions hold.
$$x^4/a=y+z=b$$
$$b≡0\pmod{x'}$$
In this case,
$$a=\sum_{i=0}^3(b-y)^i(-y)^{3-i}$$
$$a≡-4y^3\pmod{x'}$$
holds. The value on the right-hand side is not $0$ and there are no common prime factors of $a$ and $b$ since $b$ is odd and $x$ and $y$ are relatively prime.
$ $Let $a_i$, $c$, $d$, $e$ and $f$ be positive integers. We suppose that $x^4/a=x+c$ holds, $d$ is the greatest common divisor of $a$ and $c$, $e=a/d$ holds, $f$ is the greatest common divisor of $b$ and $c$ and $g=b/f$ holds. At this time, $x=\sqrt[4]{defg}$ holds, $c$ is divisible by $df$ and $c$, $e$ and $g$ are prime to one another. We consider the following equation.
$$x^4=(x+c)a$$
Since $ac≡0\pmod{\sqrt[4]{defg}}$ holds, $a=a_1\sqrt[4]{eg}$ holds.
$$x^4/\sqrt[4]{eg}=(x+c)a_1$$
Since $a_1c≡0\pmod{\sqrt[4]{defg}}$ holds, $a_1=a_2\sqrt[4]{eg}$ holds. Repeating this operation, we obtain the following congruent expression.
$$a≡0\pmod{eg}$$
Because $a$ and $b$ are coprime, $g$ must be $1$ and $b=f$ holds. However, this is a contradiction since $c$ must be a multiple of $b$ in this case.
$ $From the above, there are no positive integer solutions to the equation (1) when $n$ is $4$.

Ⅲ. When $n$ is neither an odd prime nor $4$

$ $We will take account of two cases depending on the order as follows.
ⅰ. When the order includes an odd prime
$ $Let $a$ be a non-negative integer, $b$, $q_i$, $r$, $x'$, $y'$ and $z'$ be positive integers and $p_i$ be an odd prime. We suppose that $b=\prod_{i=1}^rp_i^{q_i}×2^a$ holds and the following equation holds when the order is $b$.
$$(x'^{b/p_i})^{p_i}+(y'^{b/p_i})^{p_i}=(z'^{b/p_i})^{p_i}$$
However, there are no positive integer solutions to this equation since this is the equation (1) when $x=x'^{b/p_i}$, $y=y'^{b/p_i}$, $z=z'^{b/p_i}$ and $n=p_i$ hold.

ⅱ. When the order does not include an odd prime
$ $Let $c$, $x''$, $y''$ and $z''$ be positive integers. We suppose that $c≧3$ holds. We suppose that the following equation holds when the order is $2^c$.
$$(x''^{2^{c-2}})^4+(y''^{2^{c-2}})^4=(z''^{2^{c-2}})^4$$
Though, no positive integer solutions exist to this equation since this is the equation (1) when $x=x''^{2^{c-2}}$, $y=y''^{2^{c-2}}$, $z=z''^{2^{c-2}}$ and $n=4$ hold.