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 $n\geqq 3$.
$$x^n+y^n=z^n\dots (1)$$
If two variables of $x$, $y$ and $z$ have a same prime as factor, then 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, $4$ or some other value.
Ⅰ. When $n$ is an odd prime
Let $a$ be a positive integer. We suppose $z\not\equiv 0(modn)$ holds and there is no loss of generality in making this supposition. 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 $b$ and $z^{\prime }$ be positive integers and $z^{\prime }$ is a prime factor of $x+y$. We suppose the following equation holds.
$$z^n/a=x+y=bz\dots (2)$$
Since $a=\sum _{i=0}^{n-1}{x}^{n-1-i}(-y{)}^i$ holds,
$$z^n/(x+y)\equiv n{x}^{n-1}(mod{z}^{\prime })$$
holds. The value on the right side is not $0$ since $x$ and $z$ are relatively prime and $z\not\equiv 0(modn)$ holds. Therefore, $a$ and $x+y$ have no common prime factors. By the equation (2),
$$(x+y)a=((x+y)/b{)}^n$$
$$a{b}^n=(x+y{)}^{n-1}$$
holds. Let $c$ be a positive integer. Since $a$ and $x+y$ are prime to each other,
$$b^n=(x+y{)}^{n-1}c$$
holds. Thus, $a=1$ holds since $ac=1$ holds. By the equations (1) and (2), $x^n+y^n=x+y$ holds. The positive solutions are $x=1$ and $y=1$ when $n\geqq 3$ 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 $a^{\prime }$ be integers and $a^{\prime }$ is a prime factor of $a$. In the case ⅱ, there is at least one prime factor $a^{\prime }$ where $z^n/a\not\equiv 0(mod{a}^{\prime })$ holds. In this instance, $x+y$ does not have a prime $a^{\prime }$ as a factor. By the way, we suppose $z^n/a=dz+e$ and $e\not\equiv 0(modz)$ hold.
$$z^n=(dz+e)a$$
$$a\equiv 0(modz)$$
Hence, $x+y=1$ holds since $x+y$ does not have any prime factors of $a$. In this case, the equation (1) has no positive integer solutions.

Ⅱ. When $n$ is $4$
We suppose $x$ is even and $y$ and $z$ are odd since there are no integer solutions to the equation (1) when $z$ is even. Let $f$ be a positive integer. We suppose the following equation holds.
$$y^4=(x+z)f$$
We will consider as follows by dividing the cases according to whether $y^4/f$ is divisible by $y$ or not.
ⅰ. When $y^4/f$ is a multiple of $y$
Let $g$ and $y^{\prime }$ be positive integers. We suppose the following equation holds.
$$y^4/f=x+z=gy\dots (3)$$
Since $f=\sum _{i=0}^3{x}^{3-i}(-z{)}^i$ holds,
$$z^4/(x+z)\equiv 4x^3(mod{y}^{\prime })$$
holds. The value on the right side is not $0$ since $x$ and $y$ are relatively prime and $y$ is odd. Therefore, $f$ and $x+z$ have no common prime factors. By the equation (2),
$$(x+z)f=((x+z)/g{)}^4$$
$$f{g}^4=(x+z{)}^3$$
holds. Let $h$ be a positive integer. Since $f$ and $x+z$ are prime to each other,
$$g^4=(x+z{)}^{3}h$$
holds. Thereby, $f=1$ holds since $fh=1$ holds. By the equations (1) and (3),
$$z^4-x^4=x+z$$
$$(x^2+z^2)(z-x)=1$$
holds. In order for this equation to have integer solutions, $x^2+z^2=1$ and $z-x=1$ must hold. The integer solutions are $(x,z)=(0,1),(-1,0)$. Therefore, the equation (1) does not have a positive solution.

ⅱ. When $y^4/f$ is not a multiple of $y$
Let $j$, $k$ and $f^{\prime }$ be integers and $f^{\prime }$ is a prime factor of $f$. In the case ⅱ, $y^4/f\not\equiv 0(mod{f}^{\prime })$ holds for at least one prime factor $f^{\prime }$. And $x+z$ does not have a prime $f^{\prime }$ as a factor. On the other hand, we suppose $y^4/f=jy+k$ and $k\not\equiv 0(mody)$ hold.
$$y^4=(jy+k)f$$
$$f\equiv 0(mody)$$
Hence, $x+z=1$ holds since $x+z$ has no prime factors of $f$ and the equation (1) has no positive integer solutions.

Ⅲ. When $n$ is not an odd prime and $n\geqq 6$ 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 $l$, $m$, $x^{\prime }$, $y^{\prime }$ and $z^{\prime }$ be positive integers and $m=\prod _{i=1}^r{p}_i^{q_i}\times 2^l$ holds. We suppose the following equation holds when the order is $m$.
$$(x^{\prime m/p_i}{)}^{p_i}+(y^{\prime m/p_i}{)}^{p_i}=(z^{\prime m/p_i}{)}^{p_i}$$
However, there are no positive integer solutions to this equation since this is the equation (1) when $x=x^{\prime m/p_i}$, $y=y^{\prime m/p_i}$, $z=z^{\prime m/p_i}$ 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 integer solution when $n=4$ holds, which was proven by Fermat. Let $s$ be an integer and $s\geqq 3$ holds. Let $x^{″}$, $y^{″}$ and $z^{″}$ be positive integers. We suppose the following equation holds when the order is $2^s$.
$$(x^{″2^{s-2}}{)}^4+(y^{″2^{s-2}}{)}^4=(z^{″2^{s-2}}{)}^4$$
Though, no positive integer solutions exist to this equation since this is the equation (1) when $x=x^{″2^{s-2}}$, $y=y^{″2^{s-2}}$, $z=z^{″2^{s-2}}$ and $n=4$ hold.

From the above, it is proved that Fermat's Last Theorem is true. (Q.E.D.)