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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
$ $If $x+y≦z$ holds, then
$$(x+y)^n≦x^n+y^n$$
holds. It becomes a contradiction. If $x+y≧2z$ holds, then $(x+y)/2≧z$ holds and
$$((x+y)/2)^n≧x^n+y^n$$
$$max(x,y)^n≧x^n+y^n$$
holds. It becomes a contradiction. Therefore, the following inequalities hold.
$$z< x+y<2z\ ...(2)$$
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 $a_i$, $c$, $d$ and $e$ be positive integers. We suppose that $z^n/a=z+c$ and $0< c< z$ hold by the equation (2), $d$ is the greatest common divisor of $a$ and $z$ and $e$ is the number obtained by dividing $z$ by the greatest common divisor of $c$ and $z$. At this time, $c$, $d$ and $e$ are prime to one another. We consider the following equation.
$$z^n=(z+c)a$$
Since $ac≡0\pmod{de}$ holds, $a=a_1de$ holds. Then,
$$z^n/(de)=(z+c)a_1$$
holds. Since $a_1c≡0\pmod{de}$ holds, $a_1=a_2de$ holds. By repeating this operation, we obtain the following congruent expression.
$$a≡0\pmod{(de)^n}$$
Because $a=d^n$ holds, $e$ must be $1$. However, this is a contradiction since $c$ must be a multiple of $z$ in this case.
$ $Secondly, we consider the case where $b≡0\pmod{n}$ holds. Let $f$, $g$, $h$ and $j$ be positive integers. We suppose that the following equations hold where $f≢0\pmod{n}$ and $h≢0\pmod{n}$ hold.
$$a=fn^g$$
$$x+y=hn^j$$
By these equations,
$$x^n+(hn^j-x)^n=fhn^{g+j}$$
$$\sum_{i=1}^n{n\choose i}(hn^j)^i(-x)^{n-i}=fhn^{g+j}$$
holds. If $g≧2$ holds, then
$$hn^{j+1}x^{n-1}≡0\pmod{n^{j+2}}$$
holds. It becomes a contradiction contrary to the condition that $x$ and $z$ are coprime. Therefore, $a$ must have one prime number $n$ as a factor. Let $k$ be an integer and the result of subtracting the exponent of $n$ in $c$ from the exponent of $n$ in $b$. When $k>0$ holds, by performing the similar calculation as above, we can find that a must be a multiple of $n^{kn}$. However, this is not proper contrary to the condition that a has one prime number $n$ as a factor. Hence, there is a contradiction in the same way as above since $c$ must be a multiple of $z$.
$ $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 in this case. Therefore, we suppose that $x$ and $z$ are odd and $y$ is even. If $x+y≦z$ holds, then
$$(x+y)^4≦x^4+y^4$$
holds. This is a contradiction. Therefore, the following inequalities hold.
$$z-y< x\ ...(3)$$
Let $a$, $b$ and $x'$ be positive integers and $x'$ be a prime factor of $x$. We suppose that the following expressions hold.
$$x^4/a=z-y=b$$
$$b≡0\pmod{x'}$$
In this case,
$$a=\sum_{i=0}^3(y+b)^iy^{3-i}$$
$$a≡4y^3\pmod{x'}$$
holds. The value on the right-hand side is not $0$ and there are no common prime factors of $a$ and $b$ since $x$ is odd and $x$ and $y$ are relatively prime.
$ $Let $a_i$, $c$ and $d$ be positive integers. We suppose that $0< b< x$ hold by the equation (3), $c$ is the greatest common divisor of $a$ and $x$ and $d$ is the number obtained by dividing $x$ by the greatest common divisor of $b$ and $x$. At this time, $b$, $c$ and $d$ are prime to one another. We consider the following equation.
$$x^4=ab$$
Since $ab≡0\pmod{cd}$ holds, $a=a_1cd$ holds. Then,
$$x^4/(cd)=a_1b$$
holds. Since $a_1b≡0\pmod{cd}$ holds, $a_1=a_2cd$ holds. By repeating this operation, we obtain the following congruent expression.
$$a≡0\pmod{(cd)^4}$$
Because $a=c^4$ holds, $d$ must be $1$. However, this is a contradiction since $b$ must be a multiple of $x$ in this case. From the above, there are no positive integer solutions to the equation (1) when $n$ is $4$.

Ⅲ. 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 $a$ 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.