Short proof of Fermat's Last Theorem

Ⅰ. When $n$ is an odd prime
$ $Let $n$ be an odd prime, $x$ and $z$ be odd integers and $y$ be a non-zero even integer. We suppose $x$, $y$ and $z$ do not include a same prime and $n≧3$ holds.
$$x^n+y^n=z^n\ ...(1)$$
Let $m$ be an even integer. We suppose $z=x+m$ holds.
$$y^n=(x+m)^n-x^n\ ...(2)$$
$$y^n≡0\ (mod\ m)$$
Let $a$ be an integer.
$$y^n=am\ ...(3)$$
By the equation (1) and Fermat's little theorem,
$$x+y≡z\ (mod\ n)$$
$$y≡m\ (mod\ n)$$
holds. Let $b$ be in integer.
$$y=bn+m\ ...(4)$$
Dividing the equation (2) by $m^n$, we obtain the following equation.
$$a/m^{n-1}=(x/m+1)^n-(x/m)^n$$
Let $c$ be an integer and $d$ be a rational number and $0< d<1$ holds.
$$c=[x/m]$$
$$d=x/m-c$$
$$a/m^{n-1}=(c+1+d)^n-(c+d)^n$$
If $m$ and $x$ have a same prime as a factor, then $y$ includes the prime and it is inconsistent with the condition that $x$ and $y$ are relatively prime. Let $m'$ be an integer and $m'$ is one of $m$'s prime factors. Let $e$ be an integer and $e≢0\ (mod\ m')$ holds. We suppose two equations as below.
$$x=cm+e$$
$$e/m=d$$
By the equations (2),
$$am=(cm+e+m)^n-(cm+e)^n$$
$$a=\sum_{i=1}^{n}((cm+e+m)^{n-i}(cm+e)^{i-1})$$
$$a≡ne^{n-1}\ (mod\ m')\ ...(5)$$
holds.
ⅰ. When $a≡0\ (mod\ m')$ holds for all $m'$
$ $In this case, $n≡0\ (mod\ m')$ for all $m'$ and $n$ is a multiple of $rad(m)$. However, it becomes a contradiction since $n$ is an even contrary to the definition of $n$.

ⅱ. When $a≢0\ (mod\ m')$ holds for some $m'$
$ $By the congruent expression (5), $n≢0\ (mod\ m')$ holds in this case. Let $B(m')$ and $M(m')$ be integers and these are the numbers of prime factor $m'$ in $b$ and $m$ respectively. If $B(m')>0$ holds, then the number of prime factor $m'$ in $y$ is $Min(B(m'), \ M(m'))$ by the equation (4). At this time, the number in $y^n$ is $Min(B(m'), \ M(m'))×n$ and equals to $M(m')$. It becomes inconsistent when $n≧3$ holds. Therefore, $b≢0\ (mod\ m')$ must hold. By the equations (3) and (4),
$$(bn+m)^n=am$$
$$bn≡0\ (mod\ m')$$
holds. However, it becomes a contradiction since $b≢0\ (mod\ m')$ and $n≢0\ (mod\ m')$ must hold.
From the above, Fermat's Last Theorem is true in the case Ⅰ. When $x$ and $y$ are odd integers and $z$ is an even, it can be proven by changing the definitions of $y$ and $z$ mutually and by converting variables from $y$ to $-z$ and $z$ to $-y$ when n is an odd. If n is an even, converting odd n to even.

Ⅱ. When $n$ is an even and $n≧6$ holds
$ $We take account of two cases depending on the order as follows.
ⅰ. When the order includes an odd prime
$ $Let $f$, $f'$ and $g$ be positive integers. We suppose $f$ is an odd, $f≧3$ holds and $f'$ is one of $f$'s prime factors. Let $X$, $Y$ and $Z$ be integers. We suppose the following equation holds when the order is $f2^g$.
$$(X^{f×2^g/{f'}})^{f'}+(Y^{f×2^g/{f'}})^{f'}=(Z^{f×2^g/{f'}})^{f'}$$
However, there are no non-zero solutions to this equation since this is the equation (1) when $x=X^{f×2^g/{f'}}$, $y=Y^{f×2^g/{f'}}$, $z=Z^{f×2^g/{f'}}$ and $n=f'$ hold.

ⅱ. When the order does not include an odd prime
$ $We consider this case by reducing it to the case $n=4$ as proven by Fermat. Let $h$ be an integer and $h≧3$ holds. Let $X'$, $Y'$ and $Z'$ be integers. We suppose the following equation holds when the order is $2^h$.
$$(X'^{2^{h-2}})^4+(Y'^{2^{h-2}})^4=(Z'^{2^{h-2}})^4$$
Though, no non-zero solutions exist to this equation since this is the equation (1) when $x=X'^{2^{h-2}}$, $y=Y'^{2^{h-2}}$, $z=Z'^{2^{h-2}}$ and $n=4$ hold.
Therefore, there are no non-zero 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.)