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 $n\geqq 3$ 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)
$y^n\equiv 0(modm)$
Let $a$ be an integer.
$y^n=am$ ...(3)
By the equation (1) and Fermat's little theorem,
$x+y\equiv z(modn)$
$y\equiv m(modn)$
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 $m^{\prime }$ be an integer and $m^{\prime }$ is one of $m$'s prime factors. Let $e$ be an integer and $e\not\equiv 0(mod{m}^{\prime })$ holds. We suppose two equations as below.
$x=cm+e$
$e/m=d$
By the equations (2),
$am=(cm+e+m{)}^n-(cm+e{)}^n$
$a=\sum _{i=1}^n((cm+e+m{)}^{n-i}(cm+e{)}^{i-1})$
$a\equiv n{e}^{n-1}(mod{m}^{\prime })$ ...(5)
holds.
ⅰ. When $a\equiv 0(mod{m}^{\prime })$ holds for all $m^{\prime }$
In this case, $n\equiv 0(mod{m}^{\prime })$ for all $m^{\prime }$ 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 $a\not\equiv 0(mod{m}^{\prime })$ holds for some $m^{\prime }$
By the congruent expression (5), $n\not\equiv 0(mod{m}^{\prime })$ holds in this case. Let $B(m^{\prime })$ and $M(m^{\prime })$ be integers and these are the numbers of prime factor $m^{\prime }$ in $b$ and $m$ respectively. If $B(m^{\prime })>0$ holds, then the number of prime factor $m^{\prime }$ in $y$ is $Min(B(m^{\prime }),M(m^{\prime }))$ by the equation (4). At this time, the number in $y^n$ is $Min(B(m^{\prime }),M(m^{\prime }))\times n$ and equals to $M(m^{\prime })$. It becomes inconsistent when $n\geqq 3$ holds. Therefore, $b\not\equiv 0(mod{m}^{\prime })$ must hold. By the equations (3) and (4),
$(bn+m{)}^n=am$
$bn\equiv 0(mod{m}^{\prime })$
holds. However, it becomes a contradiction since $b\not\equiv 0(mod{m}^{\prime })$ and $n\not\equiv 0(mod{m}^{\prime })$ 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 $n\geqq 6$ holds
We take account of two cases depending on the order as follows.
ⅰ. When the order includes an odd prime
Let $f$, $f^{\prime }$ and $g$ be positive integers. We suppose $f$ is an odd, $f\geqq 3$ holds and $f^{\prime }$ is one of $f$'s prime factors. Let $X$, $Y$ and $Z$ be integers. We suppose the following equation holds when the order is $f{2}^g$.
$(X^{f\times 2^g/f^{\prime }}{)}^{f^{\prime }}+(Y^{f\times 2^g/f^{\prime }}{)}^{f^{\prime }}=(Z^{f\times 2^g/f^{\prime }}{)}^{f^{\prime }}$
However, there are no non-zero solutions to this equation since this is the equation (1) when $x=X^{f\times 2^g/f^{\prime }}$, $y=Y^{f\times 2^g/f^{\prime }}$, $z=Z^{f\times 2^g/f^{\prime }}$ and $n=f^{\prime }$ 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 $h\geqq 3$ holds. Let $X^{\prime }$, $Y^{\prime }$ and $Z^{\prime }$ be integers. We suppose the following equation holds when the order is $2^h$.
$(X^{\prime 2^{h-2}}{)}^4+(Y^{\prime 2^{h-2}}{)}^4=(Z^{\prime 2^{h-2}}{)}^4$
Though, no non-zero solutions exist to this equation since this is the equation (1) when $x=X^{\prime 2^{h-2}}$, $y=Y^{\prime 2^{h-2}}$, $z=Z^{\prime 2^{h-2}}$ 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.)