Optimized 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 equation for $n>2$.
$$x^n+y^n=z^n\ ...(1)$$
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 are prime to one another. 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 or $4$.

When $n$ is an odd prime

$ $Let $a$, $b$, $b_i$, $d$, $f$ and $g$ be positive integers and $c$ and $e$ be non-negative integers. We suppose that $a=x+y$ and $b=z^n/a$ hold. If $x=1$ and $y=1$ hold, then $z^n=2$ holds and there are no integer solution to the equation (1) in this case. Therefore, $a>2$ holds.
$ $We consider the case where $z^n/b$ is not divisible by $z$. We suppose that the following expressions hold.
$$z^n/b=cz+d$$
$$d≢0\pmod{z}$$
At this time, $z^n=(cz+d)b$ holds. Since $bd≡0\pmod{z}$ holds, $b=b_1z$ and $z^{n-1}=(cz+d)b_1$ hold. Since $b_1d≡0\pmod{z}$ holds, $b_1=b_2z$ holds. By repeating this operation, we obtain the congruent expression $b≡0\pmod{z^n}$. In this case, $b=z^n$ and $a=1$ hold. It becomes a contradiction contrary to the condition $a>2$. Hence, $z^n/b$ is divisible by $z$. Since $z^n=bcz$ and $z^n=ab$ hold, $x+y=cz$ holds.
$ $We consider the case where $z^n/a$ is not divisible by $z$. We suppose that the following expressions hold.
$$z^n/a=ez+f$$
$$f≢0\pmod{z}$$
In the similar manner as above, we obtain the congruent expression $a≡0\pmod{z^n}$. In this case, $a=z^n$ and $b=1$ hold. By the equation (1),
$$z^n-(z^n-x)^n=x^n$$
$$\sum_{i=0}^{n-1}z^i(z^n-x)^{n-1-i}=1$$
holds. Since $\sum_{i=0}^{n-1}z^iy^{n-1-i}>n$ holds obviously, this equation does not hold. Hence, $z^n/a$ is divisible by $z$ and $z^n=aez$ holds. Since $ab=aez$ holds, $b=ez$ holds. Since $b=\sum_{i=0}^{n-1}(-y)^ix^{n-1-i}$, $b≡0\pmod{z}$ and $x+y≡0\pmod{z}$ hold,
$$nx^{n-1}≡0\pmod{z}$$
holds. Therefore, $z=n$ holds since $x$ and $z$ have no common prime factors. Since $z^n=ab$ holds and $b≠1$ holds as described above, $b$ is a power of $n$. We suppose that $b=n^g$ and $g< n$ hold. By the equation (1), the following equation holds.
$$x^n+(n^{n-g}-x)^n=n^n$$
$$\sum_{i=1}^{n}{}_n\mathrm{C}_i(n^{n-g})^i(-x)^{n-i}=n^n$$
$$\sum_{i=1}^{n}{}_n\mathrm{C}_i(n^{n-g})^{i-1}(-x)^{n-i}=n^g$$
If $g>1$ holds, then
$$nx^{n-1}≡0\pmod{n^2}$$
holds. This is a contradiction contrary to the condition that $x$ and $z$ are coprime. Thus, $g=1$ holds. The following equation holds for $g=1$.
$$x^n+(n^{n-1}-x)^n=n^n$$
$$(x/n)^n+(n^{n-2}-x/n)^n=1$$
If $x/n≧1$ holds, then the left side of this equation is greater than $1$ since $y/n>0$ holds. If $x/n<1$ holds, then the left side is greater than $1$ since $n^{n-2}-x/n>2$ holds for $n>2$. Therefore, this equation does not hold.
$ $From the above, it is proved that there are no positive integer solutions to the equation (1) for $n>2$. (Q.E.D.)

When $n$ is 4

$ $By the equation (1), we consider the following equation.
$$x^4+y^4=z^4\ ...(2)$$
If $x$ and $y$ are odd, then the equation (2) has no positive integer solutions. Therefore, we suppose that $x$ and $z$ are odd and $y$ is even. Let $a$ and $b$ be positive integers. Since $z^4/(x+y)=\sum_{i=0}^3y^i(-x)^{3-i}$ and $z^4$ is divisible by $x+y$, we suppose that $x+y=a$ and $z^4=ab$ hold. By the equation (2),
$$x^4+(a-x)^4=ab$$
$$2x^4≡0\pmod{a}$$
holds. Since $a$ is odd, $x^4≡0\pmod{a}$ holds. It becomes a contradiction contrary to the condition that $x$ and $z$ are relatively prime. From the above, there are no positive integer solutions to the equation (2) for $n=4$. (Q.E.D.)