Optimized proof of Fermat's Last Theorem

When $n$ is an odd prime

Let $x$, $y$, $y'$ and $z$ be positive integers and $y'$ is a prime factor of $z-x$. Let $p$ be an odd prime. We suppose $x$, $y$ and $z$ are each prime to one another and $y≢0\ (mod\ p)$ holds. We consider the following equation.
$$x^p+y^p=z^p\ ...(1)$$
$$y^p=(z-x)\sum_{i=0}^{p-1}z^ix^{p-1-i}$$
Since $z>x$ and $z-x≡0\ (mod\ y')$ hold,
$$y^p/(z-x)≡px^{p-1}\ (mod\ y')$$
holds. The value on the right side is $0$ since $x$ and $y$ are relatively prime and $y≢0\ (mod\ p)$ holds. Let $a$ and $b$ be integers. We suppose $y^p=(ay+b)(z-x)$ and $b≢0\ (mod\ y)$ hold since $y^p/(z-x)$ is not a multiple of $y$.
$$z-x≡0\ (mod\ y)$$
Therefore, $y^p=z-x$ holds since $y^p/(z-x)$ does not have any prime fators of $y$. Then, it becomes a contradiction since $\sum_{i=0}^{p-1}z^ix^{p-1-i}=1$ holds. From the above, there are no integer solutions to the equation (1) for $x>0$, $y>0$ and $z>0$. (Q.E.D.)

When $n=4$ holds

Let $x$, $y$, $y'$ and $z$ be positive integers and $y'$ is a prime factor of $z-x$. We suppose $x$, $y$ and $z$ are each prime to one another. We consider the following equation.
$$x^4+y^4=z^4\ ...(2)$$
We suppose $x$ is even and $y$ and $z$ are odd since there are no integer solutions to this equation when $z$ is even. Since $z>x$ and $z-x≡0\ (mod\ y')$ hold,
$$y^4/(z-x)=\sum_{i=0}^3z^ix^{3-i}$$
$$y^4/(z-x)≡4x^3\ (mod\ y')$$
holds. The value on the right side is $0$ since $x$ and $y$ do not have common prime factors and $y$ is odd. Let $c$ and $d$ be integers. We suppose $y^4=(cy+d)(z-x)$ and $d≢0\ (mod\ y)$ hold since $y^4/(z-x)$ is not divisible by $y$.
$$z-x≡0\ (mod\ y)$$
Hence, $y^4=z-x$ holds since $y^4/(z-x)$ has no prime factors of $y$. And this is not proper since $\sum_{i=0}^3z^ix^{3-i}=1$ holds. From the above, there are no integer solutions to the equation (2) for $x>0$, $y>0$ and $z>0$. (Q.E.D.)

Conclusion

By the proof when $n$ is an odd prime and when $n$ is $4$ as above, when $n≧3$ holds, $x^n+y^n=z^n$ does not have an integer solution for $x>0$, $y>0$ and $z>0$.