Optimized proof of Fermat's Last Theorem

When $n$ is an odd prime

$$Let $x$, $y$, $y^{\prime }$ and $z$ be positive integers and $y^{\prime }$ is a prime factor of $z-x$. Let $p$ be an odd prime. We suppose $x$, $y$ and $z$ are each prime to one another and $y\not\equiv 0(modp)$ holds. We consider the following equation.
$$x^p+y^p=z^p...(1)$$
$$y^p=(z-x)\sum _{i=0}^{p-1}{z}^i{x}^{p-1-i}$$
Since $z>x$ and $z-x\equiv 0(mod{y}^{\prime })$ hold,
$$y^p/(z-x)\equiv p{x}^{p-1}(mod{y}^{\prime })$$
holds. The value on the right side is not $0$ since $x$ and $y$ are relatively prime and $y\not\equiv 0(modp)$ holds. Let $a$ and $b$ be integers. We suppose $y^p=(ay+b)(z-x)$ and $b\not\equiv 0(mody)$ hold since $y^p/(z-x)$ is not a multiple of $y$.
$$z-x\equiv 0(mod{y}^p)$$
Therefore, $y^p=z-x$ holds since $y^p/(z-x)$ is an integer. Then, it becomes a contradiction since $\sum _{i=0}^{p-1}{z}^i{x}^{p-1-i}=1$ 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

$$Let $x$, $y$, $y^{\prime }$ and $z$ be positive integers and $y^{\prime }$ is a prime factor of $z-x$. We suppose $x$, $y$ and $z$ are each prime to one another and $x$ is even and $y$ and $z$ are odd since there are no integer solutions to this equation when $z$ is even. We consider the following equation.
$$x^4+y^4=z^4...(2)$$
$$y^4/(z-x)=\sum _{i=0}^3{z}^i{x}^{3-i}$$
Since $z>x$ and $z-x\equiv 0(mod{y}^{\prime })$ hold,
$$y^4/(z-x)\equiv 4x^3(mod{y}^{\prime })$$
holds. The value on the right side is not $0$ since $x$ and $y$ do not have common prime factors and $y$ is odd. Let $c$ and $d$ be integers. We suppose $y^4=(cy+d)(z-x)$ and $d\not\equiv 0(mody)$ hold since $y^4/(z-x)$ is not divisible by $y$.
$$z-x\equiv 0(mod{y}^4)$$
Hence, $y^4=z-x$ holds since $y^4/(z-x)$ is an integer. And this is not proper since $\sum _{i=0}^3{z}^i{x}^{3-i}=1$ 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 $n\geqq 3$ holds, $x^n+y^n=z^n$ does not have an integer solution for $x>0$, $y>0$ and $z>0$.