Simple 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. If the left side is a multiple of $y^{\prime }$, then $y^{\prime }=p$ must hold since $x$ and $y$ are relatively prime. $z-x$ and $y^p/(z-x)$ do not have common prime factors since $y\not\equiv 0(modp)$ holds. Let $a$ and $b$ be positive integers. We suppose $a$, $b$ and $p$ are each prime to one another and $y^p=ab$ and $z-x=a$ hold. By the equation (1),
$$x^p+ab=(x+a{)}^p$$
holds. According to Fermat's little theorem,
$$x+ab\equiv x+a(modp)$$
holds.
$$(b-1)a\equiv 0(modp)$$
$$b\equiv 1(modp)$$
Let $c$ and $d$ be integers. We suppose $x=bc+d$ and $0<d<b$ hold.
$$(bc+d{)}^p+ab=(bc+d+a{)}^p$$
$$(d+a{)}^p-d^p\equiv 0(modb)$$
Let $e$ be an integer and
$$(d+a{)}^p-d^p=be...(2)$$
holds.
$$d+a-d\equiv be(modp)$$
$$a\equiv e(modp)$$
Let $f$ be an integer and $e=fp+a$ holds. By the equation (2),
$$a\sum _{i=0}^{p-1}(d+a{)}^i{d}^{p-1-i}=be...(3)$$
holds. Let $g$ be an integer. $e=ag$ holds since $a$ and $b$ are relatively prime.
$$ag=fp+a$$
$$(g-1)a=fp$$
$$g\equiv 1(modp)...(4)$$
On the other hand, by the equation (3),
$$\sum _{i=0}^{p-1}(d+a{)}^i{d}^{p-1-i}/p=bg/p$$
$$d^{p-1}\equiv bg/p(moda)$$
holds. $g\equiv 0(modp)$ must hold since $b$ does not have $p$ as a factor. It becomes a contradiction contrary to the congruent expression (4). 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. We consider the following equation.
$$x^4+y^4=z^4...(5)$$
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 $z-x\equiv 0(mod{y}^{\prime })$ hold,
$$y^4/(z-x)=\sum _{i=0}^3{z}^i{x}^{3-i}$$
$$y^4/(x-z)\equiv 4x^3(mod{y}^{\prime })$$
holds. If the right side has $y^{\prime }$ as a factor, then $y^{\prime }$ must be $2$ since $x$ and $y$ are relatively prime. $z-x$ and $y^4/(z-x)$ do not have common prime factors since $y$ is odd. Let $a$ and $b$ be odd positive integers. We suppose $a$, $b$ and $p$ are each prime to one another and $y^4=ab$ and $z-x=a$ hold. By the equation (1),
$$x^4+ab=(x+a{)}^4$$
holds. Let $c$ and $d$ be integers. We suppose $x=bc+d$ and $0<d<b$ hold.
$$(bc+d{)}^4-d^4=(bc+d+a{)}^4$$
$$(d+a{)}^4-d^4\equiv 0(modb)$$
Let $e$ be an integer and
$$(d+a{)}^4-d^4=be...(6)$$
holds. According to Fermat's little theorem,
$$(d+a{)}^2-d^2\equiv be(mod2)$$
holds.
$$be\equiv a^2(mod2)$$
$$e\equiv 1(mod2)$$
Let $f$ be an integer and $e=2f+1$ holds. By the equation (6),
$$a\sum _{i=0}^3(d+a{)}^i{d}^{3-i}=be...(7)$$
holds. Let $g$ be an integer. Since $a$ and $b$ are relatively prime, $e=ag$ holds.
$$ag=2f+1$$
$$g\equiv 1(mod2)...(8)$$
On the other hand, by the equation (7),
$$\sum _{i=0}^3(d+a{)}^i{d}^{3-i}/4=bg/4$$
$$d^3\equiv bg/4(moda)$$
holds. $g\equiv 0(mod4)$ must hold since $b$ is odd. However, it becomes a contradiction contrary to the congruent expression (8). From the above, there are no integer solutions to the equation (5) 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 $z>0$.