Simple 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. If the left side is a multiple of $y'$, then $y'=p$ must hold since $x$ and $y$ are relatively prime. $z-x$ and $y^p/(z-x)$ do not have common prime factors since $y≢0\ (mod\ p)$ 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,
$$x+ab≡x+a\ (mod\ p)$$
holds.
$$(b-1)a≡0\ (mod\ p)$$
$$b≡1\ (mod\ p)$$
Let $c$ and $d$ be integers. We suppose $x=bc+d$ and $0< d< b$ hold.
$$(bc+d)^p+ab=(bc+d+a)^p$$
$$(d+a)^p-d^p≡0\ (mod\ b)$$
Let $e$ be an integer and
$$(d+a)^p-d^p=be\ ...(2)$$
holds.
$$d+a-d≡be\ (mod\ p)$$
$$a≡e\ (mod\ p)$$
Let $f$ be an integer and $e=fp+a$ holds. By the equation (2),
$$a\sum_{i=0}^{p-1}(d+a)^id^{p-1-i}=be\ ...(3)$$
holds. Let $g$ be an integer. $e=ag$ holds since $a$ and $b$ are relatively prime.
$$ag=fp+a$$
$$(g-1)a=fp$$
$$g≡1\ (mod\ p)\ ...(4)$$
On the other hand, by the equation (3),
$$\sum_{i=0}^{p-1}(d+a)^id^{p-1-i}/p=bg/p$$
$$d^{p-1}≡bg/p\ (mod\ a)$$
holds. $g≡0\ (mod\ p)$ 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 $x$, $y$, $y'$ 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.
$$x^4+y^4=z^4\ ...(5)$$
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 $z-x≡0\ (mod\ y')$ hold,
$$y^4/(z-x)=\sum_{i=0}^3z^ix^{3-i}$$
$$y^4/(x-z)≡4x^3\ (mod\ y')$$
holds. If the right side has $y'$ as a factor, then $y'$ 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.
$$(bc+d)^4-d^4=(bc+d+a)^4$$
$$(d+a)^4-d^4≡0\ (mod\ b)$$
Let $e$ be an integer and
$$(d+a)^4-d^4=be\ ...(6)$$
holds. According to Fermat's little theorem,
$$(d+a)^2-d^2≡be\ (mod\ 2)$$
holds.
$$be≡a^2\ (mod\ 2)$$
$$e≡1\ (mod\ 2)$$
Let $f$ be an integer and $e=2f+1$ holds. By the equation (6),
$$a\sum_{i=0}^3(d+a)^id^{3-i}=be\ ...(7)$$
holds. Let $g$ be an integer. Since $a$ and $b$ are relatively prime, $e=ag$ holds.
$$ag=2f+1$$
$$g≡1\ (mod\ 2)\ ...(8)$$
On the other hand, by the equation (7),
$$\sum_{i=0}^3(d+a)^id^{3-i}/4=bg/4$$
$$d^3≡bg/4\ (mod\ a)$$
holds. $g≡0\ (mod\ 4)$ 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 $n≧3$ holds, $x^n+y^n=z^n$ does not have an integer solution for $x>0$, $y>0$ and $z>0$.