Simple proof of Fermat's Last Theorem

When $n$ is an odd prime

$ $Let $a$, $x$, $y$, $z$ and $z'$ be positive integers. We suppose that $z'$ is a prime factor of $x+y$, $x$, $y$ and $z$ are each prime to one another. We consider the following equation.
$$x^n+y^n=z^n\ ...(1)$$
$$z^n=(x+y)a$$
Since $a=\sum_{i=0}^{n^1}y^i(-x)^{n-1-i}$ and $x+y≡0\pmod{z'}$ hold,
$$a≡nx^{n-1}\pmod{z'}$$
holds. If $x+y≡0\pmod{n}$ holds, then the right side of this expression is $0$ for $z'=n$. Therefore, $a$ and $x+y$ have a common prime factor $n$ when $x+y≡0\pmod{n}$ holds. Conversely, if $x+y≢0\pmod{n}$ holds, then the right side of this expression is not $0$ since $x$ and $z$ are relatively prime.
$ $We take account the case where $x+y≡0\pmod{n}$ holds. Let $b$ and $c$ be positive integers. We suppose that $a≡0\pmod{n}$ and $b≢0\pmod{n}$ hold, $a$ and $b$ are coprime, $c$ is a power of $n$ or $1$, and $x+y=bc$ and $z^n=abc$ hold. We suppose that $c>1$ holds. By the equation (1),
$$x^n+(bc-x)^n=abc\ ...(2)$$
$$(x^n+(bc-x)^n)/c=ab$$
$$(a-1)b≡0\pmod{n}$$
$$a≡1\pmod{n}$$
holds. It becomes a contradiction contrary to the congruent expression $a≡0\pmod{n}$.
$ $Next, we give consideration to the case where $x+y≢0\pmod{n}$ holds. By the equation (2), for $c=1$,
$$x^n-(x-b)^n=ab\ ...(3)$$
$$\sum_{i=0}^{n-1}x^i(x-b)^{n-1-i}/a=1$$
$$nx^{n-1}/a≡1\pmod{b}$$
holds. Let $d$ be a positive integer. We suppose that $d$ is the value of the left side of this congruent expression.
$$x^{n-1}=ad/n$$
Since $x$ and $z$ are relatively prime, $a=n$ must hold. By the equation (3),
$$x^n-(x-b)^n=bn$$
holds. By Fermat's little theorem,
$$b≡0\pmod{n}$$
holds. This is a contradiction contrary to the congruent expression $b≢0\pmod{n}$. From the above, there are no positive integer solutions to the equation (1). (Q.E.D.)

When $n=4$ holds

$ $Let $x$, $y$, $y'$ and $z$ be positive integers. We suppose that $y'$ is a prime factor of $x+z$ and $x$, $y$ and $z$ are each prime to one another. We consider the following equation.
$$x^4+y^4=z^4\ ...(4)$$
We suppose that $x$ is even and $y$ and $z$ are odd since there are no integer solutions to this equation when $x$ and $y$ are odd. Since $y^4/(x+z)=\sum_{i=0}^3z^i(-x)^{3-i}$ and $x+z≡0\pmod{y'}$ hold,
$$y^4/(x+z)≡4z^3\pmod{y'}$$
holds. Since $y$ and $z$ are coprime and $y$ is odd, $x+z$ and $y^4/(x+z)$ have no common prime factors. Let $a$ and $b$ be odd positive integers. We suppose that $a$ and $b$ are each prime to one another and $x+z=a$ and $y^4=ab$ hold. By the equation (4),
$$(a-x)^4-x^4=ab$$
$$\sum_{i=0}^3(x-a)^ix^{3-i}/b=1$$
$$4x^3/b≡1\pmod{a}$$
holds. Since $b=\sum_{i=0}^3z^ix^{3-i}$ holds, $b>4$ holds. Hence, $x^3≡0\pmod{b}$ must hold. However, it becomes a contradiction contrary to the condition that $x$ and $y$ are relatively prime. From the above, there are no positive integer solutions to the equation (4). (Q.E.D.)

Conclusion

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