I would appreciate if someone could check my attempt in proving the Fermat's Last Theorem for prime exponents greater than $2$.
Firstly, let's prove a couple of lemmas which state that sum or difference of two non-equal odd $p$-th powers cannot be a $p$-th power for prime $p>2$.
Lemma 1:
The difference of two $p$-th powers of non-equal odd positive integers is not a $p$-th power of an integer for prime $p>2$.
Take prime $p>2$, and odd $z,x \in\Bbb Z^+, \; z>x$. Let $a=z-x$. Obviously $a$ is even. $$z=a+x$$ $$z^{\ p}=\sum_{i=0}^p \binom{p}{i}a^{\ p-i}x^{\ i}=x^{\ p}+a\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i}$$ $$z^{\ p}-x^{\ p}=a\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i}$$ Assume that the right side is a $p$-th power of an integer.
**** And here comes my erroneous logic: ****
:(
Then, since $x$ is not divisible by $a$ (because $x$ is odd, and $a$ is even), it must be: $$\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i} \equiv 0\;(mod\;a^{\ p-1})$$
$$\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i} \equiv 0\;(mod\;a)$$ $$px^{\ p-1}+\sum_{i=0}^{p-2} \binom{p}{i}a^{\ p-1-i}x^{\ i} \equiv 0\;(mod\;a)$$ $$px^{\ p-1}+a\sum_{i=0}^{p-2} \binom{p}{i}a^{\ p-2-i}x^{\ i} \equiv 0\;(mod\;a)$$ $$px^{\ p-1} \equiv 0\;(mod\;a)$$ However, since both $p>2$ and $x$ are odd, then $px^{\ p-1}$ is also odd, so it cannot be divisible by $a$ which is even. We reached a contradiction so Lemma 1 is proven.
Lemma 2:
The sum of two $p$-th powers of non-equal odd positive integers is not a $p$-th power of an integer for prime $p>2$.
Proof for the sum of two odd $p$-th powers not being a $p$-th power is almost the same. Start with odd $x,y \in\Bbb Z^+$, set $b=x+y$, and raise $x=b-y$ to the p-th power. If the sum $x^{\ p}+y^{\ p}$ were the p-th power, then $py^{\ p-1}$ would have to be divisible by $b$, which cannot be because $py^{\ p-1}$ is odd, while $b$ is even.
Fermat's Last Theorem: (from Wikipedia)
No three positive integers $x, y$, and $z$ can satisfy the equation $x^n+y^n=z^n$ for any integer value of $n$ greater than two.
Proof of the Fermat's Last Theorem for prime exponents greater than 2:
To prove the Fermat's Last Theorem for prime exponents greater than $2$, we assume that there are such $x,y,z$ and then rewrite all $3$ possible types of sums of odd and even $p$-th powers in such a way to get to a contradiction with the one of the above Lemmas.
Assume that for prime $p>2$ there exist $x,y,z \in\Bbb Z^+$, all three different, for which $$x^{\ p}+y^{\ p}=z^{\ p}$$
A) Assume that $x$ is even and $y$ is odd, both positive integers.
First note that then $z$ must be odd too.
We can rewrite it as $$z^{\ p}-y^{\ p}=x^{\ p}$$ however, since both $z$ and $y$ are odd then this is in contradiction with Lemma 1.
Therefore, combinations where one of $x,y$ is even and the other is odd are not possible.
B) Assume that both $x$ and $y$ are odd positive integers, then $x^{\ p}+y^{\ p}=z^{\ p}$ is in direct contradiction with Lemma 2.
Therefore, combinations where both $x$ and $y$ are odd are not possible.
C) Assume both $x$ and $y$ are even, then we can repeatedly divide them with $2$, until one, or both of them gets odd, which reduce this case to A) or B).
Therefore, combinations where both $x,y$ are even are also not possible.
Since neither of A), B), or C) hold, then there are no such $x,y,z \in Z^+$ to satisfy the equation $x^{\ p}+y^{\ p}=z^{\ p}$ for prime $p>2$ and $x,y,z$ all different.
