Simple Proof of Fermat's Last Theorem.

[Mariga]

\begin{align*}
a^2+b^2&=c^2\\
c^2-a^2&=b^2\\
c^2-a^2&=(c-a)(c+a)\\
b^2&=(c-a)(c+a)
\end{align*}
This is possible iff there exists integers $k$, $x_1$, and $x_2$ such that $c-a=kx_1^2$ and $c-a=kx_2^2$ to give}
$$b^2=k^2x_1^2x_2^2$$
For $n>2$,
\begin{align*}
a^n+b^n&=c^n\\
c^n-a^n&=b^n\\
c^n-a^n&=(c-a)(c^{n-1}+c^{n-2}a+c^{n-3}a^2+\cdots +ca^{n-2}+a^{n-1})\\
b^n&=(c-a)(c^{n-1}+c^{n-2}a+c^{n-3}a^2+\cdots +ca^{n-2}+a^{n-1})
\end{align*}
This would hold iff
$$b^n=(kx_1^n)(kx_2^n)(kx_3^n)\cdots (kx_n^n)$$
So that
$$b^n=k^nx_1^nx_2^nx_3^n\cdots x_n^n$$
However,$(c^{n-1}+c^{n-2}a+c^{n-3}a^2+\cdots +ca^{n-2}+a^{n-1})$ doesn't have common like terms and hence cannot be factored into a product of $n-1$ terms. Hence,
$$b^n\neq(kx_1^n)(kx_2^n)(kx_3^n)\cdots (kx_n^n)$$
thus
$$a^n+b^n\neq c^n$$

[Guest]

Your proof is not complete in the least... Keep working on it until it makes sound sense.

[Guest]

What does it make incomplete?