<s>[INST] Are there solutions to FLT which are linearly independent over $\mathbb{Z}$ Specifically, I would like to know if there is some $R$, where


*

*$R$ is a ring with unity

*$\mathbb{Z} \subseteq R$

*there are $x,y,z \in R$ and a prime $p \in \mathbb{Z}$ such that $x^p + y^p + z^p = 0$

*$ax+by+cz=0 \implies a=b=c=0$ for $a,b,c \in \mathbb{Z}\quad$ (linear independence over $\mathbb{Z}$)


If so, I'd be happy to see an example.
This non-example in $M_2(\mathbb{Z})$ lead me to me ask this question:
Let
$$
x=
\begin{pmatrix}
1 & 25 \\
0 & 1 \\
\end{pmatrix}
\!,\;y=
\begin{pmatrix}
-1 & 0 \\
5 & -1 \\
\end{pmatrix}
\,\text{and}\;z=
\begin{pmatrix}
0 & -5 \\
-1 & 0 \\
\end{pmatrix}
\!\begin{matrix}\\.\end{matrix}
$$
Then $x^5+y^5+z^5=0$, but also $x+y+5z=0$.
I realize that this phenomenon (of solutions to an equation being linearly dependent over $\mathbb{Z}$) is probably not FLT specific, so if someone wants to share something on the general phenomenon I'd be happy, too.
 [/INST] We can take
$$
x=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix},\;
y=\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix},\;
z=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{pmatrix},\;
$$
then we have $x^2=y^2=z^2=0$, so that $x^2+y^2=z^2$ for $p=2$, and $x,y,z$ are linearly independent. Also $x^p=y^p+z^p$ for all $p\ge 3$.
</s>