<s>[INST] Proving $x = y$ or $x = -y$ when $x^n = y^n$ and $n$ is even I'm currently going through Spivak's calculus, and after a lot of effort, i still can't seem to be able to figure this one out.
The problem states that you need to prove that $x = y$ or $x = -y$ if $x^n = y^n$
I tried to use the formula derived earlier for $x^n - y^n$ but that leaves either $(x-y) = 0$ or $(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$ and i'm not sure how to proceed from there.
 [/INST] Let $n=2p$. For convenience let us denote $y=a$. From the algebraic identities
\begin{eqnarray}
x^{2p}-a^{2p} &=&(x-a)\sum_{k=0}^{2p-1}a^{k}x^{2p-1-k}, \tag{1} \\
\sum_{k=0}^{2p-1}a^{k}x^{2p-1-k} &=&(x+a)\sum_{k=0}^{p-1}a^{2k}x^{2p-2-2k},\tag{2}
\end{eqnarray}
we conclude that
\begin{equation}
x^{2p}-a^{2p}=(x-a)(x+a)\sum_{k=0}^{p-1}a^{2k}x^{2p-2-2k}. \tag{3}
\end{equation}
Since for $a\neq 0$ the polynomial $\sum_{k=0}^{p-1}a^{2k}x^{2p-2-2k}$ on the right-hand side of (3) has no real
roots, it follows that the equation $x^{2p}-y^{2p}=0$ is equivalent to $(x-y)(x+y)=0$, thus proving that if $x^{n}=y^{n}$ and $ n $ is even, then $x=y$ or 
$x=-y$.
The identities $(1)$ and $(2)$ can be justified by applying Ruffini's Rule twice: for identity $(1)$
$$
\begin{array}{c|cccccccc}
  & 1 & 0 & 0 & \ldots  & 0 & 0 &  & -a^{2p} \\ 
a &    & a  & a^2 & \ldots  & a^{2p-2} & a^{2p-1} &  & a^{2p} \\ 
\hline
&   1 & a & a^{2} & \ldots  & a^{2p-2} & a^{2p-1} & | & 0
\end{array}
$$
\begin{equation*}
x^{2p}-a^{2p}=(x-a)(x^{2p-1}+ax^{2p-2}+a^{2}x^{2p-3}+\cdots
+a^{2p-2}x+a^{2p-1}),
\end{equation*}
and for identity $(2)$
$$
\begin{array}{c|cccccccc}
 & 1 & a & a^{2} & a^{3} & \ldots  & a^{2p-2} &  & a^{2p-1} \\ 
-a &   & -a & 0 & -a^{3} & \ldots  & 0 &  & -a^{2p-1} \\ 
\hline
  & 1 & 0 & a^{2} & 0 & \ldots  & a^{2p-2} & | & 0
\end{array}
$$
$x^{2p-1}+ax^{2p-2}+\cdots +a^{2p-2}x+a^{2p-1}$
$$=(x+a)(x^{2p-2}+a^{2}x^{2p-4}+a^{4}x^{2p-6}+\cdots +a^{2p-4}x^{2}+a^{2p-2})$$
</s>