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.
 A: 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 $(2)$
$$
\begin{array}{c|cccccccc}
 & 1 & a & a^{2} & a^{3} & \ldots  & a^{2p-2} &  & a^{2p-1} \\ 
-a &   & -a & 0 & -a^{3} & \ldots  & 0 &  & -a^{2p-1} \\ 
\hline
  & 1 & 0 & a^{2} & 0 & \ldots  & a^{2p-2} & | & 0
\end{array}
$$
$x^{2p-1}+ax^{2p-2}+\cdots +a^{2p-2}x+a^{2p-1}$
$$=(x+a)(x^{2p-2}+a^{2}x^{2p-4}+a^{4}x^{2p-6}+\cdots +a^{2p-4}x^{2}+a^{2p-2})$$
A: The factorization of $x^k-y^k$ is known to be $(x-y)(x^{k-1}+x^{k-2}y+x^{k-3}y^2+\cdots y^{k-1})$, as you can verify by direct multiplication (all but two terms cancel in pairs).
Then $x^{2k}-y^{2k}=(x^2-y^2)(x^{2k-2}+x^{2k-4}y^2+x^{2k-6}y^4+\cdots y^{2k-2})$.
As all terms have an even exponent in the second factor, the latter cannot equal zero, and 
$$x^{2k}=y^{2k}\iff x^2=y^2.$$
A: We have that $x\mapsto x^n\colon \mathbb R_{\geq 0} \to \mathbb R_{\geq 0}$ is strictly increasing function and thus injective. Now,
$$x^n = y^n \implies |x|^n = |y|^n \implies |x| = |y| \implies x=\pm y\stackrel{\text{$n$ is even}}\implies x^n = y^n$$ therefore, $x^n = y^n\iff x =\pm y$.
A: I'm assuming you are assuming we are dealing with real numbers.  (The result is not true in complex numbers as $i^4 = 1^4 = (-1)^4  = (-i)^4$).
Now has it been shown to your satisfaction that every $b > 0$ has a unique positive $n-th$ root? Well... okay, let's do some basics
Okay, suppose $0 < |x| < |y|$ then $|x|^n < |y|^n$.  (And similarly if $0 < |y| < |x|)$.  
[Because: if $|x| < |y|$ then $|x|^2 = |x||x| < |x||y| < |y||y| = |y|^2$ and inductively $|x|^{n-1} < |y|^{n-1} \implies |x|^{n-1}|x| < |y|^{n-1}|y|$]
Thus $|x|^n = |y|^n$ if and only if $|x| = |y|$.
Then as $x = \pm |x|$ and $y = \pm |y|$ so $x^n = (\pm 1)^n|x|^n$ (and same thing for $y^n$).  
As $n$ is even $(\pm 1)^n = 1$.
So $x^n = y^n \iff |x| = |y| \iff x = \pm y$.
===
I haven't actually shown that if $b > 0$ that there actually exists an $x$ such that $x^n = b$.  But rereading the problem I see I don't have to.  The thing though is to realize if there is such a positive $x$ it is unique.  Which I can do by showing $x \ge \lt y; x>0;y>0 \implies x^n \ge \lt y^n$.
