This is an excercise from Spivak's Calculus. Show that if $x^n=y^n$ and $n$ is odd, then $x=y$. Hint: first explain why it suffices to consider only the case x and y greater than 0, then show that x smaller than y or greater y are both impossible.
I try to prove this by using induction. The base case $n=1$ is correct. Assume that $n=2k+1$ is true, prove that $n=2k+3$ is also true. But now I don't know how to proceed.
Notice that the function $f(x)=x^n$ is bijective for odd values of $n$ therefore $x^n=y^n$ leads to $x=y$
It seems unlikely that induction is a good approach here.
If $x$ and $y$ have opposite signes, then obviously you can't have $x^n=y^n$. Now, let us assume that $x,y\geqslant0$. The case in which $x,y\leqslant0$ is similar. Note that$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-1}).$$So, if $x^n=y^n$, then $x=y$ or $x^{n-1}+x^{n-2}y+\cdots+xy^{n-1}=0$. But the last equality cannot hold, unless $x=y=0$ (in which case $x=y$), since $x,y\geqslant0$.
And if $x,y\leqslant0$,\begin{align}x^n=y^n&\implies(-x)^n=(-y)^n\\&\implies-x=-y\\&\implies x=y.\end{align}
If $n$ is odd, then you can factorize the expression $x^n-y^n$ in $$(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+... +y^{n-1})$$ which gives that $$x-y=0$$ since $x$ and $y$, both are positive.
For fun:
$n \in \mathbb{Z^+}$, odd .
1) $x=y=0$ is a solution.
2) Let $(x,y) \not =(0,0)$.
$x^n-y^n=0$;
$(\frac{x}{y})^n =1$, implies
$\frac{x}{y}=1$,
since $n \in \mathbb{Z^+}$, $n$ odd.
