If $x^n=y^n$ and $n$ is odd then $x=y$

Here, we suppose that $x,y\in\mathbb{R}$ and that $x^n=y^n$, where $n$ is odd. I want to prove that $x=y$.

Maybe we can use that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$

So, it suffices to show that if $x^n=y^n$ then $x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}\neq 0$

Any hint?

Solution 1:

Since $n$ is odd, $x^n=y^n$ implies that $x$ and $y$ have the same sign. We may therefore assume that they are both strictly positive. But then $x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}$ is strictly positive and so cannot be zero.

Solution 2:

Since this is tagged "calculus" you may want to show that $f(x) = x^n$ is increasing using the derivative of $f$.

EDIT: the tag has since been changed.

Solution 3:

Hint: Verify that for $x=0$, $y=0$ is the only solution and trivially $x=y$ for this case. Now let $z = \frac yx, x \neq 0$.

Then $y^n - x^n = x^n(z^n - 1)$

Zeroes of the LHS are zeroes of the RHS.

Meaning that $z^n - 1 = 0$.

Consider the $n$-th roots of unity for odd $n$.

Can you continue from there?