When do we have $(a-b)^n=(a+b)^{-n}$?

Question

Here, $a, b$ are positive reals, and $n$ is a positive integer. It seems that the binomial theorem for negative exponents gives some solutions. But what is the exact condition for the parameters.

Answer 1

As pointed out in the comments this is equivalent to $$(a^2-b^2)^n=1$$ So we need either $a^2-b^2=1$, $n=0$ and $a^2-b^2\ne0$ or $a^2-b^2=-1$ and $n=2k$ where $k\in\mathbb{N}_0$.

Answer 2

The $n$ is (mostly) irrelevant

$(a-b)^{-n} = (\frac 1{(a-b)})^n$

So $(a+b)^n = (\frac 1{a-b})^n$. Now $a + b >0$ so $\sqrt[n]{(a+b)^n} = a+b> 0$ and $\sqrt[n]{(\frac 1{a-b})^n} = |\frac 1{a-b}| > 0$.

So $a+b = |\frac 1{a-b}|$. and $(a+b)|(a-b)| = |(a+b)(a-b)| = 1$.

So $a^2 - b^2 = \pm 1$ and that's really it.

We can be more explicit that $a =\sqrt{b^2\pm 1}$ (and vice versa: $b =\sqrt {a^2 \mp 1}$.

....

If $n$ is odd then we know $a > b$ and $a =\sqrt{b^2 - 1}$ but if $n$ is even we don't know that.