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.
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3Hint: Multiply both sides by $(a+b)^n$ – Sorfosh Aug 04 '19 at 18:36
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1$$1=(a^2-b^2)^n$$ Use https://math.stackexchange.com/questions/384090/find-all-real-numbers-x-for-which-frac8x27x12x18x-frac76/384094#384094 – lab bhattacharjee Aug 04 '19 at 18:38
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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 $k\in\mathbb{N}_0$.
Peter Foreman
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Note that the Question specifies $n$ to be a positive integer, obviating the awkward solution case. – hardmath Aug 04 '19 at 18:50
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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: $b =\sqrt {a^2 \mp 1}$.
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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.
fleablood
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