Formula for a graphing a half circle: understanding the relationship between odd $n$ exponents (of $\sqrt{2^{n}}$) and the powers of 2

Question

I'm sorry, I'm not exactly sure how to phrase this question, but roughly the gist is:

$f_{m,n}(x)=\sqrt{(\sqrt{2^{2m+1}}+2^{n+1})(\sqrt{2^{2m+1}}-2^{n+1})}$

and it's relationship between odd powers of 2, and the powers of 2, e.g.

at $(\sqrt{2^3} + 2^1)(\sqrt{2^3} - 2^1)$ you get 2
at $(\sqrt{2^5} + 2^2)(\sqrt{2^5} - 2^2)$ you get 4
at $(\sqrt{2^7} + 2^3)(\sqrt{2^7} - 2^3)$ you get 8

etc

I'm just trying to understand the intuition behind this relationship (sorry for any poor communication on my part).

Answer 1

This is difference of two squares $(a+b)(a-b)=a^2-b^2$: $$(2^{\frac{2k+1}{2}}+2^k)(2^{\frac{2k+1}{2}}-2^k)=2^{2k+1}-2^{2k}=2^{2k}$$