<s>[INST] Finding a root of a degree 5 polynomial 
There are positive integers $m$ and $n$ such that $m^2 −n = 32$ and $$\sqrt[5]{m+\sqrt{n}} + \sqrt[5]{m-\sqrt{n}}$$ is a real root of the
polynomial $$x^5 −10x^3 + 20x −40$$ Find $m+n$.

Okay I write the other $4$ roots as $x_{1}, x_{2}, x_{3}, x_{4}$ and get $\sqrt[5]{m+\sqrt{n}} + \sqrt[5]{m-\sqrt{n}} = - (x_{1} + x_{2} + x_{3} + x_{4})$. After doing stuff with Vieta's, I get
$(x_{1} + x_{2} + x_{3} + x_{4})x_{1} x_{2} x_{3} x_{4} = -40$
$\frac{1}{x_{1}} +\frac{1}{x_{2}}+ \frac{1}{x_{3}}+ \frac{1}{x_{4}}- \frac{1}{x_{1} + x_{2} + x_{3} + x_{4}} = \frac{1}{2}$
$\sum_{1\le{i}<j\le{4}}{x_{i}x_{j}} - (x_{1} + x_{2} + x_{3} + x_{4})^2 =-10$
$(x_{1} + x_{2} + x_{3} + x_{4})(\sum_{1\le{i}<j\le{4}}{x_{i}x_{j}}) + x_{1} x_{2} x_{3} x_{4}(\frac{1}{x_{1}} +\frac{1}{x_{2}}+ \frac{1}{x_{3}}+ \frac{1}{x_{4}}) = 0$.
Okay and letting $\sqrt[5]{m+\sqrt{n}}$ be $a$ and $\sqrt[5]{m-\sqrt{n}}$ be $b$,
$a^5 + b^5 = 2m$
$ab =2$, from $m^2-n=32$
and ultimately I write $a^5 + b^5 = (a+b)((a+b)^4 -6(a+b)^2 +4)$. I guess I can just get $a+b$ from those Vieta's equations up there... but it's a bit tedious maybe and I think I might be missing something that makes this whole question easier. Can someone help please? Thanks.
 [/INST] Your second idea is better $$(a+b)^5 = a^5+b^5 + 5ab(a^3+b^3)+10a^2b^2(a+b)$$
So $$x^5 = 2m + 10(a+b)\Big((a+b)^2-6\Big)+40(a+b)$$ or $$x^5+2m+10x^3-60x+40x$$
so $$x^5-10x^3+20x -2m =0$$ and thus $m=20$ so ...
</s>