Let $f(x)$ be a polynomial of degree $n$ with rational coefficients whose zeroes are nonnegative real numbers: $x_1, \dots, x_n\geq 0$.
General question. Does there exist a simple expression for the number
$$r=\sqrt{x_1} + \sqrt{x_2} + \dots + \sqrt{x_n}$$
in terms of coefficients of $f(x)$ ?
Remark. Let $s_k = x_1^k + x_2^k + \dots + x_n^k$ be the sum of $k$-th powers of zeros of $f(x)$. It is clear that for nonnegative integer $k$, $s_k$ can be expressed as a polynomial of the coefficients of $f(x)$ (cf. Newton-Girard formulae).
The number $r$ (and even its square $r^2$) represent a zero of a polynomial whose coefficients are polynomials in $s_k$:
$n=1:\qquad r^2 – s_1 = 0$
$n=2:\qquad r^4 – 2 s_1 r^2 + (2 s_2 – s_1^2) = 0$
$n=3:\qquad 3 r^8 – 12 s_1 r^6 + (6 s_1^2 + 12 s_2) r^4 + (- 20 s_1^3 + 72 s_1 s_2 – 64 s_3) r^2 + (3 s_1^4 – 12 s_1^2 s_2 + 12 s_2^2) = 0$
$n=4:\qquad 9 r^{16} – 72 s_1 r^{14} + (180 s_1^2 + 72 s_2) r^{12} + \dots = 0$
Unfortunately the degree of these polynomials grows as $2^n$, and in general it seems it cannot be made much smaller.
Specific question. Now, let us consider a special case of $f(x)$ satisfying the identity $f(y^2)=g(y)\cdot g(-y)$ where $g(y)$ is a polynomial with rational coefficients. Then each $\sqrt{x_i}$ is a zero of either $g(y)$ or $g(-y)$. Can we find a simple expression for $r$ in this case?
Best Answer
I don't think that anything better than what you figured out already can be done. From the algebraic point of view, you cannot distinguish the $2^n$ siblings of $r$, corresponding to the different choices of the signs of the square roots. So you get a degree $2^n$ polynomial where $r$ is one of the roots. I believe it is not possible to take positivity into account.