Prove product of conjugate of roots is integer Consider $n$ given integer and the term
$\sqrt{x_1}+\sqrt{x_2}+...+\sqrt{x_n}$
There are $n-1$ plus signs in the middle.
Now consider the remaining $2^{n-1}-1$ possibilities by flipping some or all of the signs.
$$\sqrt{x_1} \pm \sqrt{x_2} \pm ... \pm \sqrt{x_n}$$
i want to prove the product of all the terms is integer
Can someone help me prove (or disprove!) this? I have gotten success on n=2,3 but only by brute force (some)
Finally I’m wondering if this can be generalised to all roots (not just square roots). The root “power” can be assumed to be the same since you can always take lcm of power. So are similar terms to multiply by for cbrt(a)+cbrt(b)+cbrt(c)?
Thank you for reading and helping.
 A: This can be deduced from the following well-known fact. If $$A(x)=\prod_{k=1}^{m}(x-\alpha_k)\in\mathbb{Q}[x],\qquad B(x)=\prod_{k=1}^{n}(x-\beta_k)\in\mathbb{Q}[x]$$ (with $\alpha_1,\ldots,\alpha_m,\beta_1,\ldots,\beta_n\in\mathbb{C}$, say), then $\prod\limits_{j=1}^{m}\prod\limits_{k=1}^{n}(x-\alpha_j-\beta_k)\in\mathbb{Q}[x]$. This holds because this polynomial is equal (up to sign) to the resultant of $A(x-y)$ and $B(y)$ considered as polynomials in $y$ (with coefficients in $\mathbb{Q}[x]$).
By induction (on $m$), if $P_k(x)=\prod\limits_{j=1}^{d_k}(x-\alpha_{k,j})\in\mathbb{Q}[x]$ for $1\leqslant k\leqslant m$, then $$P(x):=\prod_{j_1=1}^{d_1}\ldots\prod_{j_m=1}^{d_m}(x-\alpha_{1,j_1}-\ldots-\alpha_{m,j_m})\in\mathbb{Q}[x].$$
Let's apply this to $m=n-1$ and $P_k(x)=x^2-(x_{k+1}/x_1)$ (assuming, of course, that $x_1\neq 0$). The product in question is equal to $x_1^{2^{n-2}}P(1)$, hence it is rational (as we've shown $P(1)$ is). On the other hand, it is a product of algebraic integers, hence an algebraic integer, and finally is an (ordinary rational) integer.
This can be generalized to higher-degree roots, with $\pm$ replaced by (complex) roots of unity.

Here is an elementary approach for square roots. We prove that, for $n>1$, $$P_n(x):=\prod_{s_2,\ldots,s_n\in\{\pm 1\}}(x+s_2\sqrt{x_2}+\ldots+s_n\sqrt{x_n})$$ is a polynomial (in $x$) with integer coefficients, and $P_n(-x)=P_n(x)$ (hence $P_n(x)=Q_n(x^2)$ for some $Q_n(x)\in\mathbb{Z}[x]$, and the given product is $P_n(\sqrt{x_1})=Q_n(x_1)\color{blue}{\in\mathbb{Z}}$), using induction on $n$. The claim holds for $n=2$, as $(x+\sqrt{x_2})(x-\sqrt{x_2})=x^2-x_2$. Note that, for $n>2$, $$P_n(x)=P_{n-1}(x+\sqrt{x_n})P_{n-1}(x-\sqrt{x_n}).$$ So, if $P_{n-1}(-x)=P_{n-1}(x)$, then $P_n(-x)=P_n(x)$. Now suppose $P_{n-1}(x)\in\mathbb{Z}[x]$. We have $$P_{n-1}(x+y)=A(x,y^2)+yB(x,y^2)$$ for some $A(x,y),B(x,y)\in\mathbb{Z}[x,y]$ (in fact any polynomial in $\mathbb{Z}[x,y]$ can be written in this form). Then $P_{n-1}(x-y)=A(x,y^2)-yB(x,y^2)$ and $P_n(x)=A^2(x,x_n)-x_n B^2(x,x_n)\in\mathbb{Z}[x]$.
A: Here's an elementary version of the argument from field automorphisms that I hinted at in comments. First, notation: denote $$P(y_1, y_2, \ldots, y_n) = \prod_{(i_2, \ldots, i_n) \in \{-1, 1\}^{n-1}} (y_1 + i_2 y_2 + \cdots + i_n y_n).$$
The product that you want to prove is an integer is thus $P(\sqrt{x_1}, \ldots, \sqrt{x_n})$. Let's also introduce the symbol $\sqrt[N]{\cdot}$ to mean the negative square root; for example, $\sqrt[N]{4} = -2$. Note that $\sqrt{x} \sqrt{x} = \sqrt[N]{x} \sqrt[N]{x} = x$; that is, $\sqrt{\cdot}$ and $\sqrt[N]{\cdot}$ behave identically in certain algebraic simplifications. (You'll see in a bit why a separate symbol for this is useful.)
If we expand the product $P := P(\sqrt{x_1}, \ldots, \sqrt{x_n})$ and simplify, we get a sum of terms of the form $C \sqrt{x_a} \sqrt{x_b} \cdots \sqrt{x_k}$, where the indices $1 \leq a < b < \cdots < k \leq n$ are distinct and the coefficient $C$ is an integer. It's also possible, of course, to have terms with no radicals at all, just an integer coefficient.
What happens if we compute $P_k := P(\sqrt{x_1}, \ldots, \sqrt{x_{k-1}}, \sqrt[N]{x_k}, \sqrt{x_{k+1}}, \ldots, \sqrt{x_n})$ instead, with $\sqrt[N]{x_k}$ replacing $\sqrt{x_k}$? Two things.
First, the value of the product is unchanged: $P = P_k$ for all $k \in \{1, \ldots, n\}$. This is obvious for $k \geq 2$, because the product for $P$ goes over the positive and negative square roots. For $k = 1$, every term in the product for $P$ is the negative of a distinct term in $P_1$ (for example, $\sqrt{x_1} + \sqrt{x_2} - \sqrt{x_3}$ is the negative of $\sqrt[N]{x_1} - \sqrt{x_2} + \sqrt{x_3}$), and each product has an even number of terms, so $P = P_1$.
Second, because of what we said earlier about algebraic simplifications, every term in the expanded $P$ that involves $\sqrt{x_k}$ is matched by a term in the expanded $P_k$ that is identical, including the sign of the integer coefficient, except that $\sqrt[N]{x_k}$ replaces $\sqrt{x_k}$. That is, the total value of the terms involving a radical of $x_k$ in the simplified $P_k$ is the negative of their value in the simplified $P$, and the other terms are unchanged.
But the overall values of $P$ and $P_k$ are equal! The only way this is possible is if the terms involving $\sqrt{x_k}$ all cancel. Repeating this logic for all indices $k$ gives the result that no radicals at all can be left in $P$ once it is fully simplified, so $P$ must be an integer.

As metamorphy points out, this result generalizes to higher-order radicals, with the coefficients of the radicals ranging not just over $+1$ and $-1$ but over all roots of unity of the appropriate order. That is, if $x_1, \ldots, x_n$ are integers and $\omega_k := e^{2 \pi i/n}$ is a primitive $k$th root of unity, then $$\prod_{(i_2, \ldots, i_n) \in \{0, \ldots, k-1\}^{n-1}} \left(\sqrt[k]{x_1} + \omega_k^{i_2} \sqrt[k]{x_2} + \cdots + \omega_k^{i_n} \sqrt[k]{x_n}\right)$$ is an integer. The elementary argument is closely analogous to the $k = 2$ case, though slightly more complicated. I'll provide it for completeness.
As before, write $P$ for this product, and write $P_{i, j}$ for the product that results when $\sqrt[k]{x_i}$ is replaced by $\omega_k^j \sqrt[k]{x_i}$ for some $0 \leq j < k$ (trivially, $P = P_{i, 0}$). You can easily show that $P$ is real (complex conjugation permutes the terms in the product and leaves the product's value unchanged) and that $(\forall i, j) P = P_{i, j}$.
Furthermore, after expansion and simplification, every term in $P$ can be written as a sum of terms of the form $$C \sqrt[k]{x_\alpha^{e_\alpha}} \sqrt[k]{x_\beta^{e_\beta}} \cdots \sqrt[k]{x_\kappa^{e_\kappa}}$$ where the coefficient $C$ is an element of the ring $\mathbb{Z}[\omega_k]$; that is, $C = c_0 + c_1 \omega_k^1 + \cdots + c_{k-1} \omega_k^{k-1}$ where all $c_i \in \mathbb{Z}$. The coefficient of the corresponding term in $C \omega_k^{e_i j}$ where $e_i$ is the power of $\sqrt[k]{x_i}$ that appears in the term. As $1 + \omega_k + \omega_k^2 + \cdots + \omega_k^{k-1} = 0$, all terms involving $\sqrt[k]{x_i}$ in the sum $P_{i, 0} + P_{i, 1} + \cdots + P_{i, k-1} = kP$ cancel.
From this, we conclude that the expansion of $P$ has no radicals, so $P \in \mathbb{Z}[\omega_k]$. It remains to show that $P \in \mathbb{Z}$. This can be proved by Galois theory: $P$ is is invariant under the automorphisms $\omega_k \mapsto \omega_k^j$ of $\mathbb{Q}[\omega_k]/\mathbb{Q}$, so it must be in $\mathbb{Q}$; it is also an algebraic integer, and the only rational algebraic integers are in $\mathbb{Z}$. It's also possible to get this result without Galois theory by showing that $P$ is a polynomial with integer coefficients in $x_1, \ldots, x_n$ using the result that the beginning of metamorphy's answer mentions
