Independence of radicals: First-principles proof of special case Reposting from MathStackexchange, original post is here, but got no answer.
I've known this problem for a long time:
Problem. Show that the number $\alpha=\sqrt{1} + \sqrt{2} + \ldots + \sqrt{n}$ is irrational for $n\geq 2$.
but I haven't been able to find a solution from first principles (in the sense of a high-school math olympiad kind of proof, not using advanced theory; so for example you can observe using the theory of algebraic integers that if $\alpha$ is rational, it must be integer, but I would consider that too heavy of an apparatus). I was wondering if anybody knows one/can come up with one? 
Solutions that use some heavier theory, but not too much, are also welcome.
Update: I'm aware of solutions proceeding by Galois theory, etc. but my reason to believe this has an elementary solution is that it was in a rather interesting and high-quality list of high-school olympiad preparation problems that I found on a math forum. 
 A: Upon the OP's request, I share the proof (using only basic field theory) of the following more general result.
Theorem. Let $K$ be a field of characteristic different from $2$. Let $x_1,\dots,x_n$ be arbitrary elements in an arbitrary field extension of $K$ such that the square of each $x_i$ lies in $K$, the sum $x_1+\dots+x_n$ lies in $K$, and no subsum of $x_1+\dots+x_n$ vanishes (including the full sum). Then each $x_i$ lies in $K$.
Proof. We proceed by induction on $n$. For $n=1$ the statement is trivial. Now we assume that $n>1$ and the statement holds with $n-1$ in place of $n$. Let $x_1,\dots,x_n$ satisfy the conditions of the theorem. If any $x_i$ lies in $K$, then by the induction hypothesis all the other $x_j$'s lie in $K$, and we are done. So we can assume that no $x_i$ lies in $K$. The sum $x_2+\dots+x_n$ lies in the field extension $K(x_1)$, hence by the induction hypothesis each $x_i$ lies in this field extension. That is, $x_i=a_i+b_ix_1$ for some $a_i,b_i\in K$. By our assumption, each $b_i$ is nonzero. Furthermore, $x_i^2=(a_i^2+b_i^2x_1^2)+2a_ib_ix_1$ lies in $K$, whence $2a_ib_ix_1$ also lies in $K$. This forces that $a_i=0$, because otherwise $2a_ib_i\neq 0$ and $x_1=(2a_ib_ix_1)/(2a_ib_i)\in K$, a contradiction. We conclude that $x_i=b_ix_1$. As a result, $x_1+\dots+x_n=(b_1+\dots+b_n)x_1$ is a nonzero element of $K$, hence $b_1+\dots+b_n$ is also a nonzero element of $K$. But then $x_1\in K$, a contradiction again.
Corollary. Let $a_1,\dots,a_n$ be positive rational numbers such that the sum $\sqrt{a_1}+\dots+\sqrt{a_n}$ is rational. Then each term $\sqrt{a_i}$ is rational.
Proof. Apply the theorem with $K:=\mathbb{Q}$ and $x_i:=\sqrt{a_i}$. Note that no subsum of $x_1+\dots+x_n$ vanishes (including the full sum), because the $x_i$'s are positive.
