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.

Update: This has been cross-posted at mathoverflow here, where it already has one nice answer!

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    $\begingroup$ Here is a solution using Galois theory. Here is a related version that uses concepts of field extensions. Neither at olympiad level technology, so I am refraining from voting to close as a duplicate. $\endgroup$ – Jyrki Lahtonen Jul 5 '16 at 5:59
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    $\begingroup$ Because it is easy to show that $\alpha$ is a zero of a monic polynomial with integer coefficients, the rational root test implies that if $\alpha$ is rational, it is an integer. $\endgroup$ – Jyrki Lahtonen Jul 5 '16 at 6:08
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    $\begingroup$ I once saw a completely elementary proof of the fact that square-roots of primes are linearly independent over $\mathbb{Q}$, but I can't recall where, and I think that it isn't enough to get the desired result. $\endgroup$ – user21820 Jul 5 '16 at 6:17
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    $\begingroup$ Most of the Galois theory that is discussed in the linked answer is totally unnecessary to answer the actual question. All we need is an automorphism of $\mathbb Q(\alpha)$ that sends $\sqrt{2}\mapsto -\sqrt{2}$, say. Quite possibly, this can be done entirely by hand. $\endgroup$ – user138530 Jul 5 '16 at 6:17
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    $\begingroup$ I gave a proof using only basic field theory here. $\endgroup$ – user380533 Oct 19 '16 at 20:52

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