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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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amakelov
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    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. – Jyrki Lahtonen Jul 05 '16 at 05:59
  • Thank you! But I'm looking for something more elementary – amakelov Jul 05 '16 at 06:02
  • @JyrkiLahtonen: right, I don't think it's a duplicate, I've edited in my reasons to believe the question has to have an answer. – amakelov Jul 05 '16 at 06:07
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    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. – Jyrki Lahtonen Jul 05 '16 at 06:08
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    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. – user21820 Jul 05 '16 at 06:17
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    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. –  Jul 05 '16 at 06:17
  • I agree with @ChristianRemling in that most of that Galois theory is not needed. Familiarity with such arguments may point at a way of an elementary solution though. I recall having solved some IMO training problems by writing basic properties of algebraic integers in high school language. Not sure that $\sqrt2\mapsto-\sqrt2$ is best though. It will affect many terms. May be $\sqrt p\mapsto -\sqrt p$ where $p$ is the largest prime $\le n$? That may be easier to control? But, to prove that you get a well-defined map needs the other linked question, or does it? – Jyrki Lahtonen Jul 05 '16 at 06:21
  • @JyrkiLahtonen: $\sqrt{2}$ wasn't thought through, though perhaps it doesn't matter. We secretly know that the $\sqrt{\prod p_j}$, with $p_j$ some of the primes $\le n$, form a basis of $\mathbb Q(\alpha)$, but I think we can proceed without using this. Clearly these elements span $\mathbb Q(\alpha)$, and we make a basis out of this by removing elements as needed, and then we define a map by mapping the linear combinations as planned. I might be missing something, but right now I think this will work with no theory whatsoever needed. –  Jul 05 '16 at 06:40
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    I gave a proof using only basic field theory here. –  Oct 19 '16 at 20:52

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