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I had recently met a question that asked to show that $a\sqrt{2}+b\sqrt{3}+c=0$ has no solutions over the rationals (except for $0$) for $a,b$ and $c$. To show this I moved $c$ to the other side then squared both sides. Then the equation re-arranges into $\sqrt{6}=\frac{c^2-2a^2-3b^2}{2ab}$, which is a contradiction as one side is irrational and the other side is rational. I was wondering if I could extend this to $n$ terms, i.e does $a_1\sqrt{1}+a_2\sqrt{2}+...+a_n\sqrt{n}=0$ have any rational solutions for any $a_1,a_2,...,a_n$ and how would I prove that if it did? This time I can't just square both sides because there are too many terms to isolate so I was wondering if anyone could help (dis)prove this statement that would be great.

Pen and Paper
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    $2\times \sqrt 1 -\sqrt 4 = 0$. – lulu Apr 11 '22 at 13:29
  • @lulu What if I excluded all perfect squares from $n$? – Pen and Paper Apr 11 '22 at 13:31
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    See https://math.stackexchange.com/q/30687/42969 for a general statement. – Martin R Apr 11 '22 at 13:31
  • @Quippy, you have to exclude numbers with a nontrivial square factor. – lhf Apr 11 '22 at 13:32
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    $2\sqrt 3 - \sqrt {12}=0$. Study the case of primes, as proposed by @MartinR. – lulu Apr 11 '22 at 13:32
  • @MartinR I haven't learnt linear algebra so I can't understand anything in that link, if $a_1..a_n$ were all primes, would there be a more elementary proof? – Pen and Paper Apr 11 '22 at 13:35
  • I think that the proofs of the more general cases require a bit of theory of extension fields. At the very least, those pieces of theory make the arguments simpler. Like here. Sorry about advertising my own proof. Field theory makes it more transparent why square roots of square-free integers are linearly independent (you don't need them to be primes). – Jyrki Lahtonen Apr 11 '22 at 13:43
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    It may be better to check out this. It implies that the square roots of all the square-free integers are linearly independent over $\Bbb{Q}$. The reason is that those square roots all live in a suitable extension field of the prescribed type. The argument in my answer in another linked question requires Galois theory, which is one level deeper. – Jyrki Lahtonen Apr 11 '22 at 13:48
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    Does this answer your question? Proving that $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ for distinct primes $p_i$.. If you ask a question on rational solutions to equations then usually an answer without abstract or linear algebra, or even without number theory cannot be expected. – Dietrich Burde Apr 11 '22 at 15:07
  • Since you know already much about calculus I wonder why you say you haven't learned linear algebra? – Dietrich Burde Apr 11 '22 at 15:10
  • @JyrkiLahtonen, the notation has completely gone over my head, could you put into simple words what $\left(\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right)=2^n$ means. I've never studied set theory so I don't know where to start to understand your answer. COuld you just explain really briefly what some of the notation means so I can research it in my own time? – Pen and Paper Apr 11 '22 at 21:21

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