Except for the obvious cases $n=0,1$, are there any values of $n$ such that $\sum_{k=1}^n\sqrt k$ is an integer? How does one even approach such a problem? (This is not homework  just a problem I thought up.)

1@RahulNarain If you look, you'll notice that my question is related to, but not the same as the link you have suggested as a duplicate. (Although it may be possible to prove this statement from that one...) – Mario Carneiro Jul 12 '13 at 18:35

1It's a special case of that one (after you eliminate the square roots of perfect squares, which don't change the integerness of the sum). – Jul 12 '13 at 18:46

1@RahulNarain Apologies; My first reading of the question made me think that it was about the sum of only two roots. However, as demonstrated by Jyrki Lahtonen below, my question has a decidedly simpler solution (and I don't see a solution along these lines in the answers to the linked question), and even if this question is a simple case of a more complex result, future users will be better served by Jyrki's answer than those on the other page. This is why I vote to reopen. By analogy, is Euler's elementary proof of FLT for $n=3$ irrelevant since the full proof is now known? – Mario Carneiro Jul 12 '13 at 18:58

1Fair enough. Voting to reopen – Jul 12 '13 at 19:22

1[Related](http://math.stackexchange.com/q/442470/462). – Andrés E. Caicedo Jul 13 '13 at 07:13

Even $ \sqrt k+\sqrt m$ is never integer, not square number k,m – Takahiro Waki Mar 22 '18 at 12:38
1 Answers
No, it is not an integer.
Let $p_1=2<p_2<p_3<\cdots <p_k$ be all the primes $\le n$. It is known that $$K=\mathbb{Q}(\sqrt{p_1},\sqrt{p_2},\ldots,\sqrt{p_k})$$ is a Galois extension of the rationals of degree $2^k$. The Galois group $G$ is an elementary abelian 2group. An automorphism $\sigma\in G$ is fully determined by a sequence of $k$ signs $s_i\in\{+1,1\}$, $\sigma(\sqrt{p_i})=s_i\sqrt{p_i}$, $i=1,2,\ldots,k$.
See this answer/question for a proof of the dimension of this field extension. There are then several ways of getting the Galois theoretic claims. For example we can view $K$ as a compositum of linearly disjoint quadratic Galois extensions, or we can use the basis given there to verify that all the above maps $\sigma$ are distinct automorphisms.
For the sum $S_n=\sum_{\ell=1}^n\sqrt{\ell}\in K$ to be a rational number, it has to be fixed by all the automorphisms in $G$. This is one of the basic ideas of Galois correspondence. But clearly $\sigma(S_n)<S_n$ for all the nonidentity automorphisms $\sigma\in G$, so this is not the case.
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How does the $k$ of the first paragraph relate to the $n$ of the second? – Mario Carneiro Jul 12 '13 at 18:36


14It looks like it would be sufficient to just consider the automorphism $\sqrt2\mapsto\sqrt2$, since each sum for $n\ge2$ contains $\sqrt2$. This automorphism satisfies $\sigma(\sqrt k)\le\sqrt k$ and $\sigma(\sqrt2)<\sqrt2$, so $\sigma(S_n)
– Mario Carneiro Jul 12 '13 at 18:43 
4Yes, Mario. That suffices. Well spotted! My argument actually proves that the sum $S_n$ is a primitive element, i.e. $K=\mathbb{Q}(S_n)$. – Jyrki Lahtonen Jul 12 '13 at 18:46

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1Beautiful. I was expecting something arithmetic. Reading that made my night. – Alex Petzke Jul 12 '13 at 23:10

2The arithmetic nightmares are hidden in that result about the Galois group of $K/\mathbb{Q}$. I'm just picking cherries. – Jyrki Lahtonen Jul 13 '13 at 05:29