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.)
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asked Jul 12, 2013 at 18:22
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1$\begingroup$ @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...) $\endgroup$– Mario CarneiroJul 12, 2013 at 18:35
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1$\begingroup$ It's a special case of that one (after you eliminate the square roots of perfect squares, which don't change the integer-ness of the sum). $\endgroup$– user856Jul 12, 2013 at 18:46
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2$\begingroup$ @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? $\endgroup$– Mario CarneiroJul 12, 2013 at 18:58
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1$\begingroup$ Fair enough. Voting to reopen $\endgroup$– user856Jul 12, 2013 at 19:22
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1$\begingroup$ Related. $\endgroup$– Andrés E. CaicedoJul 13, 2013 at 7:13
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1 Answer
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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 2-group. 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 non-identity automorphisms $\sigma\in G$, so this is not the case.
answered Jul 12, 2013 at 18:32
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16$\begingroup$ It 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 $\sigma(S_n)<S_n$ and $\sigma$ is a nonidentity automorphism that does not fix $S_n$, so $S_n\notin\Bbb Q$. Is this correct? $\endgroup$ Jul 12, 2013 at 18:43
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5$\begingroup$ Yes, Mario. That suffices. Well spotted! My argument actually proves that the sum $S_n$ is a primitive element, i.e. $K=\mathbb{Q}(S_n)$. $\endgroup$ Jul 12, 2013 at 18:46
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2$\begingroup$ Beautiful. I was expecting something arithmetic. Reading that made my night. $\endgroup$ Jul 12, 2013 at 23:10
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2$\begingroup$ The arithmetic nightmares are hidden in that result about the Galois group of $K/\mathbb{Q}$. I'm just picking cherries. $\endgroup$ Jul 13, 2013 at 5:29