55
$\begingroup$

Related: Can a sum of square roots be an integer?

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.)

$\endgroup$
  • 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 Carneiro Jul 12 '13 at 18:35
  • 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). $\endgroup$ – Rahul Jul 12 '13 at 18:46
  • 1
    $\begingroup$ @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 Carneiro Jul 12 '13 at 18:58
  • 1
    $\begingroup$ Fair enough. Voting to reopen $\endgroup$ – Rahul Jul 12 '13 at 19:22
  • 1
    $\begingroup$ Related. $\endgroup$ – Andrés E. Caicedo Jul 13 '13 at 7:13
60
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ How does the $k$ of the first paragraph relate to the $n$ of the second? $\endgroup$ – Mario Carneiro Jul 12 '13 at 18:36
  • 9
    $\begingroup$ 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? $\endgroup$ – Mario Carneiro Jul 12 '13 at 18:43
  • 1
    $\begingroup$ Yes, Mario. That suffices. Well spotted! My argument actually proves that the sum $S_n$ is a primitive element, i.e. $K=\mathbb{Q}(S_n)$. $\endgroup$ – Jyrki Lahtonen Jul 12 '13 at 18:46
  • 1
    $\begingroup$ Nifty argument! $\endgroup$ – Steven Stadnicki Jul 12 '13 at 18:52
  • 1
    $\begingroup$ Beautiful. I was expecting something arithmetic. Reading that made my night. $\endgroup$ – Alex Petzke Jul 12 '13 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.