Is $\sqrt1+\sqrt2+\dots+\sqrt n$ ever an integer?

Question

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.)

Answer 1

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.