Can $\sum_{n=2}^{k} \sqrt{n}$ be rational?

Question

It is easy to see that $\sqrt{2}$ and $\sqrt{2}+\sqrt{3}$ are irrational. So $\sqrt{2}+\sqrt{3} + \sqrt{4}$ is irrational. But what about $\sqrt{2}+\sqrt{3} + \sqrt{4} + \sqrt{5}$? I suspect that $$\sum_{n=2}^{k} \sqrt{n}$$ is always irrational, is it true and is there a simple way to proof that?

Answer 1

$\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ cannot be rational. Let $p$ be the greatest prime in $[2,n]$. By quadratic reciprocity and Dirichlet's theorem there is some huge prime $P\equiv 1\pmod{4}$ such that $p$ is a quadratic non-residue $\!\!\pmod{P}$, while the primes less than $p$ are quadratic residues $\!\!\pmod{P}$. It follows that $\sqrt{2}+\sqrt{3}+\ldots+\sqrt{n}$ does not belong to $\mathbb{F}_{P}$ but it belongs to a quadratic extension of $\mathbb{F}_P$. In particular it cannot be rational.