# For integer $n>1$, can $\sum_{k=1}^{n}\sqrt{k}$ be a rational number? Can it be an integer? [duplicate]

I know that the sum of two (or more) irrational numbers can be rational. For example, both $$\sqrt{2}$$ and $$1-\sqrt{2}$$ are irrational numbers, but their sum is rational.

Also I know that $$\sqrt{m}$$ is either integer or an irrational number for any natural number $$m$$. In other words, it can not be a rational number unless it is an integer.

I have no problem to prove the rationality/irrationality of the square root of any natural number. Using "direct proof" for rational numbers, and using "proof by contradiction" for irrational numbers.

I could not combine what I know together to answer the following questions:

Can $$\sum_{k=1}^{n}\sqrt{k}=\sqrt{1}+\sqrt{2}+\sqrt{3}+\dots+\sqrt{n}$$ be rational for any integer $$n>1$$? If yes, then what is the least such $$n$$?

Can it be an integer for any integer $$n>1$$? If yes, then what is the least such $$n$$?

Note: If it can not be an integer, does not mean it can not be rational.

Any help would be really appreciated.

• You can show that $\sqrt1+\cdots+\sqrt n$ is an algebraic integer (i.e., is the root of a monic polynomial with integer coefficients). Consequently it cannot be rational without being an integer. – Barry Cipra Sep 14 at 21:42

We first prove that if $$x,y\in\mathbb{Q}$$ are such that $$\sqrt{x},\sqrt{y}\notin\mathbb{Q}$$, then$$\sqrt{x}+\sqrt{y}\notin\mathbb{Q}$$. Let us suppose the contrary, we have $$\sqrt{x}=\frac{(\sqrt{x}+\sqrt{y})^2+x-y}{2(\sqrt{x}+\sqrt{y})}\in\mathbb{Q}$$ which is not. Let $$n\geqslant 2$$ and $$S_n=\sum_{k=1}^n{\sqrt{k}}$$, if $$S_n-\sqrt{2}\in\mathbb{Q}$$ then $$S_n=(S_n-\sqrt{2})+\sqrt{2}\notin\mathbb{Q}$$. If $$S_n-\sqrt{2}\notin\mathbb{Q}$$, by the lemma $$S_n=(S_n-\sqrt{2})+\sqrt{2}\notin\mathbb{Q}$$. In particular $$S_n$$ is not an integer for all $$n\geqslant 2$$.
• How are you applying the lemma? You need $S_n - \sqrt 2$ to be of the form $\sqrt x$. – Eric M. Schmidt Sep 14 at 21:09
• the actual proof is not that trivial and hints toward it are linked in the comments above (it requires some Galois theory though); not sure if there is an elementary solution for $n \ge 5$ (not an ad hoc one where you just take the rational conjugates, build the minimal polynomial and show it has no integral solutions) – Conrad Sep 14 at 21:55