TITLE: Finite set of numbers whose powers sum up to irrational number
QUESTION [6 upvotes]: It is well-known that $e/\sqrt{2}$ is irrational.
Indeed, if it was rational, i.e. $p/q$ then $e^2/2 =p^2/q^2.$ Thus, $q^2e^2=2p^2,$ which would imply that $e$ is a root of $q^2x^2=2p^2.$
Now my question is: Does there exist a finite number of complex numbers $a_1,...,a_N$ all different from zero such that for every $n \in \mathbb N$
$$\sum_{i=1}^N a_i^n \in \mathbb Q \frac{e}{\sqrt{2}}?$$

REPLY [8 votes]: Such numbers do not exist. Indeed, let $p_n$ (resp. $e_n$) be the $n$-th power sum (resp. elementary symmetric polynomial) of the $a_i$'s. Since $e_n=0$ for $n>N$, Newton's identities show that
$$\sum_{m_1 + 2m_2 + \cdots + nm_n = n \atop m_1 \ge 0, \ldots, m_n \ge 0} \prod_{i=1}^n \frac{(-p_i)^{m_i}}{m_i ! i^{m_i}}=0,\qquad n>N.$$
Here $p_i^{m_i}$ is a rational multiple of $(e/\sqrt{2})^{m_i}$. As $e/\sqrt{2}$ is transcendental, the contribution of the $n$-tuples $(m_1,m_2,\dotsc,m_n)$ with fixed $m_1+m_2+\dotsb+m_n$ is zero. There is only one $n$-tuple with $m_1+m_2+\dotsb+m_n=n$, namely $(n,0,\dotsc,0)$, which shows that $p_1=0$. Using this information, we can restrict the sum to $m_1=0$. By choosing $n>N$ to be even, a similar reasoning yields that $p_2=0$. By choosing $n>N$ to be divisible by $3$, we can infer that $p_3=0$, and so on. So all the $p_i$'s are zero, which then yields (again by Newton's identities) that all the $a_i$'s are zero.