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 $n \in \mathbb N$


$$\sum\_{i=1}^N a\_i^n \in \mathbb Q \frac{e}{\sqrt{2}}?$$



Finite set of numbers whose powers sum up to irrational number
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](https://en.wikipedia.org/wiki/Newton%27s_identities#Expressing_elementary_symmetric_polynomials_in_terms_of_power_sums) 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.


