Sums of powers of algebraic numbers that are rational $\newcommand\Q{\mathbb Q}$I have a series of questions related to sum of $k$-powers of algebraic numbers (it is actually only one question that I try to weaken/strengthen the conditions). Suppose we know the following
$$\sum_{i=1}^n a_i^k \in \Q$$
for algebraic number $a_i$'s for which not all of them are in $\Q$. Then I ask


*

*If $n=k=2$  and $\Q(a_1,a_2)$ is real then is it true that $[\Q(a_1,a_2):\Q]$ is even?

*Same question as 1. except we assume the general condition $n\geq 2$ and $\Q(a_1,\dots,a_n)$ is real

*Same question as 1. except we assume any $k\geq 2$ and ask if $[\Q(a_1,a_2):\Q]$ is divisible by $k$

*Same question as 2. except we assume any $k\geq 2$ and ask if $[\Q(a_1,\dots, a_n):\Q]$ is divisible by $k$


5-8. We ask the same questions as 1-4. without assuming $\Q(a_1,\dots,a_n)$ is real.
Probably the answers are trivial (for instance if 1. is already not true). And I think 5-8. is generally easier than 1-4 (I feel that for 5-8. all of them are generally false).
 A: $\newcommand\Q{\mathbb Q}$For sums of squares the answer is no. Take any real number field $K$ that has odd degree over $\Q$. Take a non rational number $x\in K$. Then you can find a natural number $m$ such that $m>x^2$ with respect to any ordering of $K$. By Artin's theorem it follows that $m-x^2$ is a sum of squares over $K$ (this is the smallest preordering of $K$). So $m$ is the sum of squares in $K$ for which one of the summands is $x^2$ with $x$ not in $\Q$ but $[K:\Q]$ is odd. 
A: (1) seems to be false in quite a strong way. Suppose $n = k = 2$ and you even insist that the sum is equal to $1$, e.g.
$$x^2 + y^2 = 1.$$
Then you can make $K = \mathbf{Q}(x,y)$ be any field you want, and certainly of odd degree and totally real. Here is the construction. Given any number field $K$, by the primitive element theorem one can write $K = \mathbf{Q}(t)$ for some $t$. Then write
$$x = \frac{2t}{t^2+1}, \qquad y = \frac{t^2-1}{t^2+1}.$$
Then $x^2 + y^2 = 1$, and $x,y \in \mathbf{Q}(t)$, so $\mathbf{Q}(x,y) \subseteq \mathbf{Q}(t)$. Finally, note that
$$t = \frac{1+x+y}{1+x-y} \in \mathbf{Q}(x,y),$$
and so $K = \mathbf{Q}(t) = \mathbf{Q}(x,y)$.
