Prove that every non-abelian group of order $8$ has a $2$-dimensional irreducible character all of whose values are integers. 
Prove that every non-abelian group of order $8$ has a $2$-dimensional irreducible character all of whose values are integers.

I get that if $G$ is non abelian it must have an irreducible representation $>1$ and this implies that
$$8=1^2+1^2+1^2+1^2+2^2$$
is the only possibility. Hence unique $2$-dimensional irreducible character,
I cannot see how to show that all its values will be integers. I feel this question may need some Galois theory that I am not very familiar with.
EDIT: I have the following hints.
$1)$ If $\chi$ is a character then so is $\sigma(\chi)$ for every automorphism $\sigma$ of $\mathbb{C}$. 
$2)$ If $\alpha \in \mathbb{C}$ is fixed by every automorphism of $\mathbb{C}$ then $\alpha \in \mathbb{Q}$.  
 A: If $\chi_1, \ldots , \chi_4$ are the degree $1$ characters, and $\chi_5$ is the degree $2$ character, then what is $\chi_1 + \chi_2 + \chi_3+ \chi_4 + 2\chi_5$?
A: Consider complex conjugation as an automorphism of $\mathbb C$. The complex conjugate of a character is also a character. You only have one character of degree $2$, so that implies something about the values.
Do you know that the values of a character are algebraic integers? Here, they are the sums of eigenvalues of matrices which satisfy $M^4=I$ (the orders of elements in a non-abelian group of order $8$ all divide $4$), and therefore satisfy $\lambda^4=1$. So if you don't know a general theorem about algebraic integers, you can simply list the possible sums of two eigenvalues, and see what the possibilities are.
A: There only exist two non-abelian groups of order 8, they are $Q_8$ and $D_8$, you can compute the character table of each of them but spoiler: they both have a 2 dimensional irreducible character (the last character of the table)
Since there are only two such groups I think is more effective to just compute the tables.
