## Find an integer multiple of a given algebraic number which is an algebraic integer

Let $\theta = \omega/9 + \sqrt{2}/5$, where $\omega$ is a root of $(x^7-1)/(x-1)$. Find a rational integer $k$ such that $k\theta$ is an algebraic integer.

Recall that sums and products of algebraic integers are algebraic integers.

Note that $\omega$ is an algebraic integer, as is $\sqrt{2}$, and thus $5\omega$ and $9\sqrt{2}$ are algebraic integers. So $45\theta = 5\omega + 9\sqrt{2}$ is an algebraic integer.