Non-monic polynomial with roots on the unit circle When setting up to prove Dirichlet's Unit Theorem, we show that all roots of unity in a number field K are algebraic integers. Further, if all conjugates of $x \in \mathcal{O}_K$ have modulus 1 then $x$ is a root of unity. (Here $\mathcal{O}_K$ is the ring of algebraic integers in K).
So the first question is can we find $x \in K\setminus\mathcal{O}_K$ such that $x$ and all of its conjugates have modulus 1 (and therefore cannot be roots of unity).
This is equivalent to finding a non-monic irreducible polynomial $p(x)\in \mathbb{Z}[X]$ such that all of its roots are on the unit circle. Now I have found an example of this $5-6x^2+5x^4$ from a question on math overflow.
So the real question is how might you come up with such an example (is it just trial and error?) The only other observation I have made is that in such a polynomial with coefficients $a_i \in \mathbb{Z}$ then $a_{n-k}=a_k$ for all $k \in  \{0, 1, 2, ..., n\}$. This is due to the fact that $\overline{z}=1/z$ for $z \in \mathbb{C}, |z|=1$. 
 A: Here's an easy way to generate examples. Let $\alpha$ be a totally real algebraic number all of whose conjugates  $\sigma \alpha$ have absolute value strictly less than $2$ (these are easy to find --- take any totally real algebraic number and divide by a big integer). Let us also insist that $\alpha$ is not itself an algebraic integer. Let $\beta$ be a solution to the equation:
$$\beta + \frac{1}{\beta} = \alpha.$$
All the conjugates of the RHS are, by assumption, real numbers in the interval $(-2,2)$. This forces all the conjugates of $\beta$ to be complex numbers of absolute value one --- which is what you want. Moreover, $\beta$ will not be a root of unity, since otherwise $\alpha$ would be an algebraic integer (it turns out that the converse is also true --- all algebraic integers whose conjugates all of absolute value less than $2$ have the form $\zeta + \zeta^{-1}$.) Writing the conjugates of $\alpha$ as $\sigma \alpha$, then $\beta$ is a root of the polynomial
$$f(x) =\prod_{\sigma} (x^2 - \sigma \alpha x + 1)$$
Once you clear denominators, you get the desired polynomial. It will be irreducible, because, by looking at infinite places $[\mathbf{Q}(\beta):\mathbf{Q}(\alpha)] = 2$. This is essentially the unique way to produce such polynomials --- if all the conjugates of $\beta$ have absolute value $1$, then all the conjugates of
$$ \alpha = \beta + \frac{1}{\beta}$$
will be real and in $(-2,2)$. Your example corresponds to $\alpha = 4/\sqrt{5}$, each of whose conjugates is approximately $1.78885\ldots < 2$. I guess the simplest example is to take $\alpha = 1/2$, and the resulting polynomial is $2 x^2 -  x + 2$.
