Connection with golden ratio? Consider the following problem:

Let $p\in\mathbb{Z}[x]$ be a polynomial with integer coefficient.
  Suppose that the leading coefficient is 1, all roots are real
  and in $(0, 3)$. Find all possible roots of $p$.

The roots 1 and 2 are abvious, but there are more possible roots: the roots of
$x^2-3x+1$,
$$\frac{3+\sqrt{5}}{2}\quad\text{ and }\quad\frac{3-\sqrt{5}}{2}.$$
Other possible roots I didn't find, but what is interesting (at least for me)
I notice that the above non-integer roots have the form
$$\frac{3+\sqrt{5}}{2} = 1+\phi\quad\quad\frac{3-\sqrt{5}}{2} = 2-\phi,$$
where
$$\phi = \frac{\sqrt{5}+1}{2}$$
is the golden ratio constant.


*

*Is the set of all possible roots $\{1, 1+\phi, 2-\phi, 2\}$ ?

*(if the answer of question 1 is positive) Why the golden ratio constant appears in this problem ?


This answer is only for degree $2$.
There are only four pairs $(a,b)=(-2,1),(-3,1),(-3,2),(-4,4).$ (Maybe you want to eliminate $(-2,1),(-4,4)$, though.)
If $x^2+ax+b=0\ (a,b\in\mathbb Z)$, then we have
$$0\lt \frac{-a-\sqrt{a^2-4b}}{2}\lt 3,\ \ 0\lt\frac{-a+\sqrt{a^2-4b}}{2}\lt 3$$
$$\iff 0\lt -a-\sqrt{a^2-4b}\lt 6,\ \ 0\lt -a+\sqrt{a^2-4b}\lt 6\tag1$$
Since we have $0\lt -2a\lt 12\iff -6\lt a\lt 0$, 
$$(1)\iff a\lt -\sqrt{a^2-4b}\lt a+6,\ \ a\lt\sqrt{a^2-4b}\lt a+6$$
$$\iff a\lt -\sqrt{a^2-4b},\ \ \sqrt{a^2-4b}\lt a+6$$
$$\iff -a\gt \sqrt{a^2-4b},\ \ \sqrt{a^2-4b}\lt a+6$$
$$\iff (-a)^2\gt a^2-4b,\ \ a^2-4b\lt (a+6)^2\iff b\gt 0,\ \ b\gt -3a-9.$$
With $a^2-4b\ge 0$, we have
$$-6\lt a\lt 0,\ \ b\gt 0,\ \ b\gt -3a-9,\ \ b\le \frac{a^2}{4}.$$
Considering these conditions on $ab$ plane will give you all possible pairs $(a,b)$, which are the followings :
$$(a,b)=(-2,1),(-3,1),(-3,2),(-4,4)$$
