An interesting example is Sicherman dice: A pair of 6-sided dice, with positive number of eyes on each face that are not the classic 1..6 ones; if you throw them, the distribution of total eyes is the same as for two regular dice.
To construct them, one die is represented by:
$\begin{align*}
D(z)
&= z + z^2 + z^3 + z^4 + z^5 + z^6 \\
&= z (1 + z) (1 - z + z^2) (1 + z + z^2)
\end{align*}$
Throwing two dice gives the distribution:
$\begin{equation*}
D^2(z)
= z^2 (1 + z)^2 (1 - z + z^2)^2 (1 + z + z^2)^2
\end{equation*}$
So we want two dice, call them $D_1(z)$ and $D_2(z)$, such that $D^2(z) = D_1(z) \cdot D_2(z)$. The respective polynomials have to satisfy several conditions:
- We want that all faces have some eyes, i.e., the constant term of the polynomial has to be zero. We need to assign a $z$ factor to each.
- The coefficients have to be integers (number of faces with each number of eyes). It is a fun fact that if a polynomial with integer coefficients factors, the primitive factors have integer coefficients (from Gauss' lemma). So this isn't a restriction.
- The number of faces of the dice has to be six. This is the sum of the coefficients in the polynomial, which is just $D_i(1)$. The respective factors at $z = 1$ are $1$, $2$, $1$ and $3$. We have to give a $1 + z$ and a $1 + z + z^2$ to each, we can shuffle the $1 - z + z^2$ factors around.
Thus the only solution (except for switching the dice, and classical dice) is:
$\begin{align*}
D_1(z)
&= z (1 + z) (1 + z + z^2) \\
&= z + 2 z^2 + 2 z^3 + z^4 \\
D_2(z)
&= z (1 + z) (1 + z + z^2) (1 - z + z^2)^2 \\
&= z + z^3 + z^4 + z^5 + z^6 + z^8
\end{align*}$
This translates into dice with faces marked $\{1, 2, 2, 3, 3, 4\}$ and $\{1, 3, 4, 5, 6, 8\}$.
