, 2014

$$p(0) = -1,\ p(1) = 1,\ p(2) = p(3) = -1,\ p(4) = 7,$$
the roots of $p$ are positive real numbers that belong to the union of the intervals $(0, 1)$, $(1, 2)$ and $(3, 4)$. Hence the roots are distinct, and neither of them is an integer. But since the coefficients of $p$ are integers, neither can they be rational numbers, by the Gauss Lemma. Hence, $u, v$ are irrational. Suppose now that $u = v r$, where $r$ is rational, and $r \ne 1$. Then $v$ satisfies the pair of cubic equations:
$$v^3 - 5v^2 + 6v - 1 = 0,\ \quad r^3 v^3 - 5r^2 v^2 + 6r v - 1 = 0.$$
Eliminating the term involving $v^3$ in the usual way, we see that
$$(1 - r)(5r^2 v^2 - 6r(1 + r)v + 1 + r + r^2) = 0.$$
Since $r \ne 1$ by hypothesis, it follows that $u$ satisfies the quadratic equation
$$5u^2 = 6(r + 1)u - (1 + r + r^2),$$
whence (multiplying by $5u$ and substituting for $5u^2$)
$$\begin{aligned} 25u^3 &= 6(r + 1)5u^2 - 5(1 + r + r^2)u \\ &= (36(r + 1)^2 - 5(1 + r + r^2))u - 6(r + 1)(1 + r + r^2), \end{aligned}$$
i.e.,
$$25u^3 = (31 + 67r + 31r^2)u - 6(r + 1)(1 + r + r^2).$$
Substituting these expressions for $u^3, u^2$ into $u^3 - 5u^2 + 6u - 1 = 0$, we derive a linear equation for $u$ with non-zero rational coefficients:
$$0 = (31 - 83r + 31r^2)u - (1 + r)(6 - 19r + 6r^2).$$
This implies that $u$ is rational, which is false. This contradiction establishes the conclusion that $u/v$ is also irrational.