Irish Mathematical Olympiad, 2014

We eliminate variables to get an equation in $b$ and $p$. First,
$$a = p - \frac{2}{b} = \frac{bp - 2}{b}, \quad \text{so that} \quad \frac{1}{a} = \frac{b}{bp - 2}. \quad \text{Then}$$
$$c = p - \frac{2}{a} = p - \frac{2b}{bp-2} = \frac{bp^2 - 2p - 2b}{bp-2}, \quad \text{so that} \quad \frac{1}{c} = \frac{bp-2}{bp^2 - 2p - 2b}.$$
Finally,
$$b = p - \frac{2}{c} = p - \frac{2bp - 4}{bp^2 - 2p - 2b}.$$
Rearranging we obtain $b^2(p^2 - 2) + b(2p - p^3) + (2p^2 - 4) = 0$ that is
$$(p^2 - 2)(b^2 - pb + 2) = 0.$$
If $p^2 \ne 2$, $b$ must be a root of the quadratic function $f(x) = x^2 - px + 2$ and by symmetry so must $a$ and $c$. But this is a contradiction since $a$, $b$ and $c$ are different numbers. It follows that we must have $p^2 = 2$ and so $p$ must either be equal to $+\sqrt{2}$ or to $-\sqrt{2}$.
It is easy to check that both possibilities for $p$ lead to the solution set
$$a = \frac{pt - 2}{t}, \quad b = t, \quad c = -\frac{2p}{pt - 2}.$$
where $t \ne 0$, $\frac{2}{p}$. This solution must satisfy $abc = -2p$.

The equation $a + \frac{2}{b} = b + \frac{2}{c}$ implies $a - b = \frac{2}{c} - \frac{2}{b}$ which can be rewritten as
$$bc(a - b) = 2(b - c).$$
Because the given equations are cyclically symmetric, we also obtain
$$ab(c - a) = 2(a - b) \quad \text{and} \quad ac(b - c) = 2(c - a).$$
Multiplying these three equations and using that $(a - b)(b - c)(c - a) \ne 0$, we obtain
$$(abc)^2 = 8.$$
From $a + \frac{2}{b} = p = c + \frac{2}{a}$ we obtain $pb = ab + 2$ and $pa = ac + 2$. Subtracting these gives $p(a - b) = a(c - b)$. By cyclic symmetry we also get
$$p(b - c) = b(a - c) \quad \text{and} \quad p(c - a) = c(b - a).$$
Multiplying these three equations and using that $(a - b)(b - c)(c - a) \ne 0$, we obtain
$$p^3 = -abc.$$
This implies $p^6 = (-abc)^2 = 8$, thus $p^2 = 2$, and finally $abc = -p^3 = -2p$.