Irish Mathematical Olympiad

Suppose, for a fixed $\lambda$, a solution to the four equations exists. Then, from the first equation, $2\lambda = a + \frac{1}{a} - 2 = \frac{(a-1)^2}{a} \ge 0$, and so $\lambda \ge 0$. From the first equation in the second row, $2(1 - \lambda) = b + \frac{1}{b} > 0$ and so $\lambda < 1$. Hence, if a solution exists, then, in the first instance, $0 \le \lambda < 1$.

Furthermore, if $\lambda = 0$, then $a = 1$, $c = 1$, whence $b = 1$ also. Thus, one solution is $\lambda = 0$, and $a = b = c = 1$.

On the other hand, if $a = c$ the two equations which involve $1 - \lambda$ imply $b + \frac{1}{a} = a + \frac{1}{b}$, or equivalently, $a - \frac{1}{a} = b - \frac{1}{b}$. Because the equation $x + \frac{1}{x} = k$ has at most one positive solution, this implies $a = b$. From $a = b = c$ we easily obtain $\lambda = 0$. Hence, if $\lambda \ne 0$ we need to have $a \ne c$.

From now on we suppose $0 < \lambda < 1$. Clearly, $a, c$ satisfy the same quadratic equation $x^2 - 2(1 + \lambda)x + 1 = 0$. As $a \ne c$, they form its two roots and so $ac = 1$. Using this and the equations in the second row, we deduce that $b^2 = 1$, whence $b = 1$ and $c = 1 - 2\lambda > 0$. Plugging this into the second equation in the first row, we deduce that
$$2(1 + \lambda) = 1 - 2\lambda + \frac{1}{1 - 2\lambda} = \frac{(1 - 2\lambda)^2 + 1}{1 - 2\lambda},$$
whence $2(1 + \lambda)(1 - 2\lambda) = 2 - 4\lambda + 4\lambda^2$ which simplifies to $8\lambda^2 - 2\lambda = 0$.
Thus, $\lambda = 1/4$. For this value of $\lambda$, we obtain $a = 2, b = 1, c = 1/2$.

We consider the four variables $a$, $b$, $c$ and $\lambda$ as unknowns. Subtracting the two equations in the first row we get $a-c = \frac{1}{c} - \frac{1}{a}$. As $ac \neq 0$, this is equivalent to $(a-c)(ac-1) = 0$. If $a=c$, the two equations which involve $1-\lambda$ imply $b+\frac{1}{a} = a+\frac{1}{b}$. This can be rewritten as $(b-a)(1+\frac{1}{ab}) = 0$. As $ab > 0$, this implies $a=b$. From $a=b=c$ we easily obtain $\lambda=0$. The first equation implies then $a=1$. So we have one solution: $(\lambda, a, b, c) = (0, 1, 1, 1)$.

Assume now $a \neq c$, then $ac = 1$ and the two equations in the second row imply $b = \frac{1}{b}$, hence $b=1$. The first equation in the second row now becomes $1+\frac{1}{a} = 2(1-\lambda)$ from which we get $a = \frac{1}{1-2\lambda}$ and $c = \frac{1}{a} = 1-2\lambda$. Substituting this into the first equation yields $1-2\lambda + \frac{1}{1-2\lambda} = 2(1+\lambda)$ which simplifies to $\lambda(4\lambda-1) = 0$. For $\lambda=0$ we obtain $a=c=1$ in contradiction to our assumption. With $\lambda = \frac{1}{4}$ we get $a=2$ and so the second solution is $(\lambda, a, b, c) = (\frac{1}{4}, 2, 1, \frac{1}{2})$.