Austria competition problems, 2010

We first note that the right-hand side of the equation is odd. We therefore know that $x^2$ is odd, and therefore $x$ is odd. One of the neighbors of $x$ must therefore be divisible by $4$, and we therefore have $x = 2^p a \pm 1$ with $p > 1$ and $a$ odd.

Let us first assume $x = 2^p a + 1$. If $a > 1$, we first note that $x^2 - 1 = 2^{p+1} a (2^{p-1} a + 1)$ holds, and therefore, since $y$ must contain all odd factors of $x^2 - 1$,
$$y \geq 3(2^{p-1} a + 1) > 2^p a + 1 = x,$$
which contradicts $x > y$. We therefore have $a = 1$, and $x = 2^p + 1$. The only possible value for $y$ is then $2^{p-1} + 1$, since $b(2^{p-1} + 1) > 2^p + 1 = x$ holds for any $b > 1$. Any possible solution in this case must therefore have $x = 2^p + 1$, $y = 2^{p-1} + 1$ and therefore $z = p + 1$. If $p = 2$, we obtain $x = 5$ and $y = z = 3$, which contradicts $y > z$. For any $p > 2$ however, we obviously have $x > y$ and $y > z$ holds, since $2^{p-1} + 1 > p + 1 \Longleftrightarrow 2^{p-1} > p$ is certainly true. All tripels $(2^p + 1, 2^{p-1} + 1, p + 1)$ are therefore solutions for $p > 2$.

Now let us assume $x = 2^p a - 1$. If $a > 1$, we have $x^2 - 1 = 2^{p+1} a (2^{p-1} a - 1)$, and therefore
$$y \geq 3(2^{p-1} a + 1) = 2^p a - 1 + 2^{p-1} a - 2 > 2^p a - 1 + 2^p - 2 > x,$$
which again contradicts $x > y$, and we once again have $a = 1$. If $x = 2^p - 1$, possible values for $y$ are either $y = 2^{p-1} - 1$ or $y = 2(2^{p-1} - 1)$, since $b(2^{p-1} - 1) > 2^p - 1$ for any $b > 2$. We therefore have two further groups of solutions. In the first case, we have $x = 2^p - 1$, $y = 2^{p-1} - 1$ and therefore $z = p + 1$. If $p = 2$, we obtain $x = 3$, $y = 1$ and $z = 3$, which contradicts $y > z$. If $p = 3$, we obtain $x = 7$, $y = 3$ and $z = 4$ again contradicting $y > z$. If $p \ge 4$ however, $y > z$ holds, since $2^{p-1} - 1 > p + 1 \Longleftrightarrow 2^{p-1} > p + 2$ is certainly true. All tripels $(2^p - 1, 2^{p-1} - 1, p + 1)$ are therefore solutions for $p > 3$.

Finally, if $x = 2^p - 1$ and $y = 2^p - 2$, we have $z = p$. If $p = 2$, we obtain $x = 3$ and $y = z = 2$, which again contradicts $y > z$. If $p > 2$ however, $y > z$ holds, since $2^p - 2 > p \Longleftrightarrow 2^p > p + 2$ is again true. All tripels $(2^p - 1, 2^p - 2, p)$ are therefore also solutions for $p > 2$.

Summarizing, the solutions are given by the tripels
$$\begin{array}{l} (2^p + 1, 2^{p-1} + 1, p + 1) \quad \text{for } p > 2, \\ (2^p - 1, 2^{p-1} - 1, p + 1) \quad \text{for } p > 3, \\ \text{and } (2^p - 1, 2^p - 2, p) \quad \text{for } p > 2. \end{array}$$