Austria 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 $x^2-1$, $$y\ge 3(2^{p-1}a+1)>2^{p}a+1=x,$$ 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 $2^{p-1}+1>p+1⟺2^{p-1}>p$ 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 $$y\ge 3(2^{p-1}a+1)=2^{p}a-1+2^{p-1}a-2>2^{p}a-1+2^p-2>x,$$ 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 $2^{p-1}-1>p+1⟺2^{p-1}>p+2$ 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 $2^p-2>p⟺2^p>p+2$ 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 $$\begin{array}{l}(2^p+1,2^{p-1}+1,p+1)\text{for }p>2,\\ (2^p-1,2^{p-1}-1,p+1)\text{for }p>3,\\ \text{and }(2^p-1,2^p-2,p)\text{for }p>2.\end{array}$$