Problem 1688

1. Rearrange the given equation:
$$\sqrt{t(1-A^{-2})(1-x^{-2})(1-y^{-2})(1-z^{-2})} = (1+x^{-1})(1+y^{-1})(1+z^{-1})$$
Squaring both sides, we get:
$$t(1-A^{-2})(1-x^{-2})(1-y^{-2})(1-z^{-2}) = \left( (1+x^{-1})(1+y^{-1})(1+z^{-1}) \right)^2$$
Simplifying the right-hand side:
$$\left( (1+x^{-1})(1+y^{-1})(1+z^{-1}) \right)^2 = \left( \frac{x+1}{x} \cdot \frac{y+1}{y} \cdot \frac{z+1}{z} \right)^2 = \frac{(x+1)^2 (y+1)^2 (z+1)^2}{x^2 y^2 z^2}$$
Therefore, the equation becomes:
$$t(1-A^{-2})(1-x^{-2})(1-y^{-2})(1-z^{-2}) = \frac{(x+1)^2 (y+1)^2 (z+1)^2}{x^2 y^2 z^2}$$
Rearranging, we get:
$$t(1-A^{-2}) = \frac{(x+1)(y+1)(z+1)}{(x-1)(y-1)(z-1)}$$
Define:
$$f(x,y,z) = \frac{(x+1)(y+1)(z+1)}{(x-1)(y-1)(z-1)}$$

2. Assume for contradiction that there are infinitely many solutions:
We need to show that $t \leq 36$.

3. **Bounding $t$:**
- Note that $A = 1$ does not yield any solutions, and neither does $x = 1$, $y = 1$, or $z = 1$.
- Therefore, $1 - A^{-2} \geq \frac{3}{4}$.
- Since $f(x,y,z)$ is decreasing in all of $x, y, z$, we have:
$$f(x,y,z) \leq f(2,2,2) = \frac{(2+1)^3}{(2-1)^3} = 27$$
- Combining these, we get:
$$t \leq \frac{27}{\frac{3}{4}} = 36$$

4. **Fix $t = T$:**
Since $t$ is bounded, there exists some $T$ such that there are infinitely many solutions to the original equation with $t = T$.

5. **Finitely many solutions for fixed $A$:**
Since $f(x,y,z)$ is decreasing in each of $x, y, z$, there are only finitely many solutions for a fixed $A$. Consequently, there are solutions for infinitely many values of $A$.

6. **Bounding the median of $x, y, z$:**
- Without loss of generality, assume $x \geq y \geq z$.
- Suppose infinitely many solutions satisfy $y \leq N$ for some large integer $N$.
- Then there exist $y_0, z_0$ such that there are infinitely many solutions with $y = y_0$ and $z = z_0$.
- Since $f(x, y_0, z_0)$ is decreasing in $x$ and $T(1 - A^{-2})$ is increasing in $A$, there can be at most one solution, which is a contradiction.

7. **Bounding $f(x, y, z)$:**
- For all other solutions, taking $N$ large enough, we can note that:
$$f(x, y, z) \leq f(N, N, z) \leq \left( \frac{N+1}{N-1} \right)^2 \cdot \frac{z+1}{z-1} < c$$
for some fixed constant $c < 1$.

8. Combining the results:
- Adding back in the finitely many solutions that were disregarded, there is a constant $C < 1$ such that $f(x, y, z) < C$ for all solutions $(T, A, x, y, z)$.
- However, there must be solutions for infinitely many $A$, and:
$$T(1 - A^{-2}) = f(x, y, z) < C$$
fails for large enough $A$, which is a contradiction.

Thus, there can only be finitely many solutions to the equation, as desired.