Prove that if a, b, c, x, y, z, and $\alpha$ are natural numbers. For every given set of x, y, z, the number $\alpha$ obtained from the following equation:
$$\frac{a^2}{x^2} + \frac{b^2}{y^2} + \frac{c^2}{z^2} = \alpha$$
cannot be obtained from other combination of a, b and c. In other words, every natural number $\alpha$ that satisfies the above equation can only be found from a unique set of a, b and c.
Given $x$, $y$, $z$, let $a=Ax$, $b=By$, $c=Cz$; then the equation becomes $$A^2+B^2+C^2=n$$ There are many $n$ for which this equation has distinct sets of solutions $\{A,B,C\}$, for example, $$33=25+4+4=16+16+1$$ or $$38=36+1+1=25+9+4$$ or $$41=36+4+1=16+16+9$$ More examples at https://oeis.org/A223733