9.3. Natural numbers $a, x$ and $y$, greater than 100, are such that $y^{2}-1=$ $=a^{2}\left(x^{2}-1\right)$. What is the smallest value that the fraction $a / x$ can take?
# Answer. 2.

First solution. Rewrite the condition of the problem as $y^{2}=a^{2} x^{2}-a^{2}+1$. Notice that $y100$, if we set $a=2 x, y=a x-1=2 x^{2}-1$.

Second solution. We provide another proof of the estimate $a / x \geqslant 2$. Rewrite the equality from the condition as

$$
(a x-y)(a x+y)=a^{2} x^{2}-y^{2}=a^{2}-1
$$

The numbers $a^{2}-1$ and $a x+y$ are positive, so the number $k=a x-y$ is also positive (and natural). Then $a x+y=\frac{a^{2}-1}{k}$. Adding the two equations, we get

$$
2 a x=\frac{a^{2}-1}{k}+k=a^{2}+\frac{(k-1)\left(k-a^{2}+1\right)}{k} \leqslant a^{2}
$$

since $1 \leqslant k \leqslant a^{2}-1$. Thus, $2 a x \leqslant a^{2}$, which means $a / x \geqslant 2$.