2. For each $a \in \mathbb{Z}$, consider the set $S_{a}=\left\{\left.x \in \mathbb{R}\left|\frac{1}{x^{2}-a^{2}} \in \mathbb{Z},\right| x|\neq| a \right\rvert\,\right\}$.

a) Determine the values of $a$ for which the set $S_{a}$ contains the number $\sqrt{2}$;

b) Determine $a \in \mathbb{Z}$ for which $S_{a} \cap \mathbb{Q} \neq \varnothing$.
Subject 2. For each $a \in \mathbb{Z}$, consider the set $S_{a}=\left\{\left.x \in \mathbb{R}\left|\frac{1}{x^{2}-a^{2}} \in \mathbb{Z},\right| x|\neq| a \right\rvert\,\right\}$.

a) Determine the values of $a$ for which the set $S_{a}$ contains the number $\sqrt{2}$;

b) Determine $a \in \mathbb{Z}$ for which $S_{a} \cap \mathbb{Q} \neq \varnothing$.

Prof. Petre Simion and Prof. Victor Nicolae, Bucharest

| Details of solution | Associated grading |
| :--- | :--- |
| a) We have $\sqrt{2} \in S_{a}$ equivalent to the fact that there exists $k \in \mathbb{Z}^{*}$ such that $\frac{1}{2-a^{2}}=k$. |  |
| We deduce that $a^{2}+\frac{1}{k}=2 \in \mathbb{Z}$. This means that $\frac{1}{k} \in \mathbb{Z}$, so $k \in\{ \pm 1\}$. | $\mathbf{2 p}$ |
| For $k=-1$, we get $a^{2}=3$, a contradiction. |  |
| For $k=1$, we get $a^{2}=1$, from which we obtain $a \in\{ \pm 1\}$. |  |
| b) If $a \neq 0$, assume that $a^{2}+\frac{1}{k}=\frac{a^{2} k+1}{k}=\frac{m^{2}}{n^{2}}$, where $m$ and $n$ are natural numbers | $\mathbf{3 p}$ |


| relatively prime. Since the numbers $k$ and $a^{2} k+1$ are relatively prime, it follows that $n^{2}=k$ and |  |
| $a^{2} k+1=m^{2}$, i.e., $(a n)^{2}+1=m^{2}$, a contradiction. For $a \neq 0$, we have $S_{a} \cap \mathbb{Q}=\varnothing$. |  |
| For $a=0$, we get $x= \pm \sqrt{\frac{1}{k}}, k \in \mathbb{Z}_{+}$. For example, for $x=\frac{1}{2}$, we get $k=4$, | $\mathbf{2 p}$ |
| so $\frac{1}{2} \in S_{0} \cap \mathbb{Q}$. Therefore, $a=0$. |  |