## Task 3 - 260933

If a real number $a$ is given, then each real number $x$ is assigned a number $y$, namely

$$
y=\frac{x^{3}+x^{2}+a x+1}{x^{2}+1}
$$

(A) Determine, if $a=-3$ is given, two integers $x$ whose assigned numbers $y$ are also integers!

(B) Determine a real number $a$ for which the following statement (*) holds!

(*) If the number $a$ is given, then there are infinitely many integers $x$ whose respective assigned numbers $y$ are also integers.

(C) Investigate whether there is any other real number $a$ besides the one determined in (B) for which the statement (*) holds!
For every $a \in \mathbb{R}$

$$
y=\frac{x^{3}+x+x^{2}+1+(a-1) x}{x^{2}+1}=x+1+\frac{a-1}{x^{2}+1}
$$

(A): For $a=-3$, we have $y=x+1+\frac{-4}{x^{2}+1}=x+1-\frac{4}{x^{2}+1}$. Thus, for every integer $x$, the number $y$ is an integer if and only if $x+1-y=\frac{4}{x^{2}+1}$ is an integer, meaning $x^{2}+1$ is a divisor of 4. This is true, for example, for $x_{1}=0$ and $x_{2}=1$. Indeed, $y(0)=1$ and $y(1)=0$.

(B)/(C): For a statement $\left({ }^{*}\right)$ to hold for some $a$, it must be that for infinitely many integers $x$, $y-(x+1)=\frac{a-1}{x^{2}+1}$ is an integer. Since $x^{2}+1$ is also an integer, it follows that $a-1=(y-(x+1)) \cdot\left(x^{2}+1\right) \in \mathbb{Z}$ and $a-1$ must be divisible by $x^{2}+1$.

In particular, $a-1$ must be divisible by infinitely many different integers, which is only satisfied by the integer 0, so $a=1$. For $a=1$, $y=x+1$ is an integer for every integer $x$, and in particular, for infinitely many $x$. Therefore, $a=1$ is the only real number that satisfies $\left(^{*}\right)$.