1. Find all pairs of integers $a, b$, for which the following equality holds:

$$
\frac{a^{2}+1}{2 b^{2}-3}=\frac{a-1}{2 b-1}
$$

(Pavel Novotný)
Solution. Let's assume $a \neq 1$, so we can rewrite the given equation as

$$
\frac{a^{2}+1}{a-1}=\frac{2 b^{2}-3}{2 b-1}
$$

While the numerator of the fraction on the left side is positive, the numerator of the second fraction is negative only for $b \in\{-1,0,1\}$.

For $b=-1$, we get a quadratic equation $3 a^{2}-a+4=0$ after rearrangement, which has no real solutions, and similarly for $b=0$: the equation $a^{2}-3 a+4=0$ also has no real solutions.

For $b=1$, we get the equation $a^{2}+a=a(a+1)=0$, which has two solutions $a \in\{0,-1\}$. This gives us two pairs $(0,1)$ and $(-1,1)$ that satisfy the given equation.

Let's further assume that $2 b^{2}-3>0$, and determine which natural numbers can divide the fractions in (1).

If a number $n$ divides $a^{2}+1$ and $a-1$, it must also divide $a^{2}+1-(a+1)(a-1)=2$. Similarly, if a number $n$ divides $2 b^{2}-3$ and $2 b-1$, it must also divide $(2 b-1)(2 b+1)-2\left(2 b^{2}-3\right)=5$. Therefore, there are exactly four possibilities to achieve equality of the fractions in (1):

(i) $a^{2}+1=2 b^{2}-3$ and $a-1=2 b-1$; substituting $a=2 b$ into the first equation gives the relation $4 b^{2}+1=2 b^{2}-3$, which is not valid for any real $b$.

(ii) $a^{2}+1=2\left(2 b^{2}-3\right)$ and $a-1=2(2 b-1)$; substituting $a=4 b-1$ into the first equation, after rearrangement, we get the quadratic equation $3 b^{2}-2 b+2=0$, which has no real solutions.

(iii) $5\left(a^{2}+1\right)=2 b^{2}-3$ and $5(a-1)=2 b-1$; substituting $a=\frac{1}{5}(2 b+4)$ into the first equation, after rearrangement, we get the quadratic equation $3 b^{2}-8 b-28=0$, which has the integer solution $b=-2$. This corresponds to $a=0$.

(iv) $5\left(a^{2}+1\right)=2\left(2 b^{2}-3\right)$ and $5(a-1)=2(2 b-1)$; substituting $a=\frac{1}{5}(4 b+3)$ into the first equation, we get the quadratic equation $b^{2}-6 b-16=0$ with two integer roots $b=-2$ and $b=8$, corresponding to the values $a=-1$ and $a=7$.

Thus, the problem is satisfied by a total of five integer pairs $(a, b)$:

$$
(0,1),(-1,1),(0,-2),(-1,-2),(7,8)
$$