
1. Find all pairs of integers $a, b$ such that

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

(Pavel Novotný)


Solution. Obviously $a \neq 1$, thus we can rewrite the equation as

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

The numerator of the fraction on the left is positive, the numerator on the right is negative just for $b \in\{-1,0,1\}$.

For $b=-1$ we get $3 a^{2}-a+4=0$, which has no real solution.

Similarly for $b=0$ we get $a^{2}-3 a+4=0$ which has no real solution either.

For $b=1$ we get $a^{2}+a=a(a+1)=0$, with solutions $a \in\{0,-1\}$. Thus pairs $(0,1)$ and $(-1,1)$ are solutions of the problem.

Let us further assume $2 b^{2}-3>0$, and let us find out with which numbers we can reduce the fractions in (1).

If some integer $n$ divides both $a^{2}+1$ and $a-1$, it divides $a^{2}+1-(a+1)(a-1)=2$ as well. Similarly if $n$ divides both $2 b^{2}-3$ and $2 b-1$, it divides $(2 b-1)(2 b+1)-$ $2\left(2 b^{2}-3\right)=5$.

Thus there are four possibilities to fulfill the equation in (1).

(i) $a^{2}+1=2 b^{2}-3$ and $a-1=2 b-1$, which has no real solution.

(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 we get $3 b^{2}-2 b+2=0$, with no real solutions.

(iii) $5\left(a^{2}+1\right)=2 b^{2}-3$ and $5(a-1)=2 b-1$, with solution $a=0, b=-2$.

(iv) Finally $5\left(a^{2}+1\right)=2\left(2 b^{2}-3\right)$ and $5(a-1)=2(2 b-1)$ with solutions $a=-1$, $b=-2$ and $a=7, b=8$.

Thus there are five solutions of the problem:

$$
(0,1),(-1,1),(0,-2),(-1,-2),(7,8) \text {. }
$$
