62nd Czech and Slovak Mathematical Olympiad

Obviously $a\ne 1$, thus we can rewrite the equation as $$\frac{a^2+1}{a-1}=\frac{2b^2-3}{2b-1}(1)$$ 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 $3a^2-a+4=0$, which has no real solution. Similarly for $b=0$ we get $a^2-3a+4=0$ 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 $2b^2-3>0$, 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 $2b^2-3$ and $2b-1$, it divides $(2b-1)(2b+1)-2(2b^2-3)=5$. Thus there are four possibilities to fulfill the equation in (1). (i) $a^2+1=2b^2-3$ and $a-1=2b-1$, which has no real solution. (ii) $a^2+1=2(2b^2-3)$ and $a-1=2(2b-1)$; substituting $a=4b-1$ into the first equation we get $3b^2-2b+2=0$, with no real solutions. (iii) $5(a^2+1)=2b^2-3$ and $5(a-1)=2b-1$, with solution $a=0,b=-2$. (iv) Finally $5(a^2+1)=2(2b^2-3)$ and $5(a-1)=2(2b-1)$ with solutions $a=-1,b=-2$ and $a=7,b=8$. Thus there are five solutions of the problem: $$(0,1),(-1,1),(0,-2),(-1,-2),(7,8).$$