Problem 2

For $b>1$, let $k$ be the positive integer quotient. Clearing denominators gives $X^2-2kb^2X+k(b^3-1)=0$, whose one root is $a$. By Vieta, the other root is $a'=2kb^2-a=k(b^3-1)/a$. The first expression shows that $a'$ is an integer, and the second shows $a'>0$ because $k>0$ and $b^3-1>0$. The same polynomial identity gives $a'^2=k(2a'b^2-b^3+1)$, so the new denominator is positive and its quotient is again $k$.