(a,b)=(9k^2,9k) where k is a positive integer. proof) ab^2+9 l a^2b+b ab^2+9 l a^2b^2+b^2 ab^2+9 l a^2b^2+9a-9a+b^2 so ab^2+9 l b^2-9a However b^2-9a<b^2<ab^2<ab^2+9 so b^2-9a=0 and a= b^2/9 a has to be a positive integer so b is a multiple of 3. when b=3k a=k^2 and plugging that into the first equation you get: k/3 meaning k also has to be a multiple of 3 in order for it to be an integer. so b=9t, a=9t^2

parmenides51 wrote: Determine all pairs $(a, b)$ of positive integers for which $\frac{a^2b+b}{ab^2+9}$ is an integer number. I claim that $(a,b)=(9t^2,9t)$ where $t \in \mathbb{N}$ $ab^2+9 | a^2b+b$ $ab^2+9 | a^2b^2+b^2$ $ab^2+9 | a^2b^2+9a-9a+b^2$ $ab^2+9 | b^2-9a$ But $b^2-9a<ab^2+9$ So we have, $b^2-9a=0$ Thus, $b=9t, a=9t^2$ as mentioned earlier.

MathInfinite wrote: parmenides51 wrote: Determine all pairs $(a, b)$ of positive integers for which $\frac{a^2b+b}{ab^2+9}$ is an integer number. I claim that $(a,b)=(9t^2,9t)$ where $t \in \mathbb{N}$ $ab^2+9 | a^2b+b$ $ab^2+9 | a^2b^2+b^2$ $ab^2+9 | a^2b^2+9a-9a+b^2$ $ab^2+9 | b^2-9a$ But $b^2-9a<ab^2+9$ So we have, $b^2-9a=0$ Thus, $b=9t, a=9t^2$ as mentioned earlier. $(a,b)=(32,1),(73,1),(22,2)$ also satisfies the equation