MHB Find the sum of all values of positive integer a

[anemone]

For a pair of positive integers $(a,\,b)$, $Q(a,\,b)$ is defined by

$Q(a,\,b)=\dfrac{a^2b+2ab^2-5}{ab+1}$.

Let $(a_1,\,b_1),\,(a_2,\,b_2),\,\cdots, (a_n,\,b_n)$ be all pairs of positive integers such that $Q(a,\,b)$ is an integer. Calculate $\displaystyle \sum_{i=1}^n a_i$,

 

[Opalg]

Spoiler

If $ab+1$ divides $a^2b + 2ab^2 - 5$ then it also divides $(a+2b)(ab+1) - (a^2b + 2ab^2 - 5) = a+2b+5$.

So suppose that $a+2b+5 = k(ab+1)$ for a positive integer $k$. Then $k^2ab - ka - 2kb = 5k - k^2$. Therefore $$(ka-2)(kb-1) = 2 + 5k - k^2.$$ If $k=1$ then $(a-2)(b-1) = 6$. The four possible factorisations of $6$ give solutions $(a,b) = (3,7),\, (4,4),\, (5,3),\, (8,2)$.

If $k=2$ then $(2a-2)(2b-1) = 8$, or $(a-1)(2b-1) = 4$, giving only one solution $(a,b) = (5,1)$ (because $2b-1$ must be odd).

If $k=3$ then $(3a-2)(3b-1) = 8$, giving solutions $(1,3)$ and $(2,1)$.

If $k=4$ then $(4a-2)(4b-1) = 6$, or $(2a-1)(4b-1) = 3$, giving the solution $(1,1)$.

If $k=5$ then $(5a-2)(5b-1) = 2$, which has no solutions in positive integers.

If $k\geqslant6$ then $2+5k-k^2$ is negative, so there can be no more solutions.

So in total there are eight pairs of positive integers for which $Q(a,b)$ is an integer, namely $$(a,b) = (1,1),\ (1,3),\ (2,1),\ (3,7),\ (4,4),\ (5,1),\ (5,3),\ (8,2).$$ The sum of their $a$-coordinates is $\displaystyle\sum_{i=1}^8 a_i = 1+1+2 +3 +4 +5 +5 +8 = 29.$