I came across the following problem. Find two integers, $u\_{n}$ and $v\_{n}$, such that 


$$a\_{n}=4u\_{n}v\_{n}+(6n-1)v\_{n}+(6n-1)u\_{n}+8n^{2}-4n$$
divides 
$$b\_{n}=-(2n-1)u\_{n}v\_{n}-(2n^{2}-3n)v\_{n}-(2n^{2}-3n)u\_{n}+8n^{2}.$$


For example, for $n=2$, the choices $u\_{2}:=-2$ and $v\_{2}:=0$ yield $a\_{2}=2$ and $b\_{2}=36$.



A problem of divisibility
If $a\_n\mid b\_n$, then $a\_n$ also divides
$$Q := (2n-1)a\_n + 4b\_n = (2n+1)^2 (u\_n + v\_n+4n).$$


We will blatantly require $a\_n = Q$. Notice that
$$a\_n-Q = 4u\_nv\_n - 2(2n^2-n+1)u\_n - 2(2n^2-n+1)u\_n - 8n(2n^2+n+1) = (2u\_n-(2n^2-n+1))(2v\_n-(2n^2-n+1)) - (n+1)^2(2n+1)^2.$$ 
Thus, it's enough to take $d\mid (n+1)^2(2n+1)^2$ and set
$$u\_n = \frac{1}{2}(d + 2n^2-n+1),$$
$$v\_n = \frac{1}{2}(\frac{(n+1)^2(2n+1)^2}{d} + 2n^2-n+1).$$


In particular, we can always take $d=(n+1)(2n+1)$ and get
$$u\_n = v\_n = 2n^2+n+1.$$


**UPDATE**. As pointed out in the comments, the above construction works only for odd $n$. For an even $n$, we have $Q/a\_n\equiv 3\pmod{4}$ and thus we can require that $a\_n = -Q$. Then
$$0 = a\_n+Q = 2u\_nv\_n + 2(2n^2+5n)u\_n + 2(2n^2+5n)u\_n + 8n^2(2n+3) = 
(2u\_n + (2n^2+5n))(2v\_n + (2n^2+5n)) + n^2(2n+1)^2.$$
Again, let $d\mid \frac{n^2(2n+1)^2}4$ and set
$$u\_n = d - \frac{2n^2+5n}2,$$
$$v\_n = \frac{n^2(2n+1)^2}{4d} - \frac{2n^2+5n}2.$$
E.g., taking $d=\frac{n}{2}$, we get 
$$u\_n = -n^2-2n\qquad\text{and}\qquad v\_n=n(2n^2+n-2),$$
for which $a\_n = -2n^3(2n+1)^2$ divides $b\_n=n^4(2n+1)^2$.


