<s>[INST] A problem of divisibility 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 $b_{2}=36$.
 [/INST] 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 $b_n=n^4(2n+1)^2$.
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