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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 $b_{2}=36$.

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    $\begingroup$ I think I'd phrase the question as, find three integers $u,v,n$ such that $4uv+(6n-1)(u+v)+8n^2-4n$ divides $-(2n-1)uv-(2n^2-3n)(u+v)+8n^2$. $\endgroup$
    – Gerry Myerson
    Commented Jan 8, 2018 at 16:32
  • $\begingroup$ @GerryMyerson: I think the OP wants a pair $(u,v)$ for any integer $n\geq 1$. This is why he denoted it by $(u_n,v_n)$. $\endgroup$
    – GH from MO
    Commented Jan 8, 2018 at 16:33
  • $\begingroup$ @GHf, you may be right, I was just guessing at what OP wanted. $\endgroup$
    – Gerry Myerson
    Commented Jan 8, 2018 at 16:34
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    $\begingroup$ @GerryMyerson: Well, he gave an example for $n=2$, see the last sentence. $\endgroup$
    – GH from MO
    Commented Jan 8, 2018 at 16:43

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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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  • $\begingroup$ @ Max Alekseyev: But you then get $a_{n}= 2(n+1)(2n+1)^{3} $ and $b_{n }= -(n-1)(n+1)(2n+1)^{3}$ and there is no possibility of division unless $n$ is odd. $\endgroup$
    – Safwane
    Commented Jan 9, 2018 at 13:06
  • $\begingroup$ @ZeraouliaElhadj: Indeed, $a_n=Q$ guarantee only that $a_n\mid 4b_n$, and divisibility $a_n\mid b_n$ in this case happens only for odd $n$. For even $n$, we have $Q/a_n\equiv 3\pmod{4}$, and thus we can similarly require $3a_n=Q$ instead of $a_n=Q$. $\endgroup$
    – Max Alekseyev
    Commented Jan 9, 2018 at 15:17
  • $\begingroup$ @ Max Alekseyev: It seems that the property of division does not holds true for all $n$ unless one requires some congruance. $\endgroup$
    – Safwane
    Commented Jan 9, 2018 at 15:40
  • $\begingroup$ @ZeraouliaElhadj: See update. $\endgroup$
    – Max Alekseyev
    Commented Jan 9, 2018 at 18:25

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