Let $a,b,c,d$ be pairwise relatively prime positive integers such that the divisibility relations $$ (kc-d) \mid \bigl((2k^2+1)b^2-2kab-(k^2-1)a^2\bigr) $$ and $$ (c-kd) \mid \bigl((k^2+2)b^2-2kab+(k^2-1)a^2\bigr) $$ hold for every nonzero integer $k$.
QUESTION: Can any of $a,b,c,d$ be determined or characterized (even partially) from this information alone?
If it helps, I’m trying to prove $c=d=1$.
\begin{cases} \begin{align} (2k^2+1)b^2-2kab-(k^2-1)a^2 &= (kc-d)q, \\ (k^2+2)b^2-2kab+(k^2-1)a^2 &= (c-kd)r. \end{align} \end{cases}
\begin{align} a &= ((k+1)s+(k-1)p)((k-1)p^2+2(k^3-k^2-1)ps+(k+1)(k^2+2)s^2), \\[0.5em] b &= ((k+1)s+(k-1)p)(-(k-1)^2p^2+2k(k-1)ps+k^2(k^2-1)s^2, \\[0.5em] c &= (2k^4-5k^3+7k^2-5k+1)p^2+(4k^4+2k^3-4k^2-2k-4)ps \\ &\hspace{2em} +(k^5+k^4+k^3+3k^2+4k+2)s^2, \\[0.5em] d &= (k^4-k^3+3k^2-5k+2)p^2+(6k^4-4k^3-2k^2-4)ps \\ &\hspace{2em} +(k^6-k^5-k^4+3k^3+4k^2+4k+2)s^2, \\[0.5em] q &= (k-1)^3p^4-2(2k+1)(k-1)^2sp^3-2k(k^2-1)(2k^2-2k-1)p^2s^2 \\ &\hspace{2em} -2(k^2-2)(k+1)^2ps^3+(k^4-2k^3-2)(k+1)^3s^4, \\[0.5em] r &= -((k-1)p^2-2kps+(k+1)(k^2+2)s^2)((k+1)s+(k-1)p)^2. \end{align}