Prove that there is such integer $a, b, c, d$ for prime number $p$ which satisfies $p|\Big(n^4-1-(n^2+an+b)(n^2+cn+d)\Big)$ for all $n \in \mathbb{N}$.
My attempt: \begin{align} &n^4-1-(n^2+an+b)(n^2+cn+d) = n^4-1-(n^4+cn^3+dn^2+an^3+acn^2+adn + bn^2+bcn+bd) \\ &= -(a+c)n^3-(ac+b+d)n^2-(ad+bc)n-bd-1 \equiv 0 (\mod p). \\ &\therefore (a+c)n^3+(ac+b+d)n^2+(ad+bc)n+bd \equiv -1 (\mod p). \end{align}
Since $n^4 - 1 = (n^2 - 1)(n^2 + 1)$, this suggests using $b = -1$ and $d = 1$. Along with $a = ep$ and $c = gp$ for any integers $e$ and $g$, so those terms are congruent to $0$ modulo $p$, this gives
$$\begin{equation}\begin{aligned} f(n) & = n^4 - 1 - (n^2 + an + b)(n^2 + cn + d) \\ & = n^4 - 1 - ((n^2 - 1) + epn)((n^2 + 1) + gpn) \\ & = n^4 - 1 - ((n^4 - 1) + gpn(n^2 - 1) + epn(n^2 + 1) + egp^2n^2) \\ & = p(-gn(n^2 - 1) - en(n^2 + 1) - egpn^2) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Somewhat more simply, using congruences, we get
$$\begin{equation}\begin{aligned} f(n) & \equiv n^4 - 1 - (n^2 + an + b)(n^2 + cn + d) \\ & \equiv n^4 - 1 - ((n^2 - 1) + epn)((n^2 + 1) + gpn) \\ & \equiv n^4 - 1 - (n^2 - 1)(n^2 + 1) \\ & \equiv 0 \pmod{p} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Both equations show $p \mid f(n)$ for all $n \in \mathbb{N}$. Note a basic case of $e = g = 0$ gives $f(n) = 0$.
Update: Similar to $a$ and $c$ being any integers congruent to $0$ modulo $p$, we could also have $b \equiv -1 \pmod{p} \; \to \; b = -1 + hp$, and $d \equiv 1 \pmod{p} \; \to \; d = 1 + ip$, for any integers $h$ and $i$. In addition, the $-1$ and $1$ between $b$ and $d$ could be switched. I didn't include this originally to keep the calculations somewhat shorter & simpler, especially for \eqref{eq1A}, plus the question only asks to show there's at least one solution of a set of integers instead of to determine all possible ones.