Hard work pays off

Let $\space x^4 - nx + 63 = \begin{cases} (x+a)(x^3+bx^2+cx + d) & ...(1) \\(x^2+ax+b)(x^2+cx+d) & ...(2) \end{cases}$

Case 1:

$\begin{aligned} x^4 - nx + 63 & = (x+a)(x^3+bx^2+cx + d) \\ & = x^4 + (a+b) x^3 + (ab+c) x^2 +(ac+d)x + ad \end{aligned}$

Equating the coefficients:

$\begin{cases} a+b = 0 & \Rightarrow b = -a \\ ab + c = 0 & \Rightarrow c = a^2 \\ ac+d = -n & \Rightarrow n = - a^3 - d \\ ad = 63 & \Rightarrow n = - a^3 -\dfrac{63}{a} \end{cases}$

We note that $a$ must be divisor of $63$ and that $n < 0$ for $a > 0$ . Therefore, the possible $n$ are:

\(\begin{array} {} a= -1 & \Rightarrow n = 64 & a = -3 & \Rightarrow n = 48 & a = -7 & \Rightarrow n = 352 \\ a = -9 & \Rightarrow n =736 & a = -21 & \Rightarrow n = 9264 & a = -63 & \Rightarrow n = 250048 \end{array} \)

Case 2:

$\begin{aligned} x^4 - nx + 63 & = (x^2+ax+b)(x^2+cx+d) \\ & = x^4+(a+c)x^3 + (b + d + ac)x^2 + (ad+bc)x +bd \end{aligned}$

Equating the coefficients:

$\begin{cases} a+c = 0 & \Rightarrow c = -a \\ b + d + ac = 0 & \Rightarrow a^2 = b + d \\ ad+bc = -n & \Rightarrow n = a(b-d) \text{ for } b > d \\ bd = 63 \end{cases}$

$bd = 63 \Rightarrow \begin{cases} b = 63 & d = 1 & \Rightarrow a = \sqrt{64} = 8 & n = 8(63-1) = 496 \\ b = 21 & d = 3 & \Rightarrow \color{#D61F06}{a = \sqrt{24}} & \color{#D61F06}{\text{Rejected: non-integer n}} \\ b = 9 & d = 7 & \Rightarrow a = \sqrt{16} = 4 & n = 4(9-7) = 8 \end{cases}$

Therefore, the smallest positive integer $n = \boxed{8}$