2010 AIME II Problem 6

Try splitting into two quadratics $(x^2 + px + q)(x^2 - px + r);$ matching coefficients gives $q + r = p^2$ and $qr = 63$

Look for factor pairs of $63$ whose sum is a perfect square, and also compare with the case of a linear factor, whose root must divide $63$

If there is a linear factor, then some integer $b$ is a root, so $b^4 - nb + 63 = 0$ and $n = b^3 + \frac{63}{b},$ forcing $63$ to be divisible by $b$ and $b \gt 0.$ The smallest value is $48,$ at $b = 3.$

Otherwise the polynomial splits into two quadratics, which we may take monic; since the $x^3$ coefficient vanishes, they have the form $$\begin{aligned} &(x^2 + px + q)(x^2 - px + r) \\ &= x^4 + (q + r - p^2)x^2 \\ &\quad {}+ p(r - q)x + qr. \end{aligned}$$ Matching coefficients gives $q + r = p^2,$ $qr = 63,$ and $n = p(q - r).$ The factor pairs of $63$ with square sum are $\{7, 9\}$ (sum $16,$ so $p = 4$) and $\{1, 63\}$ (sum $64,$ so $p = 8$), giving $n = 4 \cdot 2 = 8$ or $n = 8 \cdot 62 = 496.$

The smallest positive value overall is $n = 8;$ indeed $(x^2 + 4x + 9)(x^2 - 4x + 7)$ $= x^4 - 8x + 63.$