Instruction: Find the smallest positive integer $n$ with the property that the polynomial $x^4 - nx + 63$ can be written as a product of two nonconstant polynomials with integer coefficients.

Output:
The given polynomial has degree $4,$ so either it is the product of a linear term and a cubic term, or it is the product of two quadratic terms. Also, we may assume that both terms have leading coefficient $1.$

In the first case, the linear term must be of the form $x-a,$ so the polynomial must have an integer root $a.$ That is, $a^4-na + 63 = 0$ for some integer $a.$ Since $n > 0,$ this is impossible when $a \le 0,$ so we must have $a > 0.$ Then \[n = \frac{a^4+63}{a} = a^3 + \frac{63}{a}.\]Testing various positive divisors of $63,$ we see that $n$ is minimized for $a=3,$ giving $n = 3^3 + \frac{63}{3} = 27 + 21 = 48.$

In the second case, let \[x^4 - nx + 63 = (x^2+ax+b)(x^2+cx+d)\]for some integers $a, b, c, d.$ Comparing the $x^3$ coefficients on both sides shows that $a+c=0,$ so $c=-a.$ Then, comparing the $x^2$ coefficients, we get \[b + ac + d = 0 \implies b + d = a^2.\]We also have $bd = 63,$ looking at the constant terms. The only possibilities for $(b, d)$ are $(b, d) = (1, 63), (7, 9).$ Then the corresponding values of $a$ are $a =\pm 8, \pm4,$ giving $n = \pm 496, \pm 8.$ Therefore the least value for $n$ is $\boxed{8}.$


Instruction: The parabola with equation $y=ax^2+bx+c$ and vertex $(h,k)$ is reflected about the line $y=k$. This results in the parabola with equation $y=dx^2+ex+f$.  Express $a+b+c+d+e+f$ in terms of $k.$

Output:
The equation of the original parabola can be written as
\[y = a(x - h)^2 + k.\]The equation of the reflected parabola is then
\[y = -a(x - h)^2 + k.\]Thus,
\[ax^2 + bx + c + dx^2 + ex + f = 2k.\]Setting $x = 1,$ we get $a + b + c + d + e + f = \boxed{2k}.$