Proof. Put \( f_n(x) = x^3 + nx + 2 \), for \( n \in \mathbb{Z} \). For \( f_n(x) \) to be reducible, there must be a linear factor over \( \mathbb{Z} \); ie., \( f_n(x) = x^3 + nx + 2 = (x - \xi)(x^2 + ax + b) \), with \( \xi, a, b \in \mathbb{Z} \). Multiplying out the right side and equating coefficients for the powers of \( x \): \[ \begin{align*} x^2 &: \;\; 0 = a - \xi, \\ x &: \;\; n = b - a \xi, \\ 1 &: \;\; 2 = - \xi b. \end{align*} \] This reduces to \( a = \xi, \; n = b - a^2, \; ab = -2 \). The rational roots theorem says that the only possibilities for \( \xi \) are \( \xi = \pm 1, \pm 2 \), no matter the value of \( n \). Map out the values for \( a \) and \( b \) for each possible value of \( \xi \): \begin{array}{c|c} \xi & a & b & n = b - a^2 \\ \hline 1 & 1 & -2 & -3 \\ -1 & -1 & 2 & 1 \\ 2 & 2 & -1 & -5 \\ -2 & -2 & 1 & -3 \end{array} The three values for \( n \) in the right column lead to the exceptions: \( (x-1)(x+2) \; | \; f_{-3}(x), \; \) \( x+1 \; | \; f_{1}(x), \; \) \(x-2 \; | \; f_{-5}(x). \; \) \( f_n(x) = x^3 + nx + 2 \) is reducible for these three values of \( n \), irreducible otherwise. QED.