Show that a given family of polynomials is irreducible over ZZ

Let $p(x) = x^5 - tx - 1 \in \mathbb{Z}[x]$. Show that $p(x)$ is irreducible in $\mathbb{Z}[x]$ unless $t \in \{0,2,-1\}$.

Note that if $p(x)$ is reducible in $\mathbb{Z}[x]$, then it must have either a linear or a quadratic factor.

Suppose $p(x)$ has a linear factor in $\mathbb{Z}[x]$; say $p(x) = (x+a)(x^4+bx^3+cx^2+dx+e)$, with these coefficients in $\mathbb{Z}$. Comparing coefficients, we have the following system of equations: $a+b = 0$, $c+ab = 0$, $d+ac = 0$, $e+ad = -t$, $ae = -1$. Since $a$ and $e$ are integers, the last equation yields the two cases $(a,e) = (-1,1)$ and $(a,e) = (1,-1)$. Note also that $-a$ is a root of $p(x)$, so that $t = (a^5+1)/a$. We can see that in the first case, $t = 0$, and in the second, $t = 2$. Indeed, this yields the factorizations $x^5 - 1 = (x-1)(x^4+x^3+x^2+x+1)$ and $x^5-2x-1 = (x+1)(x^4-x^3+x^2-x-1)$. For no other $t$ does $p(x)$ have a linear factor.

Now suppose $p(x)$ has a quadratic factor in $\mathbb{Z}[x]$; say $p(x) = (x^2+ax+b)(x^3+cx^2+dx+e)$. Again comparing coefficients, we have $a+c = 0$, $b+ac+d = 0$, $e+ad+bc = 0$, $ae+bd = -t$, and $be = -1$. Substituting the first two equations into the third, we see that $a$ is a rational root of $x^3-2bx+e$.

If $(b,e) = (-1,1)$, then $a$ is a rational root of $q(x) = x^3+2x-1$; by the rational root test, $a = \pm 1$. Neither of these is a root of $q(x)$, a contradiction.

If $(b,e) = (1,-1)$, then $a$ is a rational root of $q(x) = x^3-2x-1$; by the rational root theorem, $a = -1$. Then $c = 1$ and $d = 0$, and we have $t = -1$. Indeed, this corresponds to the factorization $x^5+x-1 = (x^2-x+1)(x^3+x^2-1)$. For no other $t$ does $p(x)$ have a quadratic factor.