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.

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