Show that for every positive integer \(n\), there exists an irreducible polynomial over \(\mathbb Q\) of degree \(n\) all of whose roots are real.
Solution
For \(n\ge1\), define \[ F_n(x)\coloneqq\prod_{k=1}^n(x-4k), \qquad G_n(x)\coloneqq F_n(x)+2. \]
1. The polynomial \(G_n\in\ZZ[x]\) has degree \(n\) and isirreducible over \(\QQ\).
The polynomial \(F_n\) is monic of degree \(n\). Every nonleading coefficient of \(F_n\) is divisible by \(4\), because every elementary symmetric polynomial of positive degree in \(4,8,\ldots,4n\) is divisible by \(4\). Hence every nonleading coefficient of \(G_n\) is divisible by \(2\).
The constant term of \(G_n\) is \[ 2+(-4)^n n!, \] which is congruent to \(2\) modulo \(4\). Thus its constant term is not divisible by \(4\), while the leading coefficient is \(1\) and is not divisible by \(2\). Eisenstein’s criterion at the prime \(2\) shows that \(G_n\) is irreducible over \(\QQ\).
2. For \(n=1\), the polynomial \(G_1\) has one real root.
Here \[ G_1(x)=2+(x-4)=x-2. \] Its unique root is \(2\).
3. Suppose \(n\ge2\). For \(1\le j\le n-1\), set\[ x_j\coloneqq4\left(j+\frac12\right)=4j+2. \]Then\[ \operatorname{sgn}G_n(x_j)=(-1)^{n-j}. \]
Directly, \[ F_n(x_j) =4^n\prod_{k=1}^n\left(j+\frac12-k\right). \] Exactly \(n-j\) factors in the product are negative, so \[ \operatorname{sgn}F_n(x_j)=(-1)^{n-j}. \] The two factors nearest zero have absolute value \(1/2\), while every other factor has absolute value at least \(3/2\). Therefore \[ \abs{F_n(x_j)} \ge \frac{4^n}{4} =4^{n-1} >2. \] Since \(G_n(x_j)=F_n(x_j)+2\), adding \(2\) cannot change the sign, which proves the claim.
4. For \(n\ge2\), the polynomial \(G_n\) has at least one real root ineach of the \(n\) disjoint intervals\[ (-\infty,x_1),\quad (x_1,x_2),\quad\ldots,\quad (x_{n-2},x_{n-1}),\quad (x_{n-1},\infty). \]
Step 3 shows that the signs at consecutive \(x_j\) alternate, so the intermediate value theorem gives a root in every interval \((x_j,x_{j+1})\).
Because \(G_n\) is monic of degree \(n\), \[ \lim_{x\to\infty}G_n(x)=+\infty, \qquad \operatorname{sgn}G_n(x)=(-1)^n \quad\text{for all sufficiently negative }x. \] Step 3 gives \(\operatorname{sgn}G_n(x_1)=(-1)^{n-1}\) and \(\operatorname{sgn}G_n(x_{n-1})=-1\). Thus the intermediate value theorem also gives a root in \((-\infty,x_1)\) and a root in \((x_{n-1},\infty)\).
5. All \(n\) roots of \(G_n\) are real.
2. For \(n\ge2\), step 4 gives \(n\)distinct real roots because the listed intervals are disjoint. Since\(G_n\) has degree \(n\), it has no further roots over \(\CC\).
6. For every positive integer \(n\), the polynomial\[ \boxed{G_n(x)=2+\prod_{k=1}^n(x-4k)} \]is irreducible over \(\QQ\), has degree \(n\), and has only real roots.
Irreducibility and degree are step 1, and reality of all roots is step 5.
7. Q.E.D.
Step 6 supplies the required polynomial for every \(n\ge1\).