If \(a,b\) are odd integers, prove that \[ f(x)=x^3+ax+b \] has no rational roots.
Solution
1. Any rational root of \(f\) would in fact be an integer.
The polynomial \[ f(x)=x^3+ax+b \] is monic with integer coefficients. By the rational root theorem, a rational root of a monic integer polynomial must be an integer.
2. Modulo \(2\), the polynomial \(f\) becomes\[ \overline f(x)=x^3+x+1 \]in \(\FF_2[x]\).
Because \(a\) and \(b\) are odd, \[ a\equiv1\pmod2, \qquad b\equiv1\pmod2. \] Reducing every coefficient of \(f\) modulo \(2\) therefore gives the displayed polynomial.
3. The polynomial\[ x^3+x+1 \]has no root in \(\FF_2\).
The field \(\FF_2\) has only the elements \(0\) and \(1\). Direct evaluation gives \[ 0^3+0+1=1 \] and \[ 1^3+1+1=1 \] in \(\FF_2\). Hence neither element is a root.
4. The polynomial \(f\) has no rational root.
Suppose \(r\in\QQ\) were a root. By step 1, \(r\in\ZZ\). Reducing the identity \[ f(r)=0 \] modulo \(2\) would make the residue class of \(r\) a root of \(\overline f\) in \(\FF_2\), contradicting step 3.
5. Q.E.D.
Step 4 is the required conclusion.