p(x) is a polynomial with integral coefficients. The leading coefficient, the constant term, and p(1) are all odd. Show that p(x) has no rational roots.

**Solution**

*Easy.*

Suppose there is a rational root p/q. Let p(x) = ∑_{0}^{n}a_{r}x^{r}. Then ∑a_{r}p^{r}q^{n-r} = 0. So p divides a_{0} and hence is odd, and q divides a_{n} and must also be odd. But now ∑a_{r}p^{r}q^{n-r} = a_{r} = 1 (mod 2). Contradiction.

© John Scholes

jscholes@kalva.demon.co.uk

11 Nov 1999