12th Putnam 1952

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) = ∑₀ⁿa_rx^r. Then ∑a_rp^rq^n-r = 0. So p divides a₀ and hence is odd, and q divides a_n and must also be odd. But now ∑a_rp^rq^n-r = a_r = 1 (mod 2). Contradiction.

12th Putnam 1952

© John Scholes
jscholes@kalva.demon.co.uk
11 Nov 1999