12th Putnam 1952

Problem A1

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.




Suppose there is a rational root p/q. Let p(x) = ∑0narxr. Then ∑arprqn-r = 0. So p divides a0 and hence is odd, and q divides an and must also be odd. But now ∑arprqn-r = ar = 1 (mod 2). Contradiction.



12th Putnam 1952

© John Scholes
11 Nov 1999