37th Putnam 1976 Problem A4 Let p(x) ≡ x 3 + ax 2 + bx - 1, and q(x) ≡ x 3 + cx 2 + dx + 1 be polynomials with integer coefficients. Let α be a root of p(x) = 0. p(x) is irreducible over the rationals. α + 1 is a root of q(x) = 0. Find an expression for another root of p(x) = 0 in terms of α, but not involving a, b, c, or d. Solution
: the other roots are -1/(α+1) and - (α+1)/α. Let the roots of p(x) be α, β, γ. The polynomial q(x+1) has α as one of its roots. Suppose it is different from p(x). Then by subtracting we get either a quadratic or a linear equation which also has α as a root. It cannot be a linear equation, because then α would be rational and hence p(x) would not be irreducible over the rationals. If it was quadratic, then by multiplying it by a suitable rational factor rx + s and subtracting from p(x) we would get a linear factor ux + v which also had α as a root. If this factor is zero, then we have reduced p(x) over the rationals (as (rx + s) times the quadratic). If not, then α is a root of ux + v and hence rational, so that p(x) is still reducible. So we must have that p(x) and q(x+1) are identical. So the three roots of q(x) must be α+1, β+1, γ+1. Hence we have the relations: αβγ = 1, (α+1)(β+1)(γ+1) = -1 (derived from the constant terms of p(x) and q(x) respectively). Hence α(β+γ)+(β+γ)+α+1+βγ+αβγ = -1, so (β+γ)(α+1) = - (3+α+1/α), or β+γ = - (3α+α 2 +1) / ( α(α+1) ). Hence (β-γ) 2 = (β+γ) 2 - 4βγ = (α 4 +2α 3 +3α 2 +2α+1)/ ( α 2 (α+1) 2 ). So β-γ = ± (α 2 +α+1)/ ( α(α+1) ) and hence β = -1/(α+1) or - (α+1)/α. 37th Putnam 1976 © John Scholes jscholes@kalva.demon.co.uk 23 Jan 2001