34th Putnam 1973

Problem B5

If x is a solution of the quadratic ax2 + bx + c = 0, show that, for any n, we can find polynomials p and q with rational coefficients such that x = p(xn, a, b, c) / q(xn, a, b, c). Hence or otherwise find polynomials r, s with rational coefficients so that x = r (x3, x + 1/x) / s(x3, x + 1/x).



The first part of this is somewhat confusing. Clearly we can find particular values of a, b, c for which we do not have x = p(xn, a, b, c) / q(xn, a, b, c). For example, take a = 1, b = 0, c = -2. Then x is irrational, whereas any p and q would have rational values.

One can carry out a careful analysis of what values of a, b, c are allowed, but that does not seem to be what the question is after. It seems to want something much simpler.

An easy induction shows that xn = (x p1(a, b, c) + p2(a, b, c) )/ p3(a, b, c), where p1, p2 and p3 are polynomials with integer coefficients. Just start with x2 = (- bx - c)/a and at each stage multiply through by x and substitute. Now p1 cannot be identically zero because then both solutions to the original quadratic would have the same |x|, whereas it is easy to see that this is not true for most values of a, b, c. So we may divide by p1 to get the required relation with p(xn, a, b, c) = xn p3(a, b, c) - p2(a, b, c) and q(xn, a, ,b c) = p1(a, b, c).

For the second part, it is simpler to start from scratch. (x + 1/x)2 = x2 + 1/x2 + 2. So x (x + 1/x)2 = x3 + 1/x + 2x. Hence x (x + 1/x)2 - x = x3 + (x + 1/x). That is essentially the relation we want: x = (x3 + y) / (y2 - 1), where y = x + 1/x.



34th Putnam 1973

© John Scholes
27 Jan 2001