17th USAMO 1988

Let p(x) be the polynomial (1 - x)^a (1 - x²)^b (1 - x³)^c ... (1 - x³²)^k, where a, b, ..., k are integers. When expanded in powers of x, the coefficient of x¹ is -2 and the coefficients of x², x³, ... , x³² are all zero. Find k.

Solution

Answer: 2²⁷ - 2¹¹.

We have p(x) = 1 - 2x + O(x³³). Hence p(-x) = 1 + 2x + O(x³³). Multiplying p(x)p(-x) = 1 - 2²x² + O(x³³). Now p(x) p(-x) cannot have any odd terms, so we can write it as a polynomial in x², q(x²). Hence q(x²) = 1 - 2²x² + O(x³⁴). Similarly, r(x⁴) = q(x²) q(-x²) = 1 - 2⁴x⁴ + O(x³⁶), s(x⁸) = r(x⁴) r(-x⁴) = 1 - 2⁸x⁸ + O(x⁴⁰), and t(x¹⁶) = 1 - 2¹⁶x¹⁶ + O(x⁴⁸).

Now go back to the definition of p(x). When we take p(x) p(-x), the term (1 - x)^a becomes (1 - x²)^a. All the even terms just double their exponent, so (1 - x²)^b becomes (1 - x²)^2b, (1 - x⁴)^d becomes (1 - x⁴)^2d and so on. The odd terms all keep the same exponent, so (1 - x³)^c becomes (1 - x⁶)^c and so on. Thus we get t(x¹⁶) = (1 - x¹⁶)ⁿ(1 - x³²)^16k ... . The first exponent is a sum of several exponents from p(x), but the details are unimportant. We know that t(x¹⁶) = 1 - 2¹⁶x¹⁶ + O(x⁴⁸). The x¹⁶ term can only come from (1 - x¹⁶)ⁿ, so n = 2¹⁶. Now there is no x³² term, so putting N = 2¹⁶ we have NC2 = 16k, were NC2 is the binomial coefficient N(N -1)/2 = 2³¹ - 2¹⁵. Hence k = 2²⁷ - 2¹¹.

17th USAMO 1988

© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002