17th USAMO 1988 Problem 5 Let p(x) be the polynomial (1 - x) a (1 - x 2 ) b (1 - x 3 ) c ... (1 - x 32 ) k , where a, b, ..., k are integers. When expanded in powers of x, the coefficient of x 1 is -2 and the coefficients of x 2 , x 3 , ... , x 32 are all zero. Find k. Solution
: 2 27 - 2 11 . We have p(x) = 1 - 2x + O(x 33 ). Hence p(-x) = 1 + 2x + O(x 33 ). Multiplying p(x)p(-x) = 1 - 2 2 x 2 + O(x 33 ). Now p(x) p(-x) cannot have any odd terms, so we can write it as a polynomial in x 2 , q(x 2 ). Hence q(x 2 ) = 1 - 2 2 x 2 + O(x 34 ). Similarly, r(x 4 ) = q(x 2 ) q(-x 2 ) = 1 - 2 4 x 4 + O(x 36 ), s(x 8 ) = r(x 4 ) r(-x 4 ) = 1 - 2 8 x 8 + O(x 40 ), and t(x 16 ) = 1 - 2 16 x 16 + O(x 48 ). Now go back to the definition of p(x). When we take p(x) p(-x), the term (1 - x) a becomes (1 - x 2 ) a . All the even terms just double their exponent, so (1 - x 2 ) b becomes (1 - x 2 ) 2b , (1 - x 4 ) d becomes (1 - x 4 ) 2d and so on. The odd terms all keep the same exponent, so (1 - x 3 ) c becomes (1 - x 6 ) c and so on. Thus we get t(x 16 ) = (1 - x 16 ) n (1 - x 32 ) 16k ... . The first exponent is a sum of several exponents from p(x), but the details are unimportant. We know that t(x 16 ) = 1 - 2 16 x 16 + O(x 48 ). The x 16 term can only come from (1 - x 16 ) n , so n = 2 16 . Now there is no x 32 term, so putting N = 2 16 we have NC2 = 16k, were NC2 is the binomial coefficient N(N -1)/2 = 2 31 - 2 15 . Hence k = 2 27 - 2 11 . 17th USAMO 1988 © John Scholes jscholes@kalva.demon.co.uk 21 Aug 2002