62nd Putnam 2001 Problem A3 For what integers n is the polynomial x 4 - (2n + 4) x 2 + (n - 2) 2 the product of two non-trivial polynomials with integer coefficients? Solution
: n = square or twice a square. We can write the polynomial as (x 2 - (n - 2)) 2 - 8x 2 . This cannot have any integral solutions because then we would have A 2 = 8 B 2 , and the lhs has an even number of powers of 2 and the rhs an odd number. So if the polynomial factorises, it must do so as (x 2 + ax + b)(x 2 + Ax + B). Comparing terms in x 3 , A = - a. There are now two cases: (1) a is non-zero, (2) a is zero. In case (1), comparing terms in x, we deduce B = b. Comparing constant terms, gives b = n - 2 or 2 - n. Comparing terms in x 2 gives a 2 = 2n + 4 + 2b. We cannot have b = 2 - n, for then a 2 = 8, contradicting a integral. Hence, b = n - 2 and a 2 = 4n. So n must be a perfect square. Put n = k 2 , then we can easily check that the polynomial factorises: x 4 - (2k 2 +4)x 2 + (k 2 - 2) 2 = (x 2 +2kx + k 2 - 2)(x 2 -2kx + k 2 - 2). In case (2), we have b + B = -(2n + 4), bB = (n - 2) 2 . Hence b(-b - 2n - 4) = (n - 2) 2 , so b 2 + b(2n + 4) + (n - 2) 2 = 0. Hence (b + n + 2) 2 = 8n. So n must be 2k 2 for some integer k. It is easy to check that in that case the polynomial factorises: x 4 - (4k 2 +4)x 2 + (2k 2 - 2) 2 = (x 2 - 2k 2 - 4k - 2)(x 2 - 2k 2 + 4k - 2). 62nd Putnam 2001 © John Scholes jscholes@kalva.demon.co.uk 16 Dec 2001