15th Putnam 1955 Problem A6 For what positive integers n does the polynomial p(x) ≡ x n + (2 + x) n + (2 - x) n have a rational root. Solution
: n = 1. There are no roots at all if n is even. If n = 1, then -4 is the root. So suppose n is odd and at least 3. We can use a similar argument to that showing that √2 is irrational. Suppose x = r/s with r and s having no common factor is a root. Then r n + (2s + r) n + (2s - r) n = 0. The last two terms expand and add to give terms with even coefficients, so r must be even. So set r = 2t and we have t n + (s + t) n + (s - t) n = 0. The same argument shows that t must be even. But now we have t n + 2(s n + nC2 s n-2 t 2 + ... ) = 0. Since n ≥ 3, s must be even also. Contradiction. So there are no rational roots for n odd ≥ 3. 15th Putnam 1955 © John Scholes jscholes@kalva.demon.co.uk 26 Nov 1999