For what positive integers n does the polynomial p(x) ≡ xn + (2 + x)n + (2 - x)n have a rational root.
Answer: 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 rn + (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 tn + (s + t)n + (s - t)n = 0. The same argument shows that t must be even. But now we have tn + 2(sn + nC2 sn-2 t2 + ... ) = 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
26 Nov 1999