For what positive integers n does the polynomial p(x) ≡ x^{n} + (2 + x)^{n} + (2 - x)^{n} have a rational root.

**Solution**

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 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.

© John Scholes

jscholes@kalva.demon.co.uk

26 Nov 1999