Find all real polynomials p(x) such that x is rational iff p(x) is rational.
Solution
It is necessary for all the coefficients of x to be rational. This is an easy induction on the number of coefficients. For p(0) must be rational, hence ( p(x) - p(0) )/x must be rational for all rational x and so on.
Clearly this condition is also sufficient for polynomials of degree 0 or 1.
There are obviously no quadratics, for if p(x) = ax2 + bx + c, with a, b, c rational, then p(√2 - b/2a) = 2a - b2/4a + c, which is rational.
We prove that if there are also no higher degree polynomials. The idea is to show that there is a rational value k which must be taken for some real x, but which cannot be taken by any rational x.
Suppose p(x) has degree n > 1. Multiplying through by the lcm of the denominators of the coefficients, we get p(x) = (a xn + b xn-1 + ... + u x + v)/w, where a, b, ... , w are all integers. Put x = r/s, where r and s are coprime integers, then p(r/s) = (a rn + b rn-1s + ... + u r sn-1 + v sn)/( w sn). Let q be any prime which does not divide a or w. Consider first a > 0. p(x) must assume all sufficiently large positive values. So it must in particular take the value k = m + 1/q, where m is a sufficiently large integer. So k = (mq + 1)/q. The denominator is divisible by q, but not q2 and the numerator is not divisible by q. Suppose p(r/s) = k for some integers r, s. The denominator of p(r/s) is w sn. We know that w is not divisible by q, so q must divide s. But n > 1, so q2 divides w sn. The numerator of p(r/s) has the form a rn + h s. Neither a nor r is divisible by q, so the numerator is not divisible by q. Thus no cancellation is possible and we cannot have p(r/s) = k. Thus there must be some irrational x such that p(x) = k.
If a < 0, then the same argument works except that we take k = m + 1/q, where m is a sufficiently large negative integer.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002