The polynomial p(x) = x1991 + a1990x1990 + ... + a0 has integer coefficients. Show that the equation p(x)2 = 9 has at most 1995 distinct integer solutions.
Solution
We want solutions to p(x) = ±3. We claim that if p(x) = 3 has an integer solution, then p(x) = -3 has at most 4 distinct integer solutions. For suppose a1, a2, a3, a4, a5 are distinct integer solutions to p(x) = 3. Then for some q(x) with integer coefficients we have p(x) - 3 = (x - a1)(x - a2)(x - a3)(x - a4)(x - a5)q(x). So if b is a solution to p(x) = -3, then p(b) = -3, so -6 = p(b) - 3 = (b - a1)(b - a2)(b - a3)(b - a4)(b - a5) q(b). But -6 cannot be written as the product of 5 distinct integers. (The best we can do is -6 = 1·(-1)·2·3). Contradiction.
It follows that there are at most 1991 distinct integer solutions to p(x) = ±3 (which is a stronger result than that in the question).
© John Scholes
jscholes@kalva.demon.co.uk
1 Jan 2003
Last corrected/updated 1 Jan 03