IMO 1991 SL 21

Origin: HKG

Problem

Let f(x) be a monic polynomial of degree 1991 with integer coefficients. Define g(x) = f 2(x) −9. Show that the number of distinct integer solutions of g(x) = 0 cannot exceed 1995.

Solution

The polynomial g(x) factorizes as g(x) = f(x)2 −9 = (f(x)−3)(f(x)+3). If one of the equations f(x) + 3 = 0 and f(x) −3 = 0 has no integer solutions, then the number of integer solutions of g(x) = 0 clearly does not exceed 1991. Suppose now that both f(x) + 3 = 0 and f(x) −3 = 0 have in- teger solutions. Let x1, . . . , xk be distinct integer solutions of the for- mer, and xk+1, . . . , xk+l be distinct integer solutions of the latter equa- tion. There exist monic polynomials p(x), q(x) with integer coefficients such that f(x) + 3 = (x −x1)(x −x2) . . . (x −xk)p(x) and f(x) −3 = (x −xk+1)(x −xk+2) . . . (x −xk+l)q(x). Thus we obtain (x−x1)(x−x2) . . . (x−xk)p(x)−(x−xk+1)(x−xk+2) . . . (x−xk+l)q(x) = 6. Putting x = xk+1 we get (xk+1 −x1)(xk+1 −x2) \cdot \cdot \cdot (xk+1 −xk) | 6, and since the product of more than four distinct integers cannot divide 6, this implies k \leq4. Similarly l \leq4; hence g(x) = 0 has at most 8 distinct integer solutions. Remark. The proposer provided a solution for the upper bound of 1995 roots which was essentially the same as that of (IMO74-6).