24th Putnam 1963 Problem B1 Find all integers n for which x 2 - x + n divides x 13 + x + 90. Solution
: n = 2. If n is negative or zero, then the quadratic has two real roots. But we can easily check that the other polynomial has derivative everywhere positive and hence only one real root. So n must be positive. If x 2 - x + n divides x 13 + x + 90, then x 13 + x + 90 = p(x) (x 2 - x + n), where p(x) is a polynomial with integer coefficients. Putting x = 0, we see that n must divide 90. Putting x = 1, we see that it must divide 92. Hence it must divide (92 - 90) - 2. So the only possibilities are 1 and 2. Suppose n = 1. Then putting x = 2, we have that 3 divides 2 13 + 92. But 2 odd is congruent to 2 mod 2, so 2 13 + 92 is congruent to 1 mod 3. So n cannot be 1. To see that n = 2 is possible, we write explicitly: (x 2 - x + 2) (x 11 + x 10 - x 9 - 3 x 8 - x 7 + 5 x 6 + 7 x 5 - 3 x 4 - 17 x 3 - 11 x 2 + 23 x + 45) = x 13 + x + 90. 24th Putnam 1963 © John Scholes jscholes@kalva.demon.co.uk 5 Feb 2002