Find all integers n for which x2 - x + n divides x13 + x + 90.
Solution
Answer: 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 x2 - x + n divides x13 + x + 90, then x13 + x + 90 = p(x) (x2 - 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 213 + 92. But 2odd is congruent to 2 mod 2, so 213 + 92 is congruent to 1 mod 3. So n cannot be 1.
To see that n = 2 is possible, we write explicitly: (x2 - x + 2) (x11 + x10 - x9 - 3 x8 - x7 + 5 x6 + 7 x5 - 3 x4 - 17 x3 - 11 x2 + 23 x + 45) = x13 + x + 90.
© John Scholes
jscholes@kalva.demon.co.uk
5 Feb 2002