Find all integers n for which x^{2} - x + n divides x^{13} + 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 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.

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002