67th Romanian Mathematical Olympiad

Question

a) Prove that, for every integer k, the equation x^3 - 24x + k = 0 has at most one integer solution. b) Prove that the equation x^3 + 24x - 2016 = 0 has exactly one integer solution.

Accepted Answer

a) Suppose that there exist two different integers

m

and

n

so that

m^{3} - 24 m + k = 0

and

n^{3} - 24 n + k = 0

. Subtracting the above yields

(m - n) (m^{2} + mn + n^{2} - 24) = 0

. Since

m

and

n

are different,

m^{2} + mn + n^{2} = 24

, whence

(2 m + n)^{2} + 3 n^{2} = 96

. Therefore

n^{2} \leq 32

, hence

n^{2} \in {0, 1, 4, 9, 16, 25}

. This leads to

(2 m + n)^{2} \in {96, 93, 84, 69, 48, 21}

, a contradiction.

b) The equation is

x (x^{2} + 24) = 2016

, so

x

must be a positive integer;

x = 12

is a solution. If

x < y

are positive solutions, then

x^{2} + 24 < y^{2} + 24

and

2016 = x (x^{2} + 24) < y (y^{2} + 24) = 2016

, a contradiction.