Answer to Question #348073 in Discrete Mathematics for Api

Question

The set S consists of all integers from 1 to 2007 inclusive. For how many

elements n in S is f(n)=2n³+n²-n-2 over n²-2 an integer?

Answer 1

$$\dfrac{2x^3+x^2-x-2}{x^2-2}$$

$$=\dfrac{2x(x^2-2)+(x^2-2)+3x}{x^2-2}$$

$$=2x+1+\dfrac{3x}{x^2-2}$$

Given $n\in \Z^+$

$n=1, 2(1)+1+\dfrac{3(1)}{(1)^2-2}=0, 0\in \Z$

$$n\ge 2, \dfrac{3x}{x^2-2}\ge1$$

$$\dfrac{x^2-3x-2}{x^2-2}\le0$$

$$\dfrac{x^2-3x-2}{x^2-2}\le0$$

$$x^2-3x-2=0$$

$$x_1=\dfrac{3-\sqrt{17}}{2}, x_2=\dfrac{3+\sqrt{17}}{2}$$

$$x\in(-\sqrt{2}, \dfrac{3-\sqrt{17}}{2}]\cup (\sqrt{2}, \dfrac{3+\sqrt{17}}{2}]$$

Since $n$ is positive integer, we consider $n=2,3.$

$$n=2,\dfrac{3n}{n^2-2}=\dfrac{3(2)}{(2)^2-2}=3, integer$$

$$n=3,\dfrac{3n}{n^2-2}=\dfrac{3(3)}{(3)^2-2}=\dfrac{9}{7}, is\ not \ integer$$

$n=1,n=3.$