Let $\mathbf{ x \in \mathbb{R} , x^{2014} - x^{2004} , x^{2009} - x^{2004} in \mathbb{Z} }$ . Then show that x is an integer. (Hint: First show that x is a rational number)

Discussion: $\mathbf{ x^{2014} - x^{2004} - x^{2009} + x^{2004} = x^{2014} - x^{2009} = x^{2009}(x^{5} - 1 ) }$ is an integer $\mathbf{x^{2009} - x^{2004} = x^{2004}(x^5 - 1) }$ is also an integer.
Hence ratio of those two are $\mathbf{ \frac{ x^{2009}(x^5 -1)}{x^{2004}(x^5 - 1)} = x^5 }$ is a rational.
Again $\mathbf{(x^5)^{400} = x^{2000}, x^5 - 1 }$ are rational. Also it is given that $\mathbf{ x^{2000}\cdot x^4 cdot (x^5 - 1) }$ is integer. Hence $\mathbf{ x^4 }$ is rational.
Therefore $\mathbf{ \frac{x^5}{x^4} =x }$ is rational. Since ratio of rationals is rational.

Suppose x = p/q, then $\mathbf{ \frac{p^{2014}}{q^{2014}} - \frac{p^{2004}}{q^{2004}} = k }$ where gcd(p, q) = 1 and k is an integer. $\mathbf{ p^{2014} - p^{2004}q^{10} = kq^{2014} }$
But this implies q divides p which means q = 1.

Hence x is an integer. (Proved)