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)