**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)

## One Reply to “Integer x (CMI Entrance 2014 solutions)”