INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

May 16, 2014

Integer x | CMI Entrance 2014 solutions

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)

Some Useful Links:

ISI CMI Entrance Program

Limit of square roots – Video

One comment on “Integer x | CMI Entrance 2014 solutions”

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
enter