It's not a geometry problem this time ; - D

Let x be a real number such that $x^{ 2014} -x^{ 2004} and x^{ 2009 }-x^{ 2004}$ are both integers. Show that x is an integer.

Please post your solutions here as well. I'd be interested to read your methods.

#NumberTheory #VeryInteresting #Interesting #Cmi

Easy Math Editor

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Comments

Fixing the solution of Visnhu C.

Subtract the second from the first and you'll get that $x^5(x^{2009}-x^{2004})$ is an integer

We see that $x^5$ is a rational number. That implies $x^5 - 1$ is also a rational number.

Now, $x^{2009} - x^{2004} = x^{2004}(x^5 - 1) = (x^5)^{400}*x^4*(x^5 - 1)$ implies that $x^4$ is also a rational number. Therefore $\dfrac{x^5}{x^4} = x$ is also a rational number.

Now assume $x = \frac{p}{q}$ where $p,q$ are co-prime integers. Then $x^{2009} - x^{2004} = x^{2004}(x^5 - 1) = \frac{p^{2004}}{q^{2004}}(\frac{p^5}{q^5} - 1)$

$= \frac{p^{2004}}{q^{2004}}( \frac{p^5 - q^5}{q^5}) = \frac{p^{2004}(p^5 - q^5)}{q^{2014}}$

Now, we see that $p^5 - q^5 \equiv p^5 \not \equiv 0 \pmod{q}$ . So $q$ does not divide either $p^{2004}$ or $(p^5 - q^5)$ ,. Therefore the rational number is in its lowest terms and its denominator is $q^{2014}$ . But since $x^{2009} - x^{2004}$ is an integer, $q^{2014} = 1 \implies q = 1$ . Therefore $x$ is an integer.

Also, vishnu c , is this a question from this year's paper? If so, could you post all the questions here?

Siddhartha Srivastava - 6 years, 1 month ago

Once you know that x is rational, you can just use the rational root theorem.

Otto Bretscher - 6 years, 1 month ago

Oh. I was unaware of that form of the rational root theorem. That makes my solution smaller. Thanks.

and TBH, my solution was small on paper. Writing down the justifications and calculations made it seem much bigger than it should have been.

Siddhartha Srivastava - 6 years, 1 month ago

Thanks man! I'm sorry for the mistake! Learnt my lesson.

vishnu c - 6 years, 1 month ago

Oh, I also shared a set with CMI's questions form last year's paper. This one is also from last year's paper. I'm gonna be writing this year's paper on Monday! Wish me luck! I hope I don't make this kind of mistakes in the hall.

vishnu c - 6 years, 1 month ago

The rational root theorem states that when you have a polynomial with nonzero leading coefficient and constant term, like $x^5-1$ then every rational x that is a root of this polynomial, expressed as p/q, where p and q are coprime integers satisfies these two conditions:

p divides the constant term and,
q divides the leading coefficient.

In our case, we have the equation $x^{2014}-x^{2004}=k, \quad \forall \quad k\in Z.$ So, we get that q divides 1 and p divides any integer you throw at it. So, we get that x is an integer.

vishnu c - 6 years, 1 month ago

Again, you have to be careful: We don't know at this point that $x^5-1$ is an integer.

Otto Bretscher - 6 years, 1 month ago

@Otto Bretscher – It's the middle of the night here. I can't see clearly. Good night!

vishnu c - 6 years, 1 month ago

@Vishnu C – Ok good night... to be continued. You want to apply the rational root theorem to one of the original equations, for example, $x^{2014}-x^{2004}-k=0$ , since we are told that $k$ is an integer.

Otto Bretscher - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$