# 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.

Note by Vishnu C
2 years, 11 months ago

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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?

- 2 years, 11 months 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.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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.

- 2 years, 11 months 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.

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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

- 2 years, 11 months 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.

- 2 years, 11 months ago

Comment deleted May 12, 2015

"A product of an integer and $$x^5$$ is an integer. So, $$x^5$$ is also an integer,..." Really?

- 2 years, 11 months ago

Damn it! I'm really sorry. It's rational. I got it now! Thanks!

- 2 years, 11 months ago

No problem, I'm just trying to help. Siddhartha's proof is good, but you can make it shorter by using the rational root theorem... once you know that $$x$$ is rational, you are done!

- 2 years, 11 months ago

I think I wrote what you meant by use the rational root theorem (new theorem to me. Learning everyday.) as a reply to Siddhartha's proof. I made a rationally irrational mistake :)

- 2 years, 11 months ago

An important theorem: If you have a rational root of a monic polynomial with integer coefficients, then the root must be an integer (that's just the special case we need here).

- 2 years, 11 months ago