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.

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## Comments

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TopNewestFixing the solution of Visnhu C.

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?

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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:

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.

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

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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.

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

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

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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.

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Comment deleted May 12, 2015

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

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

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