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.

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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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? – Siddhartha Srivastava · 1 year, 5 months ago

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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 · 1 year, 5 months ago

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– Otto Bretscher · 1 year, 5 months ago

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

– Vishnu C · 1 year, 5 months ago

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

– Otto Bretscher · 1 year, 5 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.Log in to reply

– Vishnu C · 1 year, 5 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.Log in to reply

– Vishnu C · 1 year, 5 months ago

Thanks man! I'm sorry for the mistake! Learnt my lesson.Log in to reply

– Otto Bretscher · 1 year, 5 months ago

Once you know that x is rational, you can just use the rational root theorem.Log in to reply

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 · 1 year, 5 months ago

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– Otto Bretscher · 1 year, 5 months ago

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

– Vishnu C · 1 year, 5 months ago

Damn it! I'm really sorry. It's rational. I got it now! Thanks!Log in to reply

– Otto Bretscher · 1 year, 5 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!Log in to reply

– Vishnu C · 1 year, 5 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 :)Log in to reply

– Otto Bretscher · 1 year, 5 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).Log in to reply