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

Let x be a real number such that x2014x2004andx2009x2004 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.

Note by Vishnu C
4 years, 7 months ago

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Fixing the solution of Visnhu C.

Subtract the second from the first and you'll get that x5(x2009x2004)x^5(x^{2009}-x^{2004}) is an integer

We see that x5 x^5 is a rational number. That implies x51 x^5 - 1 is also a rational number.

Now, x2009x2004=x2004(x51)=(x5)400x4(x51) x^{2009} - x^{2004} = x^{2004}(x^5 - 1) = (x^5)^{400}*x^4*(x^5 - 1) implies that x4 x^4 is also a rational number. Therefore x5x4=x \dfrac{x^5}{x^4} = x is also a rational number.

Now assume x=pq x = \frac{p}{q} where p,q p,q are co-prime integers. Then x2009x2004=x2004(x51)=p2004q2004(p5q51) x^{2009} - x^{2004} = x^{2004}(x^5 - 1) = \frac{p^{2004}}{q^{2004}}(\frac{p^5}{q^5} - 1)

=p2004q2004(p5q5q5)=p2004(p5q5)q2014 = \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 p5q5p5≢0(modq) p^5 - q^5 \equiv p^5 \not \equiv 0 \pmod{q} . So q q does not divide either p2004 p^{2004} or (p5q5) (p^5 - q^5) ,. Therefore the rational number is in its lowest terms and its denominator is q2014 q^{2014} . But since x2009x2004 x^{2009} - x^{2004} is an integer, q2014=1    q=1 q^{2014} = 1 \implies q = 1 . Therefore x 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 - 4 years, 7 months ago

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

Otto Bretscher - 4 years, 7 months ago

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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.

Siddhartha Srivastava - 4 years, 7 months ago

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

vishnu c - 4 years, 7 months ago

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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.

vishnu c - 4 years, 7 months ago

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The rational root theorem states that when you have a polynomial with nonzero leading coefficient and constant term, like x51x^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 x2014x2004=k,kZ.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 - 4 years, 7 months ago

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Again, you have to be careful: We don't know at this point that x51x^5-1 is an integer.

Otto Bretscher - 4 years, 7 months ago

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

vishnu c - 4 years, 7 months ago

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@Vishnu C Ok good night... to be continued. You want to apply the rational root theorem to one of the original equations, for example, x2014x2004k=0x^{2014}-x^{2004}-k=0, since we are told that kk is an integer.

Otto Bretscher - 4 years, 7 months ago

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