Waste less time on Facebook — follow Brilliant.
×

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.

Note by Vishnu C
2 years, 5 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 - 2 years, 5 months ago

Log in to reply

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 - 2 years, 5 months ago

Log in to reply

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

Otto Bretscher - 2 years, 5 months ago

Log in to reply

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

Vishnu C - 2 years, 5 months ago

Log in to reply

@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 - 2 years, 5 months ago

Log in to reply

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 - 2 years, 5 months ago

Log in to reply

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

Vishnu C - 2 years, 5 months ago

Log in to reply

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

Otto Bretscher - 2 years, 5 months ago

Log in to reply

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 - 2 years, 5 months ago

Log in to reply

Comment deleted May 12, 2015

Log in to reply

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

Otto Bretscher - 2 years, 5 months ago

Log in to reply

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

Vishnu C - 2 years, 5 months ago

Log in to reply

@Vishnu C 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!

Otto Bretscher - 2 years, 5 months ago

Log in to reply

@Otto Bretscher 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 :)

Vishnu C - 2 years, 5 months ago

Log in to reply

@Vishnu C 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).

Otto Bretscher - 2 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...