1

I joined the math contest $1$ week ago ( In Azerbaijan). There were $6$ questions. Unfortunately, I could solve $1$ question correctly.I know I can not write all the questions here, because It is against the MSE rules.I want to solve the $6th$ question that is known as the most difficult question.In fact, this question could not be solved by our teacher.

Question $6$. Find all real solutions of the equation:

$$x^5+5x^3+5x+2017=0$$

The only thing I have learned is that there are no integer solutions of this equation.Maybe, I'm wrong.

nonuser
  • 496
  • There are, in fact, no rational solutions. We can see this because the polynomial is monic, which means that any rational solution must be an integer which divides the constant term. $2017$ is prime, and therefore there are only four candidates to check: $\pm 1, \pm2017$. None of them are roots. – Arthur Nov 28 '17 at 13:57
  • @Dietrich Burde I am in a state of exhaustion right now. I feel incompetent myself. Teacher, 1 question confirms that I solved the problem correctly. – nonuser Nov 28 '17 at 14:06
  • @DietrichBurde I think OP means to say, "Unfortunately, I could solve only one question correctly." – Jaideep Khare Nov 28 '17 at 14:08
  • I 'm sorry English is my second language. – nonuser Nov 28 '17 at 14:11
  • I'm sorry, too. English is also my second language. I have erased the comment. – Dietrich Burde Nov 28 '17 at 14:11
  • 3
    This is the beauy of science ! For most of us, English is not our first language and we are able to communicate almost perfectly. Cheers to everybody. – Claude Leibovici Nov 28 '17 at 14:16
  • @DietrichBurde While that makes $-1$ a root, it seems to me that the other roots aren't as nice. But I don't know. – Arthur Nov 28 '17 at 14:20
  • @DietrichBurde. Beside $x=-1$, the roots are monstrous. Where did you find the test ? – Claude Leibovici Nov 28 '17 at 14:20
  • @ClaudeLeibovici Thank you for those excellent words. – nonuser Nov 28 '17 at 14:21
  • The test asked for integral roots only. I just could imagine, that there is a typo in the given polynmial, but I don't know. – Dietrich Burde Nov 28 '17 at 14:21
  • @Dietrich Burde Equation is correct ..because I work 10 days with equation..:) I believe we will find solution. – nonuser Nov 28 '17 at 14:26
  • Dear friends, Equation is correct... I joined the Olympiad myself.. – nonuser Nov 28 '17 at 14:28
  • Dear mathematicians, I think we should find only real root here. Clearly, the complex root does not pass the definition of the problem. – nonuser Nov 28 '17 at 14:49
  • 1
    It's a solveable quintic, see eg https://math.stackexchange.com/questions/1537069/the-trigonometric-solution-to-the-solvable-demoivre-quintic – leonbloy Nov 28 '17 at 15:05
  • Thank you everyone!! @leonbloy I understood – nonuser Nov 28 '17 at 15:39

2 Answers2

9

Hint:

Let $x=y-\frac 1y$

We get $$y^5-\frac {1}{y^5}+2017=0$$

And let $y^5=t$

We get

$$t^2+2017t-1=0$$

Community
  • 1
Zaharyas
  • 385
  • 1
    So simple .... when we know the elegant solution ! $\to +1$ – Claude Leibovici Nov 29 '17 at 09:37
  • @ClaudeLeibovici Thank you so much :) Mathematics is mother tongue all of us.As you say. :) – Zaharyas Nov 29 '17 at 10:43
  • 2
    What I am almost sure is that my English is terrible but all of us communicate. In 1978, I was giving lectures in Russia; I was speaking French and all of that was translated (it was forbidden to not go through the translators). I don't know how my blabla was translated. But during coffes breaks, the Russian guys and I were doing Q/A just with equations; no problem .... except that the translators were very, very unhappy since unable to pick anything. Cheers. – Claude Leibovici Nov 29 '17 at 10:48
0

let $$f(x) = x^5+5x^3+5x+2017$$

then $f'(x)= 5x^4+15x^2+5>0\forall x \in \mathbb{R}$ (Strictly increasing function.)

and $\lim_{x\rightarrow -\infty}f(x)\rightarrow -\infty$ and $\lim_{x\rightarrow \infty}f(x)\rightarrow \infty$ and $f(0) = 2017$

so $f'(x) = 0$ has no real roots and $f(x) = 0$ has exactly one real roots

DXT
  • 11,241