Find all real solutions of the equation: $x^5+5x^3+5x+2017=0$

Question

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.

Answer 1

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

Answer 2

Hint:

Let $x=y-\frac 1y$

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

And let $y^5=t$

We get

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