Finding the roots of $x^n+\frac{1}{x^n}=k$ 
Find  the roots of
  $$x^n+\frac{1}{x^n}=k$$
  when $n$ is an integer number and the $k$ is positive integer number.

So far I found one root which is $x=\frac{1+\sqrt{5}}{2}$ when  $n$ is even.
 A: Hint: The value $x=0$ can be eliminated immediately. So it is valid to multiply both sides of the equation by $x^n$. Doing this and moving all terms to the left side gives you
$$(x^n)^2-k(x^n)+1 = 0$$ which is a quadratic in $x^n$.
Apply the quadratic formula to get
$$x^n = \tfrac{k}{2} \pm \sqrt{\left(\tfrac{k}{2}\right)^2 - 1}$$
You should be able to find the solution(s) from here.
Note: If you are only looking for real roots, you must be careful; $k$ must be at least $2$ for the square root to exist.
A: This is a quadratic equation problem in disguise. Since we have two occurrences of $x^n$, let's make this slightly easier on ourselves by setting $y=x^n$, then we want to solve
$$y + \frac{1}{y} = k.$$
As written, it's a little difficult to solve so the most reasonable thing to do is multiply everything by $y$ so that we get only get positive powers of $y$ (which are the kinds of equations we know how to solve). This gives us
$$y^2 + 1 = ky \Longrightarrow y^2-ky+1 = 0.$$
You can solve this for $y$, then using $y=x^n$, you can find $x$.
A: Hint: $$\eqalign{x^n+\dfrac1{x^n}=k&\iff\dfrac{(x^n)^2+1}{x^n}=k\\&\iff(x^n)^2+1=k\cdot x^n  \\&\iff(\color{royalblue} {x^n})^2-k\cdot\color{royalblue}{ x^n}+1=0,}$$ this is a quadratic equation, which can be easily solved using the quadratic formula. It will give you at most two roots, say $r_1$, $r_2$. And to finish it off you only need to take the $n$-th root.
A: $x^{2n}+1=kx^{n}$
$x^{2n}-kx^{n}+1=0$
$z=x^{n}$
$z^2-kz+1=0$
Do the quadratic, then substitute $x^n$ back in for $z$.
Take the $n$th root and you have an $x=$ equation!
A: I think this question contains a good idea because the solution doesn't  depend on $k$
the first root as mentioned in question 
$$x=\frac{1+\sqrt{5}}{2}$$ when $n=$an even number (2,4,6,...)
the values of $$k=3 ,n=2$$$$k=7,n=4$$ $$k=18,n=6$$ 
the second root is $$x=\frac{3+\sqrt{5}}{2}$$ when $n$ an even or odd number 
the values of $$k=3,n=1$$ $$k=7,n=2$$ $$k=18,n=3$$ $$k=47,n=4$$ 
Now how we know these roots? we can easly find the roots if solve the equation with lower power
