TITLE: Solve $x^{2k}+(x-2)^{2k}=2k$ with $k\in \mathbb N$ and $x\in\mathbb{R}^+$
QUESTION [7 upvotes]: Solve $x^{2k}+(x-2)^{2k}=2k$  with $k\in \mathbb N$ and $x\in\mathbb{R}^+$.

I have no idea how to solve this, I just can find one solution, $x=k=1$.

REPLY [2 votes]: Assume $k\ge2$. At $x=0$, and at $x=2$, the left side is bigger than the right. At $x=1$, the right side is bigger than the left. It follows that there is a solution with $0\lt x\lt1$, and another solution with $1\lt x\lt2$. 
For $k=2$ it's a quartic equation, and there is a formula for solving those (a websearch for "quartic equation" will undoubtedly turn it up). For $k\ge3$ it's an equation of degree $6$ or higher, and it is unlikely that the solutions can be expressed in terms of integers, arithmetical operations, extractions of $n$th roots (or exponentials, logarithms, trig & inverse trig functions). You can get as close to a solution as you like by numerical methods. If you know some Calculus, there's Newton's Method.