Number of real roots of the equation $(x-a)^{2n+1}+(x-b)^{2n+1}=0$ is? 
Let $n$ be a positive integer and $0 < a < b < ∞$. The total number of real roots of the equation $(x-a)^{2n+1}+(x-b)^{2n+1}=0$ is ?

I tried it for $n=1$ and always get $1$ real root. How can I guarantee that it is $1$ only for each $n$?
 A: Let $$f(x) = (x-a)^{2n+1}+(x-b)^{2n+1}$$ 
Now $f(x)$ is an odd degree polynomial so it has at least one real root. If we prove:
$$
\frac{d}{dx}f(x) = (2n+1)\left((x-a)^{2n}+(x-b)^{2n}\right)> 0, \forall x
$$
Then we are done, as then $f$ is monotonically increasing and therefore has precisely one root.
Note that $(x-a)^{2n}+(x-b)^{2n} \ge 0$ for all $x$. Therefore, $f$, is non-decreasing. Now, we have that$$
(x-a)^{2n}+(x-b)^{2n} = 0 \longrightarrow (x-a)^{2n} = -(x-b)^{2n}
$$
Since $x\in \mathbb{R}$, equality holds only when $x=a=b$, But, by hypothesis, $b>a$, thus $\frac{d}{dx}f(x) > 0 \forall x$, and the result follows. 
A: Rewrite the equation as $(x-a)^{2n+1} = -(x-b)^{2n+1}$. As $a,b$ are arbitrary we can think of this as asking for two real numbers $u,v$ such that $u^m = -v^m$, $m$ odd. Calling  $-v$ as $w$ we want two POSITIVE real numbers $u,w$ such that $u^m= w^m$, $m$ odd. This forces $u=w$, which translates into $x-a= b-x$, and so the unique (real) solution is  $x=\frac12(b-a)$.
