If equation $f(x)=0$, and it has roots $α$ of degree m, then prove equation $h(x)=f(x+f(x))-f(x)=0$ has roots of degree $2m-1$. 
If equation $f(x)=0$, and it has roots $\alpha$  of degree $m$, then prove equation $h(x)=f(x+f(x))-f(x)=0$ has roots of degree $2m-1$.

When $\alpha$ is root of degree $m$, that means
$$
f(\alpha)=f'(\alpha)=f''(\alpha)= \ldots =f^{(m-1)}(\alpha)=0,f^m(\alpha)\neq 0
$$
I try prove $h(\alpha)=h'(\alpha)=h''(\alpha)= \ldots =h^{(2m-2)}(\alpha)=0$, $h^{(2m-1)}(\alpha)\neq 0$ but i did not find relation between $h(x)$ and $f(x)$ to prove $h^{(2m-2)}(\alpha)=0$.
 A: This is a nice application of Taylor's Theorem: If $f(x)$ is an $m$ times differentiable function at $a$ then
$$f(x)=\sum_{n=0}^{m-1} f^n(a)\frac{(x-a)^n}{n!}+f^m(c)\frac{(x-a)^m}{m!}$$
where $c$ is some value in $(x,a)$ or $(a,x)$ depending on $x$. By your conditions, we have
$$f(x)=f^m(c)\frac{(x-a)^m}{m!}$$
This then implies
$$h(x)=f(x+f(x))-f(x)=\frac{f^m(c)}{m!}(f(x)+x-a)^m-f(x)$$
$$=\frac{f^m(c)}{m!}\left[\left(\frac{f^m(c)}{m!}(x-a)^m+x-a\right)^m-(x-a)^m\right]$$
Now, recall that $f^m(x)$ is continuous and $c\in (x,a)$ (or $(a,x)$) implies
$$\lim_{x\to a}f^m(c)=f^m(a)$$
More importantly, since $f^m(a)\neq 0$ there exists some $\delta>0$ such that $|x-a|<\delta$ implies
$$\frac{1}{2}|f^m(a)|<|f^m(c)|<\frac{3}{2}|f^m(a)|$$
Define
$$Q=\max\left\{\frac{1}{2}f^m(a),\frac{3}{2}f^m(a)\right\}$$
$$q=\min\left\{\frac{1}{2}f^m(a),\frac{3}{2}f^m(a)\right\}$$
Then the statement above can be rewritten as
$$q<f^m(c)<Q$$
for all $|x-a|<\delta$. This implies that for these $x$ we have
$$\frac{q}{m!}<\frac{f^m(c)}{m!}<\frac{Q}{m!}$$
Since this quantity is never $0$, we may as well ignore it. Thus, our problem simplifies to showing that
$$g(x)=\left(\frac{f^m(c)}{m!}(x-a)^m+x-a\right)^m-(x-a)^m$$
has a root at $a$ of degree $2m-1$. This expands to
$$=\sum_{n=0}^m \binom{m}{n}\left(\frac{f^m(c)}{m!}\right)^n(x-a)^{nm+m-n}-(x-a)^m$$
$$=\sum_{n=1}^m \binom{m}{n}\left(\frac{f^m(c)}{m!}\right)^n(x-a)^{nm+m-n}+\binom{m}{0}\left(\frac{K}{m!}\right)^0(x-a)^{0\cdot m+m-0}-(x-a)^m$$
$$=\sum_{n=1}^m \binom{m}{n}\left(\frac{f^m(c)}{m!}\right)^n(x-a)^{nm+m-n}$$
To prove that this expression indeed has a root of order $2m-1$, notice that
$$\lim_{x\to a}\frac{g(x)}{(x-a)^{2m-1}}=\lim_{x\to a}\sum_{n=1}^m \binom{m}{n}\left(\frac{f^m(c)}{m!}\right)^n(x-a)^{nm+m-n-2m+1}$$
$$=\lim_{x\to a}\left[\sum_{n=2}^m \binom{m}{n}\left(\frac{f^m(c)}{m!}\right)^n(x-a)^{nm+m-n-2m+1}+m\left(\frac{f^m(c)}{m!}\right)\right]$$
Since $2\leq n\leq m$ implies
$$nm+m-n-2m+1\geq m-n+1\geq 1$$
the expression above becomes
$$\lim_{x\to a}m\left(\frac{f^m(c)}{m!}\right)=\frac{f^m(a)}{(m-1)!}\neq 0$$
which is both finite and non-zero as desired.
A: Let reduce claim to case $\alpha=0$ to shorten next record:
Let $x=\alpha$ is root of $f(x)$ of degree $m$. Consider $g(x)=f(\alpha+x)$. Then $0$ is root of $g(x)$ of degree $m$. And consider $i(x)=h(\alpha+x)$. If $x=\alpha$ is root of $h(x)$ of degree $2m-1$, then $x=0$ is root of $i(x)$ of degree $2m-1$ and vice versa. Note, that $$i(x)=f(\alpha+x+f(\alpha+x))-f(\alpha+x)=g(x+g(x))-g(x)$$
Then it is enough to prove the claim for case $\alpha=0$.
Let consider case $x=0$ is root of $f(x)$ of degree $m$:
$$h(x)=f(x+f(x))-f(x)$$
$$\lim_{x\to 0}\frac{h(x)}{x^{2m-1}}=\lim_{x\to 0}\frac{f(x+f(x))-f(x)}{x^{2m-1}}=\\
=\lim_{x\to 0}\frac{f(x+f(x))-f(x)}{f(x)}\frac{f(x)}{x^{2m-1}}=\\
=\lim_{x\to 0}\left(\lim_{t\to 0}\frac{f(x+t)-f(x)}{t}\right)\frac{f(x)}{x^{2m-1}}=\\
\lim_{x\to 0}\frac{f(x)}{x^m}\frac{f'(x)}{x^{m-1}}$$
Let $f^{(m)}(0)=C$, $C\neq 0$, then $$\lim_{x\to 0}\frac{f(x)}{x^m}=\lim_{x\to 0}\frac{f'(x)}{mx^{m-1}}=\frac{C}{m!}$$
$$\lim_{x\to 0}\frac{h(x)}{x^{2m-1}}=\frac{C}{m!} \frac{C}{(m-1)!}=\frac{C^2}{m! (m-1)!}\neq 0$$
Existence of this limit requires $h(0)=0$, $h^{(k)}(0)=0$ for $1\leq k \leq 2m-2$. Also this limit is equal to $\frac{h^{(2m-1)}(0)}{(2m-1)!}$ and this quantity is not equal to zero. Then $x=0$ is root of $h(x)$ of degree $2m-1$.
