Solve the system $\begin{cases}mz^{m+1}-(m+1)z^m+1=0\\ nz^{n+1}-(n+1)z^{n}+1=0\end{cases}$ for $z\in\mathbb{C}$ with $m\ne n\in\mathbb{N}$ 
I wish to solve
$$\begin{cases} 
mz^{m+1} - (m+1)z^m + 1 = 0 \\ \;nz^{n+1} - (n+1)z^n + 1 = 0 \end{cases}$$
for $z \in \mathbb{C}$ and fixed $m \ne n \in \mathbb{N}$.

I suspect the only solution is $z=1$ regardless of $m, n.$ In fact, I have been working on a problem that reduced to showing that $z=1$ is the only solution, so it would be very convenient if this was the case.
However, I have no idea how to proceed. Certainly, there are only a finite number of solutions to either equation, but how do we avoid the scenario where some $z \ne 1$ satisfies both equations? You can use the Euclidean algorithm if specific values of $m, n$ are given to you, but I want to solve the problem for all $m, n$ in one fell swoop. Is there any way to do this? Maybe I'm missing some obvious idea.
 A: Some thoughts
It is easy to prove that
$$mz^{m+1} - (m+1)z^m + 1 = (z-1)^2\sum_{k=1}^m kz^{k-1}$$
and
$$nz^{n+1} - (n+1)z^n + 1 = (z-1)^2\sum_{k=1}^n kz^{k-1}.$$
Let $f = \sum_{k=1}^m kz^{k-1}$ and $g = \sum_{k=1}^n kz^{k-1}$.
If we can prove that $\mathrm{Res}(f, g) \ne 0$, then $z = 1$ is the only solution.
(For the resultant of two polynomials, see: http://www2.math.uu.se/~svante/papers/sjN5.pdf)
I can not prove it currently. I can prove the special case as follows.
We prove the case when $1\le m < n$ and $\mathrm{gcd}(\lfloor \frac{m}{2}\rfloor + 1, \lfloor \frac{n}{2}\rfloor + 1) = 1$.
We will use the following auxiliary result.
Fact 1: Let $1\le M < N$ be two integers with $\mathrm{gcd}(M+1, N+1) = 1$. Let $F = 1 + z^2 + z^4 + \cdots + z^{2M}$ and $G = 1 + z^2 + z^4 + \cdots + z^{2N}$. Then $\mathrm{Res}(F, G) = 1$.
Hint: Note that $(z^2-1)F = z^{2M+2}-1$ and $(z^2-1)G = z^{2N+2}-1$.
$F$ has $2M$ roots $\xi_j = \mathrm{e}^{\mathrm{i}\frac{2j\pi}{2M+2}}, j \in \{0, 1, 2, \cdots, 2M+1\}\backslash \{0, M+1\}$.
Then, $\mathrm{Res}(F,G) = \prod_{j \in \{0, 1, 2, \cdots, 2M+1\}\backslash \{0, M+1\}} G(\xi_j)$.
Now, let $f_1 = \sum_{k=1}^{\lfloor \frac{m}{2}\rfloor} z^{2k}$ and
$g_1 = \sum_{k=1}^{\lfloor \frac{n}{2}\rfloor} z^{2k}$. By Fact 1, we have $\mathrm{Res}(f_1, g_1) = 1$.
Then, we have
\begin{align}
\mathrm{mod}(\mathrm{Res}(f, g), 2) &= \mathrm{mod}(\det (\mathrm{Syl}(f,g)), 2)\\
 &= \mathrm{mod}(\det(\mathrm{mod}(\mathrm{Syl}(f,g) , 2)), 2)\\
 &= \mathrm{mod}(\mathrm{Res}(f_1, g_1), 2)\\
 &= 1
\end{align}
which results in $\mathrm{Res}(f, g) \ne 0$. The desired result follows.
A: First, I will assume you require that $a$ and $b$ be positive integers and that you are letting $z$ vary over the reals.
Consider the derivative of the first equation:
$a(a+1)z^a - a(a+1)z^{a-1} = a(a+1)(z-1)z^{a-1}$
Say $a$ is odd. Notice it is strictly positive for $z>1$ and non positive for $z<1$. This is because $a(a+1)$ is positive and $z^{a-1}$ is non negative for any $z$ and odd $a$. Thus we may conclude the minimum is at $z=1$ and thus $z=1$ is the only solution.
Say $a$ is even, then again examining the derivative will show that the slope is negative for $0<z<1$ however positive for $z<0$. Thus another solution will occur for some negative $z$. Thus if we hope to find a solution other than $z=1$, we should pick both $a$ and $b$ to be even.
Consider z=-1, this gives $az^{a+1}-(a+1)z^a+1 = -a-(a+1)+1 = -2a<0$ (recall a is even and positive). Thus the function crosses the $z$ axis after $z=-1$.
Recall the derivative $a(a+1)(z-1)z^{a-1}$. Say we increase $a$ to a larger even value. This decreases the value of the derivative when $-1<z<0$ thus when $z$ is in the interval -1 to 0, the slope is less. Thus the root must occur earlier when a is increased. Thus for different values of a and b, the roots do not coincide. Thus the only solution is z=1.
A: I think that for $m=4, n=5$ you get the resultant of the quotients of these polynomials by $z-1$ equal $0$. So in this case the polynomials have at least 2 common roots (1 and something else).
Edit: These polynomials have root 1 of mutiplicity 2, so we need to divide by $(z-1)^2$. Then the resultant is not zero.
In any case one should compute the resultant to show that polynomials have/have not common roots. I suggest computing the resultants modulo $2,3$ or $5$. This should be not too difficult. Patterns may be like this: "If $m$ is $...\mod ...$ and $n$ is $...\mod ...$ then the resultant is not zero modulo $2$ (or $3$ or $5$).
A: COMMENT.-It is very difficult (or impossible) to explicitly solve this system and it is clear that for many values of $m$ and $n$ there will be no solution. However, there is a Sylvester theorem giving a necessary and sufficient condition for the two proposed equations to have common roots. This theorem consists in that the determinant of a matrix formed with the coefficients, associated to the system must be null (in this "Sylvester resultant" the degree with respect to the coefficients of the second equation will be $m+1$ and it will be $n+1$ in the coefficients of the first equation).
Example.-For $(m,n)=(3,2)$ put for see it better the system $$\begin{cases} 
3z^4 - 4z^3+0z^2+0z+1 = 0 \\ \;2z^3 - 3z^2+0z+1 = 0 \end{cases}$$ Then in order to have common roots we must have the equality
$$\begin{vmatrix}
3 & -4& 0 & 0&1&0&0 \\ 
0 & 3 & -4 & 0&0&1&0 \\ 
0 & 0 & 3 & -4&0&0&1 \\ 
2 & -3 & 0 & 1&0&0&0\\0&2&-3&0&1&0&0\\0&0&2&-3&0&1&0\\0&0&0&2&-3&0&1\notag
\end{vmatrix}=0$$
The calculation of this determinant will tell whether or not the system in this example has a solution.
