Suppose $m$ and $n$ are two nonnegative integers. What is the number of zeros of the polynomial $(1+z)^{m+n}-z^n$ in the unit ball $|z|<1$?
Some calculations for small values of $m$ and $n$ suggests the following formula: $$n-1+2\left\lfloor\frac{m-n+5}{6} \right\rfloor$$
Does this formula work for all values of $m$ and $n$?
I have encountered this problem a while ago when working on boundary conditions in graphene, see http://arxiv.org/abs/0710.2723 for more context. Since I liked the problem and the solution, I now give it as a test exercise to prospective PhD students, and that is perhaps how it ended here (well, I suppose I'll need to replace the problem now). The solution is given in appendix B of the reference above, I'll copy it here for the interested people.
The reason why I ask to solve this problem is because the best way to solve it is to first gain intuition by studying the problem numerically (luckily it's easy), and then to try to generalize. However mostly I see that the applicants try to apply textbook methods, namely the Rouché's theorem, which leads to a messy integral and a dead end. So let us start with studying how the equation behaves. Let us plot the solutions for a particular choice of n and m:
Indeed, as hinted by @fedja in his comment we see that the roots lie on a contour which passes through $-1/2 \pm i\sqrt{3}/2$. Now it's clear what has to be done. We rewrite the equation in polar coordinates:
$$|1+z|^{n+m} = |z|^n$$ $$(n+m)\arg(1+z) = n\arg(z) \mod 2\pi$$
The first equation requires that the roots lie on the contour shown in bold in the figure above. The second equation allows to count the roots between any two points on this contour by calculating the increase of the arguments between these two points. Using that $\arg(1+z)$ and $\arg(z)$ change by $\pm 2\pi/3$ monotonically between $-1/2 -i\sqrt{3}/2$ and $-1/2 + i\sqrt{3}/2$ and counting the multiplicities leads us to the correct answer written above.
For $n$ and $m$ divisible by $3$ there are two roots that lie on the unit circle, which was important for me in the context of the application to graphene, and which is the border case that @fedja mentioned.
Here is my solution of this problem. So we have next equation for the zeros: $$ (1+z)^{n+m}=z^n $$ We can modify it like that: $$ \Bigl(1+\frac{1}{z}\Bigr)^n \Bigl(1+z\Bigr)^m = 1 $$ Next we can mark the first factor as $re^{i\varphi}$, so the equation splits into two ones: $$ \Bigl(1+z\Bigr)^m = re^{i\varphi} \\ \Bigl(1+\frac{1}{z}\Bigr)^n = \frac{1}{r}e^{-i\varphi} $$ Let's consider the first equation. We have to extract the root: $$ 1 + z = r^{\frac{1}{m}} e^{i\frac{\varphi + 2 \pi k}{m}} $$ Next we take away $1$ and write down the module of $z$ using the condition $|z|<1$: $$ |z|^2 = |r^{\frac{1}{m}} e^{i\frac{\varphi + 2 \pi k}{m}} - 1|^2 = r^{\frac{2}{m}} - 2 r^{\frac{1}{m}} \cos \frac{\varphi + 2 \pi k}{m} + 1 < 1 $$ Finally, we divide by $r^{\frac{1}{m}}$, so: $$ r^{\frac{1}{m}} < 2 \cos \frac{\varphi + 2 \pi k}{m} $$ At the same time we know that $r^{\frac{1}{m}}<1$, so we have to understand, which inequation is stronger. Let's mark $z=\rho e^{i\psi}$ and write down the following inequation: $$ \Bigl|1 + \frac{1}{z}\Bigr|^2 = \rho^{-2} + 2\rho^{-1} \cos \psi + 1 > 1 $$ Next, $$ \rho \cos \psi + 1 > \frac{1}{2} $$ It's clear that the left part equals $\Re \sqrt[m]{re^{i\varphi}}$, so: $$ r^{\frac{1}{m}} \cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2} $$ Because of $r^{\frac{1}{m}} < 1$ we have: $$ \cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2} $$ Therefore $r^{\frac{1}{m}} < 1$ is stronger and we have $\cos \frac{\varphi + 2 \pi k}{m} > \frac{1}{2}$. Next, we perform the same procedure for the second equation, and finally obtain another inequation: $$ \cos \frac{-\varphi + 2 \pi k'}{n} < \frac{1}{2} $$ Let's transform these inequations: $$ -\frac{m}{6} < \frac{\varphi}{2\pi} + k < \frac{m}{6} \\ \frac{n}{6} < -\frac{\varphi}{2\pi} + k' < \frac{5n}{6} $$ We have to take away $\varphi$, so: $$ -\frac{m}{6} - \frac{\varphi}{2\pi} < k < \frac{m}{6} - \frac{\varphi}{2\pi} $$ And: $$ -\frac{m}{6} + \frac{n}{6} - k' < k < \frac{m}{6} + \frac{5n}{6} - k' $$ So, finally: $$ \frac{n-m}{6} < \ell < \frac{5n+m}{6} $$ where $\ell = k + k'$. The number of $\ell$'s satisfying this inequation defines the number of zeros in the unit disk.
