Roots of a polynomial inside the unit circle Let $k$ be a even positive integer. Now, consider the polynomial 
$$
p(x)=x^k-px^{k-1}-qx^{k-2}-x^{k-3}-\cdots -x-1,
$$
with $p$ and $q$ integers satisfying $q-1>p\geq 1$.
How to prove that this polynomial has k-2 roots inside the unit circle (there are two roots outside this circle by using the Descarte sign rule and the intermediate value theorem).
Any suggestion? I tried to use Rouché theorem, but without success.
 A: The conjecture is not true for the case where $k$ is odd. The smallest degree polynomial which gives contradiction is as follow:
$$x^7-x^6-3x^5-x^4-x^3-x^2-x-1,$$
Where we have $k=7$, $p=1$ and $q=3$. If the claim is true, we must have $5$ roots in the interior of the unit circle. But, three roots of the above polynomial are as follows:
$$-1, -1, 2.479561783.$$
Actually, there are infinite counterexamples for the case where $k$ is odd. Two of them are as follows:
$$x^9-x^8-3x^7-x^6-x^5-x^4-x^3-x^2-x-1$$
with three roots
$$-1.071441460, -1., 2.480930101$$
and 
$$x^{13}-x^{12}-3x^{11}-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x-1$$
with three roots
$$-1.130681001, -1., 2.481187340.$$
If we assume that the unit circle is closed (it contains its boundary), I did not have any counterexample until yet. I think in this case the claim may be true for odd and even $k$.
I will try to study the case where $k$ is even.
A: The condition $k$ is even is crucial because for such it is obvious there are only two real roots, one negative and one positive (for $x < -1$, using $y=-x$, it is easy to see the decomposition into the sum of two increasing polynomials, namely the first three terms and the rest, which start negative at -1 and go to $+\infty$ by the evenness condition on $k$, while it is obvious $p$ is negative between -1 and 0, as well as having one "large" root for positive $x$ as $p$ is a Cauchy polynomial and the positive root is largest of all roots in absolute value for such)
Since for the limiting case $q=p+1$, $x+1$ factors and we remain with a polynomial for which the Kakeya theory applies and we can conclude that it has one large positive root and all other roots inside the unit circle, a strategy for solving this would be to fix the coefficient $p>1$ and then show that for $q>p+1$, $q$ real, the corresponding polynomial has no roots on the unit circle, then use homotopy of polynomials (the existence of topological degree etc) to conclude that all such have same number roots inside/outside and then the limiting case gives us the solution since for $q=p+1$, $p>1$ we have a root outside, one on the circle and the rest strictly inside.
A: I think it should be true in the stronger assumption $q>p+2$, but I suspect it could be false for integer $p\ge1$ and $q=p+2$, and an even $k$. Let me explain why -it could be possibly useful to make either a counterexample, or a proof of the statement, at your taste.  Consider the equivalent statement, that the reciprocal polynomial $$Q(x):=-1+px+(p+2)x^2+x^3+\dots+x^k$$
has no  zeros in the closed unit disk, besides the two real zeros in the interval $(-1,1)$ (coming from $Q(-1)=1$, $Q(0)=-1$, $Q(1)=2p+k-1>0$). 
A zero $x$ of $Q$ on the unit circle would solve 
$$-1+(p-1)x+(p+1)x^2={x-z\over x-1}$$
with $z=x^{k+1}$. The linear fractional transformation on the RHS maps the unit circle $|z|=1$ to a circle of radius ${1\over |x-1|}$ passing by $0$ and $1$, namely $\partial B\Big({x\over x-1},{1\over |x-1|}\Big)$.  Now it seems there are many $x\in\partial B(0,1)$, such that the trinomial at the LHS lies on the circle $\partial   B\Big({x\over x-1},{1\over |x-1|}\Big)$. For instance $x=-1$ is one of these, but it could well be a point with a dense orbit $\{x^n\}_{n\in\mathbb{N}}$. In this case it could be possible, or quite hard to rule out, that for some even $k$  one has $z=x^k$, making $x$ a further zero of the corresponding $Q=Q_k$.
