Show that a polynomial with gaps with different roots under the following certain conditions is equal to zero I am wondering if the following conjecture is true:
Let $f$ be a polynomial of the form
$$f(x) = \sum_{k=0}^{p-1} c_k x^k
+ \sum_{k=n}^{n+q-1}c_k x^k.$$
Suppose that $y_1,\ldots,y_{p+q}$,
be pairwise different numbers,
$y_1,\ldots,y_{p+q}>0$, and
$$f(y_j)=0\qquad (1\le j\le p+q).$$
Then $f=0$, i.e.,
$c_k=0$ for every $k$.
Here is my attemp:
We denote by $\lambda$ the following integer partition (integer tuple, decreasing in the non-strict sense):
$$\lambda=(\,\underbrace{n-p,\ldots,n-p}_q\,,\,\underbrace{0,\ldots,0}_{p}\,).$$
The conditions $f(y_j)=0$, $1\le j\le p+q$, can be written as the following system of $p+q$ homogeneous linear equations for in the variables $c_0,\ldots,c_{p-1},c_n,\ldots,c_{n+q-1}$:
$$\sum_{k=1}^{p+q} y_j^{\lambda_k+p+q-k} c_{\lambda_k+p+q-k} = 0\qquad(1\le j\le p+q).$$
The determinant of this system equals
$$D=\det \bigl[ y_j^{\lambda_k+p+q-k} \bigr]_{j,k=1}^{p+q}.$$
For example for $p=2$ y $q=2$,the determinant is of the form
$$
\begin{bmatrix}
y_1^{n+1} & y_1^n & y_1 & 1
\\[1ex]
y_2^{n+1} & y_2^n & y_2 & 1
\\[1ex]
y_3^{n+1} & y_3^n & y_3 & 1
\\[1ex]
y_4^{n+1} & y_4^n & y_4 & 1
\end{bmatrix}.
$$
My next idea is to divide $D$ over the Vandermonde polynomial $\prod_{1\le j<k\le p+1}(y_j-y_k)$, but I do not know how to do it properly, and what else I should do to show that the unique solution of the system is the trivial one. Thank you in advance for you help friends.
 A: Descartes' rule of signs says that for a polynomial with real coefficients (which is not identically zero)

the number of positive roots is at most the number of sign changes in the sequence of polynomial's coefficients (omitting the zero coefficients)

Your polynomial $f(x) = \sum_{k=0}^{p-1} c_k x^k + \sum_{k=n}^{n+q-1}c_k x^k$ has at most $p+q$ non-zero coefficients, so that there are at most $p+q-1$ sign changes. Therefore, it $f$ is not identically zero, it can have at most $p+q-1$ positive zeros.
Which means that your conjecture is true.
A: Comment on the "Vandermonde" determinant. The computation of the determinants in your proposed strategy seemed challenging, so I worked a little in some particular cases. I have guessed a formula that I haven't been able to prove for all cases. In the following, I still use $p$ and $q$ as you do but I don't use the letter $n$. Instead, I changed a little your notation and I use the letter $m$ to mark the size of the gap, i.e., $m=n-p$. The case $m=0$ is the usual Vandermonde determinant, and in general $m\ge 1$.
First, some notation may be helpful to make the results less ugly. Let's define:
$$
\begin{equation}
H^{m;r}_{s}=\sum_{\substack{0\le k_1,\ldots,k_s\le m\\ k_1+\ldots+k_s=r }} y_1^{k_1}y_2^{k_2}\cdots y_s^{k_s} \;.
\tag{1}
\end{equation}
$$
Also, let's write the typical Vandermonde determinant as:
$$
\begin{equation}
G_s=\prod_{0\le i<j\le s} \left(y_j-y_i\right)\;.
\tag{2}
\end{equation}
$$
Finally, let's denote your determinant as $D^{(m)}_{p,q}$. I have been able to prove the following formulae:
$$
\begin{eqnarray}
D^{(m)}_{1,2}=-G_{3}H_{3}^{m;2m}\;, \tag{3}\\
D^{(m)}_{2,1}=-G_{3}H_{3}^{m;m}\;, \tag{4}\\
D^{(1)}_{p,q}=(-1)^{p-q}G_{p+q}H_{p+q}^{1;q}\;. \tag{5}
\end{eqnarray}
$$
I proved formulae (3) and (4) by brute algebra, i.e., computing the $3\times 3$ determinant, using factorizations of $y_j^m-y_i^n$, etc.
I proved formula (5) by adding the "missing" column plus an extra variable $y_{p+q+1}=\alpha$. Then the matrix is a typical Vandermonde matrix whose determinant is a polynomial in $\alpha$. By computing the coefficient of $\alpha^p$, I get formula (5). I think this last strategy is the best for recursively computing formulae for $m=2,3,4,\ldots$, even for making a proof by induction. However, I have been stuck in combinatorial nuisances. I am almost sure that the general result looks like
$$
\begin{equation}
D^{(m)}_{p,q}=(-1)^{N}G_{p+q}K_{p+q}^{m;mq} \;,
\tag{6}
\end{equation}
$$
where $K_{p+q}^{m;mq}$ very similar to $H_{p+q}^{m;mq}$ but (maybe) with extra positive combinatorial coefficients. $N$ is just an integer that depends on $m,p,q$.
In conclusion, this strategy will work to prove your hypothesis but it's very cumbersome and it's more a determinant problem by itself.
