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.

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.
