Somemathhere

Consider the integers $${n \choose k}$$ for $k=0,1,\ldots,n$. It turns out that the number of those which are odd is always a power of $2$. Proof. Direct application of Lucas theorem. $\displaystyle {n \choose k}$ is odd exactly when the $1's$ in the binary represenation of $k$ correspond only to $1's$ in the binary representation of $n$. If $n$ has exactly $d$ $1's$ then the number of such $k$ is $2^d$.

Consider a problem we already solved (in a weaker setting) in a previous post. Let $x_1,\ldots, x_n$ be complex numbers, such that $$x_1^k+\cdots +x_n^k=0$$ for all $k\in\{1,\ldots,n\}$. Prove that all of the numbers are zeros. Here we would investigate another approach. The idea is simple. First we construct a matrix $A$ whose characteristic polynomial has as roots $x_1,\ldots, x_n$. Then we use the fact that $\text{Tr}(A^k)=x_1^k+\cdots+x_n^k$ to successively eliminate all the coefficients, hence showing that $x_1,\ldots,x_n$ are roots of $x^n=0$. Assume $x^n-a_1 x^{n-1}-\cdots-a_{n-1}x-a_n$ is a polynomial whose roots are precisely $x_1,\ldots,x_n$. Then the matrix $$  A= \begin{pmatrix}         a_1       & a_2 & a_3 & \dots & a_{n-1} & a_n \\         1       & 0 & 0 & \dots  &0 & 0 \\      ...

$$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)$$