There are integers such that this determinant is nonzero?

Question

I was wondering if given a strictly decreasing sequence of non-negative integers $\mu_{n},\mu_{n-1},\cdots,\mu_{1}$ it is always possible to find $x_{1}=2$ and $x_{2},\cdots,x_{n}$ odd integers such that $$det\begin{pmatrix} x_{1}^{\mu_{1}} & x_{1}^{\mu_{2}} & \cdots & x_{1}^{\mu_{n}}\\ x_{2}^{\mu_{1}} & x_{2}^{\mu_{2}} & \cdots & x_{2}^{\mu_{n}}\\ \vdots & \vdots & \vdots & \vdots\\ x_{n}^{\mu_{1}} & x_{n}^{\mu_{2}} & \cdots & x_{n}^{\mu_{n}} \end{pmatrix}\not = 0$$

Answer 1

Of course it is possible. First look at the value of $\det\begin{pmatrix} x_{1}^{\mu_{1}} & x_{1}^{\mu_{2}} \\ x_{2}^{\mu_{1}} & x_{2}^{\mu_{2}} \end{pmatrix}$ as a function of $\bf x_2$. It is a polynomial, and hence may only have so many roots, so there must be quite a few integers which are not roots, and we pick any of them for our $x_2$. Then go on to $x_3$ and the corresponding $3\times3$ determinant, and repeat the same line of thought: as a function of $x_3$ it is a polynomial, hence...

You might wonder why did we need the $2\times2$ determinant. Well, that was to make sure that at least one coefficient of the next polynomial is non-zero.