Let $C\subset\mathbb{P}^n$ be a rational normal curve of degree $n$. It is know that $C$ is a set-theoretic complete intersection and that, if $n\geq 3$, is a not a scheme-theoretic complete intersection.


I would like to know if it is true that, if $n\geq 3$ then $C$ can not be written as a set-theoretic complete intersection of $n-1$ quadrics.


Indeed, if $C$ can be written as a set-theoretic complete intersection of $n-1$ quadrics then we have $2^{n-1} = kn$ for some integer $k$. That is $\frac{2^{n-1}}{n}$ is an integer.


Therefore if $\frac{2^{n-1}}{n}$ is not an integer the answer to my question should be positive. What happens if $\frac{2^{n-1}}{n}$ is an integer ?


For instance, if $n = 4$ do there exist three irreducible quadrics cutting out the rational normal curve of degree four twice ?



Rational normal curves as set-theoretic complete intersections
If $\frac{2^{n-1}}{n}$ is an integer, that is if $n$ is a power of $2$, then $C$ is actually the set-theoretic complete intersection of $n-1$ quadrics.
This is a theorem of Perron (1941); to see the defining equations see formula (3) of my paper <http://www.dima.unige.it/~torrente/RationalNormalCurves.pdf>.
Furthermore, for a general degree $n$, Gallarati and Rollero (1988) proved that $C$ can be written as the set-theoretic complete intersection of $s-1$ quadrics and $n-s$ forms of degrees $s+1,\ldots,n$, where $s=\max\{k \in \mathbb N : 2^k\le n\}$. 


