How to explain IDFT of square root of a DFT?
Since $V(\omega)V(\omega)=W(\omega)$, in the time domain we expect $v[n]*v[n]=w[n]$. That is, the convolution of $v[n]$ with itself should become $w[n]$.

$w[n]$ is in the form of a triangular signal. A triangular signal of odd length $N$ can be constructed by convolving a half-rectangular signal of length $(N+1)/2$ with itself. So $v[n]$ is a rectangular pulse with $3$ elements equal to one and the remaining $2$ elements equal to zero.