A positive definite quadratic form $q(x)=x^TQx$ (assume $Q$ is symmetric) is said to be Minkowski reduced if for all $1\leq k\leq n$ we that any $x\in \mathbb{Z}^n$ with $GCD(x_k,\ldots,x_n)=1$ has $q(x)\geq q(e_k)$ .
I wonder if for any Minkowski reduced form we can replace $GCD(x_k,\ldots,x_n)=1$ by $GCD(x_k,\ldots,x_n)\geq 1$? In other words, is there a positive definite quadratic form of any dimension that is Minkowski reduced but such that the following does not hold: for any $x\in \mathbb{Z}^n$ with $GCD(x_k,\ldots,x_n)\geq 1$ we have $q(x)\geq q(e_k)$ for all $1\leq k\leq n$?
I think I have been able to show that such an example does not exist. However, based on a few things I have read, I have a suspicion that they do exist.
The lattice of the following basis is Minkowski reduced,
$A=\begin{bmatrix}1 & 0 & 0 & 0 & 1/2\\ 0 & 1 & 0 & 0 & 1/2\\ 0 & 0 & 1 & 0 & 1/2\\ 0 & 0 & 0 & 1 & 1/2\\ 0 & 0 & 0 & 0 & 1/2 \end{bmatrix}.$
However, $e_5\in A\mathbb{Z}^n$ despite the fact that $|e_5|< a_5$ where $a_5$ is the fifth vector of $A$. This is therefore a counterexample.