In Lemma 2 of [1], Heath-Brown proves the following (I state a simplified version of a more general result):



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> Let $\Lambda \subset \mathbb{Z}^2$ be a lattice of determinant $d(\Lambda)$. Then
>  $$\# \{ (x\_1,x\_2) \in \Lambda: \max\_i |x\_i| \leq B, \gcd(x\_1,x\_2) = 1\} \leq 16\left (\frac{B^2}{d(\Lambda)} + 1\right).$$
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My question is whether this generalises to arbitrary dimensions.



> 
> Does an analogous result hold for lattices in $\Lambda \subset \mathbb{Z}^n$? Namely, is 
>  $$\# \{ (x\_1,\dots, x\_n) \in \Lambda: \max\_i |x\_i| \leq B, \gcd(x\_1,\dots,x\_n) = 1\} \leq C\_n\left (\frac{B^n}{d(\Lambda)} + 1 \right)$$
>  for some constant $C\_n$? 
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If it helps, I'm primarily interested in the case $n=3$.


Obviously I'm aware of standard lattice point counting techniques, but these usually give an error term of the shape $O(\text{boundary of the region/first successive minima})$, and I don't know how to control this in my case. So I'm just looking for uniform upper bounds where this term doesn't appear.


[1] Heath-Brown - Diophantine approximation with Square-free numbers



Counting primitive lattice points
No, there is no result in this form because in dimension 3 or higher it is allowed to have some non-first minima relatively small even when the first minimum is very small.


For example, for any $N>0 $ consider the lattice $$ \Lambda\_N = \frac 1 N \mathbb Z \times \mathbb Z \times \mathbb Z^{n-3}\times N\mathbb Z$$
with unit determinant. Then the set $$ S\_N = \left\{\left(\frac k N,1 , 0,\ldots, 0\right):\ -N\leq k\leq N\right\}$$
is a set of $2N+1$ primitive vectors of $\Lambda\_N$ contained in the unit hypercube. 


Now, to answer your question, take $B=N$ prime and the lattices $N\Lambda\_N\subset \mathbb Z^n$ to see that there is no such constant $C\_n$.


