For a convex polytope $\mathcal K$ in $\Bbb R^n$ presented by $O(n^c)$ linear inequalities is it true that for $|\mathcal K\cap \Bbb Z^n|=0$ it is necessary that at least one axis of John's ellipsoid have length smaller than $1$? 



On necessary condition for no integer points in polytope
The answer is $no$.


In the plane, the circle of diameter $d$ greater than $1$ but smaller than $\sqrt2$ and centered at $({1\over2},{1\over2})$ contains no point with integer coordinates, and the square $K\_2$ circumscribed about it with diagonals parallel to the coordinate axes contains no such point either. Let $K\_3\subset\mathbb{R}^3$ be the Cartesian product $K\_2\times[0,d]$. Then $K\_3$ contains no lattice point, the $2$-sphere inscribed in $K\_3$ is of diameter $d$, and the number of facets of $K\_3$ is $6$, since, in general, the product $K\times[0,1]$ has two more facets than $K$. Iterating the product procedure, we get $K\_n=K\_2\times[0,d]^{n-2}\subset\mathbb{R}^n$ with $2n$ facets, containing no lattice point but containing a sphere of diameter $d>1$. 


Similarly, a sequence of polytopes $K\_{k,n}$ ($k$ fixed, $n=k, k+1, k+2, \ldots$) can be constructed that have the same properties as above, with $d$ arbitrarily close to $\sqrt k$. This shows that for any constant $d$ in place of $1$ in the question asked here, the answer is still in the negative. Namely, begin with $K\_k$ to be the cross-polytope in $\mathbb{R}^k$ circumscribed about the $k$-sphere of diameter slightly smaller than $\sqrt k$ and centered at $({1\over2},{1\over2}, \ldots,{1\over2})$ and with main diagonals (the "cross") parallel to the coordinate axes, then proceed as above. The number of facets of $K\_{k,n}$ will be $2^k + 2(n-k)$, which is $O(2n)$.


