Consider the set $\mathbf{Z}^d.$ We can give $\mathbf{Z}^d$ different metrics, consider the metrics $\|x\|_p$ for $p = 1, 2, \infty,$ where $\|x\|_p$ is the $p$-norm of $\mathbf{R}^d $ restricted to $\mathbf{Z}^d$; that is $$\|(x_1, \ldots, x_d)\|_p = \left( \sum_{k = 1}^d |x_k|^p \right)^{\frac{1}{p}} (1 \leq p < \infty); \hspace{1cm} \|(x_1, \ldots, x_d)\|_\infty = \sup(x_1, \ldots, x_d).$$ Denote by $\mathrm{S}_{\mathbf{Z}^d,p}(r)$ the sphere of centre $0$ and radius $r > 0$ in $\mathbf{Z}^d$ with respect to the $p$-norm. It is clear that for $p = 1$ and $p = \infty,$ said sphere is empty when $r$ is not an integer. This is not true for $p = 2$ as the case $r = \sqrt{2}$ and $d = 2$ shows: $\|(1,1)\|_2 = \sqrt{2}.$ It is easy to see that the subset $\mathrm{J}_{d, p}$ of $\mathbf{R}$ of possible radius for which the sphere is not empty in $\mathbf{Z}^d$ with respect to the $p$-norm is countable (as stated before, it is $\mathbf{N}$ for $p = 1, \infty$).
Question. Is the set $\mathrm{J}_{d, 2}$ consisting solely of isolated points? Or does it possess a cluster value? More refined, is there an $\varepsilon > 0$ such thatany two points in $\mathrm{J}_{d, 2}$ are at least $2$-norm distance $\varepsilon$ apart? If such an $\varepsilon$ does exist, does it depend on the dimension $d$ or it can be taken universal for all the lattices $(\mathbf{Z}^d, \|x\|_2)$ simultaneously ($d \in \mathbf{N}$)?
Yes. For any interval $(a,b)$, the intersection $(a,b)\cap\mathrm{J}_{d, 2}$ contains only finitely many points, because $\mathbb Z^d$ has only finitely many points inside the (solid) ball of radius $b$.
No. With large $n$, we can make $\sqrt{n^2+1}-\sqrt{n^2}=\frac12n^{-1}+O(n^{-3})$ arbitrarily small.