It is impossible to build equilateral triangle with all vertices on lattice
(just consider rotation of any lattice point around the origin by $60^{\circ}$).Denote defect of triangle $\triangle ABC$ as $$d_{\triangle ABC} = d_A + d_B + d_C,$$ where defect of the vertex $A$ is $$d_{A}= |AB|^2 - |AC|^2.$$
The smallest defect of the triangle on lattice is $2$.
Let call such triangles "almost equilateral triangles on lattice" (AETL).
Possible combinations of side lengths of AETLs are:
$(\sqrt{n}, \sqrt{n}, \sqrt{n+1})$ and $(\sqrt{n-1}, \sqrt{n}, \sqrt{n})$, where $n\in\mathbb{N}$.
- I see only $2$ (infinite) sets of such triangles:
the set $S_1$: \begin{array}{|c|c|} \hline side \; lengths & example\; of\; triangle \; vertices\; \\ \hline \sqrt{2k^2-1}, \sqrt{2k^2-1}, k\sqrt{2} & (0,0) \; (a,b) \; (b,a) \\ \hline 1,1,\sqrt{2} & (0,0)\; (1,0)\; (0,1) \\ \sqrt{17},\sqrt{17},3\sqrt{2} & (0,0)\; (4,1)\; (1,4) \\ \sqrt{241},\sqrt{241},11\sqrt{2} & (0,0)\; (15,4)\; (4,15) \\ \sqrt{3361},\sqrt{3361},41\sqrt{2} & (0,0)\; (56,15)\; (15,56) \\ \sqrt{46817},\sqrt{46817},153\sqrt{2} & (0,0)\; (209,56)\; (56,209) \\ \cdots & \cdots \\ \hline \end{array} where $k_{j+1}=4k_j-k_{j-1}$; ($a_{j+1},b_{j+1} -$ similar way);
the set $S_2$: \begin{array}{|c|c|} \hline side \; lengths & example\; of\; triangle \; vertices\; \\ \hline k,\sqrt{k^2+1}, \sqrt{k^2+1} & (0,0) \; (a,b) \; (a,-b) \\ \hline 2,\sqrt{5},\sqrt{5} & (0,0)\; (2,1)\; (2,-1) \\ 8,\sqrt{65},\sqrt{65} & (0,0)\; (7,4)\; (7,-4) \\ 30,\sqrt{901},\sqrt{901} & (0,0) (26,15) (26,-15) \\ 112,\sqrt{12545},\sqrt{12545} & (0,0) (97,56) (97,-56) \\ 418,\sqrt{174725},\sqrt{174725} & (0,0) (362,209) (362,-209) \\ \cdots & \cdots \\ \hline \end{array} where $k_{j+1}=4k_j-k_{j-1}$; ($a_{j+1},b_{j+1} -$ similar way).
Question: is there almost equilateral triangle on lattice with sides which are not included to described sets?
My suggested approach: move one of the vertices to the origin; then draw the circle with radius $\sqrt{n}$, and consider distances between lattice points of the circle. But I'm not sure that other solutions exist at all.
