At most three squares can be formed from seven points in the plane. First proof (Hugo van der Sanden, Sep 28 2021) Fix 4 points on the standard unit square, and wlog assume this is the smallest square in the final combination. We do not assume that the remaining points have integer coordinates. There are now four possibilities. A) There's another square of the same size, sharing two points with the first. Then the 7th point can only give us a third square in two symmetrically placed ways, of side sqrt(2). No fourth is possible. B) There's another square of the same size, sharing one point with the first. Then all 7 points are accounted for. Any third square must use two points from each of the existing two squares, so again can only have a side of sqrt(2); that is possible, but there can't be a fourth. C) There's another square of a different size, sharing two points with the first; this can only have side sqrt(2). The only ways to select three of those points such that the 7th point can be added to give a square in each case reduce us to case A or B. D) There's another square of a different size, sharing one point with the first. All points are accounted for, and any additional square must share two points with each of the first two squares; but then any such third square must reduce us to case A or C. So the possibilities are: .XX XXX XX. and .X. XXX XXX Second proof (Benoit Jubin, Sep 29 2021) Another proof that a(7) = 3 (that is, a planar figure with 7 points contains at most 3 squares, and there are figures realizing this number): First, note that two different squares cannot share three vertices. If a 7-point figure contains (at least) 3 squares, then by simple counting there is a pair of squares sharing two vertices. There are only two possibilties: (1) they share an edge; (2) an edge of one of them is a diagonal of the other. None of these 6-point figures contain a third square. Therefore, if the 7-point figure contains other squares, they are formed from a 7th point, and three points in the 6-point figure (and not all three belonging to the same square of that figure, by the preliminary remark). The triples of points in these 6-point figures that are in square-position are few: two (one up to symmetry) in case (1), and three (two up to symmetry) in case (2). They lead to the two 7-point 3-square configurations (since one of the cases in (2) yields the same figure as the one from (1)). None of these two figures contain a fourth square. QED