Square In A Haystack

The lattice points of this triangular grid can be represented by integer pairs $(a,b)$ , where the distance between $(a,b)$ and $(c,d)$ by cosine rule is

$(a-c)^2 + (b-d)^2 - 2 (a-c)(b-d) \cos 60^\circ = (a-c)^2 + (b-d)^2 - (a-c)(b-d).$

To show that we cannot find 4 points which form a square, we will show that there does not exist 3 distinct points $X = (a,b), Y = (c,d), Z = (e,f)$ such that $|XY| = |YZ|$ and $XY \perp YZ$ .

Proof by contradiction. Suppose not, then by comparing lengths $|XY|$ and $|XZ|$ , we have
$\sqrt{2} = \frac{ |XZ| } {|XY| } = \frac{ (a-e)^2 + (b-f)^2 - (a-e)(b-f) } { (a-c)^2 + (b-d)^2 - (a-c)(b-d)}$ Since these points are distinct, notice that the numerator and denominator on the right-hand side are non-zero. Also, since $a, b, c, d, e$ and $f$ are integers, thus the numerator and denominator are integers. Hence, we have a representation of $\sqrt{2}$ as a rational numbers. This contradicts the fact that $\sqrt{2}$ is irrational .

For any two lattice points on this grid, a third point which forms equilateral triangle with these two points is also a lattice point. (The proof of this should be easy so I am not showing it)
Now suppose there are squares of lattice points on this plane.
Pick the smallest one.
However, we can construct a smaller one base on this square:

which leads to a contradiction to the 'smallest' assumption.
So there does not exist squares of lattice points on this plane.

Let the sides of the equilateral triangle be $a$ .

Now Consider 2 axes mutually perpendicular to each other because the angle contained by the sides of a square is 90°

Let one axis be along the sides of the triangles(X axis), while the other be along the height of the equilateral triangles(Y axis)

To go from one lattice point to other on the X axis, we must go in multiples of $a$

Therefore, one side of the square is $n.a$ for integral $n$ .

Similarly, to move from one lattice point to the other on the Y axis, we must go in multiples of $\sqrt{3}a$

Therefore, another side of the square is $m.\sqrt {3}a$ for integral $m$ .

But the sides of a square are equal, which implies $m.\sqrt{3}.a = n.a$

Therefore $\frac{n}{m}$ = $\sqrt{3}$ .

But $n$ and $m$ are integral and thus LHS is rational, however RHS is irrational.

Hence, it is impossible to embed a square on this lattice.

P.S. : For those wondering why to take two integers $m$ and $n$ , consider a rectangle with dimensions 2x3 You totally need 6 rectangles to form a square(6x6).

Let $l$ be the side of the triangle. We know that its area is $A=\dfrac{l^2\sqrt{3}}{4}$ . For any two points we choose, we can determine the square's side. Letting $n$ be the amount of triangles between the two chosen points, $nl$ will be the square's side, with $n \in \mathbb{Z}$ . If it were possible to form a square choosing any four points among the vertices, we would have that $mA=n^2l^2$ , where $n^2l^2$ is the area of the square and $m$ represents the amount of triangles inside the square and has to be, necessarily, an integer.

$m\dfrac{l^2\sqrt{3}}{4}=n^2l^2$

$m=\dfrac{n^2 4\sqrt{3}}{3}$

We can clearly see that $m$ isn't an integer and, therefore, a whole number of triangles can't fill the square.

It is not possible to form any square by taking any four points, it will lead to the formation of quadrilateral whose two adjacant angle will be ( $60°$ and $120°$ ) or $90°$ , and the resulting quadrilateral formed will be either $(rhombus$ or $parallelogram$ or $trapezium$ ) or $rectangle$

for further clarification you may ask any questions