Answer to Question #100209 in Geometry for Mark

Question

Does there exist an equilateral triangle (in the (x,y)-plane) all three of whose vertices are at lattice points?

Answer 1

Answer: no

Explanation:

The first result is the inability to find a regular triangle on the integer lattice Z² apparently was proved by E. Lucas in 1878.

At the heart of his evidence basic information from theory and now we repeat.

Theorem 2.1. The regular triangle cannot be positioned on the integer lattice Z².

Proof of (I).

Suppose that any regular triangle can be positioned on the lattice in the right way and that the origin is at one of its vertices, and the other two

vertices have coordinates (a, b) and (c, d). We can assume that four integers a, b, c, d do not have common divisors other than ± 1.

The latter follows from the fact that the points (0, 0), (a / k, b / k),

(c / k, d / k) are also vertices of a regular triangle if k is a common divisor of all four numbers.

As

a² + b² = c² + d² = (a − c)² + (b − d)²

we conclude that

a² + b² = c² + d² = 2(ac + bd)

Consequently,

a^2 + b² + c² + d² = 4(ac + bd)

that is, the sum of the squares of four numbers is divisible by 4. But then either all four numbers are even, or all are odd.

The first is impossible because these numbers, according to our choice, are mutually simple.

The second is impossible because then the formula

a² + b² = (a − c)² + (b − d)²

is not satisfied because its left part is not divisible by 4, and the right part is divisible. The contradiction occurs

and it proves the stated statement.