Equilateral triangles and rectangular grids
The other day, I encountered the following problem: “Does there exist an equilateral triangle whose vertices have integer (orthogonal Cartesian) coordinates?”. The problem is, indeed, as elementary as it looks, but in solving it, I had a few pleasant surprises; hence this note. I urge the reader to think about the problem before reading on.
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My first concern was how to characterize the notion “equilateral triangle”: all edges of the same length, all angles of the same size, or a mixture of the two. Considering that, in analytical geometry, lengths are more readily expressed than angles, I chose the first characterization, i.e. upon the triangle with with vertices (0,0), (a,b), (c,d) I imposed the requirement that its sides be of equal length. More precisely, my interest turned to integer quintuples (a,b,c,d,k) satisfying (0) ∧ (1) ∧ (2) with
| (0) | a² + b² = k |
| (1) | c² + d² = k |
| (2) | (a² − c)² + (b² − d)² = k . |
Using (0) and (1), (2) can be simplified to
| (2´) | 2·(a·c + b·d) = k . |
Seeing that k is even, we turn our attention in view of (0) and (1) to squares and factors of 2. And here we have to make a little jump. It does not suffice to observe (the correct) even.x² ≡ even.x, which for a sum of 2 squares only implies even.(x²+y²) ≡ even.x ≡ even.y. We have to reduce squares modulo 4 and use
| (3a) | x² mod 4 = 0 ≡ even.x |
| (3b) | x² mod 4 = 1 ≡ odd.x |
from which we conclude about the sum of squares
| (4a) | (x² + y²) mod 4 = 0 ≡ even.x ∧ even.y |
| (4b) | (x² + y²) mod 4 = 1 ≡ even.x ≢ even.y |
| (4c) | (x² + y²) mod 4 = 2 ≡ odd.x ∧ odd.y . |
And now we observe of an integer solution (a,b,c,d,k) of (0) ∧ (1) ∧ (2´):
k mod 4 ≠ 0
= {(2´), in particular even.k}
k mod 4 = 2
= {(0) and (1)}
(a²+b²) mod 4 = 2 ∧
(c²+d²) mod 4 = 2
= {(4c) twice}
odd.a ∧ odd.b ∧ odd.c ∧ odd.d
⇒ {arithmetic}
even.(a·c + b·d)
= {(2´)}
k mod 4 = 0 .
Having thus proved
| (5) | k mod 4 = 0 , |
we conclude on account of (0), (1), (4a)
| (6) | even.a ∧ even.b ∧ even.c ∧ even.d . |
If (a, b, c, d, k) is an integer solution of (0) ∧ (1) ∧ (2), (a/2, b/2, c/2, d/2, k/4) obviously solves that equation as well; (5) and (6) tell us now that the latter is again an integer solution. By mathematical induction we conclude that any solution of (0) ∧ (1) ∧ (2) is divisible by any power of 2 (4), and hence equal to (0,0,0,0,0), 0 being the only integer with an unbounded number of divisors. All zeros is indeed a solution, i.e. there exist equilateral triangles whose vertices have integer coordinates, but such a triangle is degenerate and its vertices coincide.
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For several reasons I was pleased with the above argument. Of course I regret the rabbit embodied by (3a)(3b), but I console myself with the thought that it is a fairly small one. The proof of k mod 4 = 0 is a nice example of proving P by constructing a weakening chain from ¬P to P, and the final part of the argument nicely exploits that only 0 has an unbounded number of divisors.
Remark The latter observation can be turned into a heuristic: a good way of showing that a Diophantine equation has no other solution than 0, it suffices to show that any solution p can be divided by some n (n>1) such that p/n is again a solution. (End of Remark).
Inks Lake State Park,
12 January 1994
prof.dr. Edsger W.Dijkstra
Department of Computer Sciences
The University of Texas at Austin
Austin, TX 78712-1188
USA