
COMMENTS

A closetoequilateral integer triangle is defined to be a triangle with integer sides and integer area such that the largest and smallest sides differ in length by unity. The first five closetoequilateral integer triangles have sides (5, 5, 6), (17, 17, 16), (65, 65, 66), (241, 241, 240) and (901, 901, 902).
Next four terms are {three sides a<=b<=c and area}: {46816, 46817, 46817, 949077360}, {174725, 174725, 174726, 13219419708}, {652080, 652081, 652081, 184120982760}, {2433601, 2433601, 2433602, 2564481115560}. Also, if we allow degenerate triangles (area 0), the first case would be {1,1,2,0}. We have 12 cases and a weak conjecture is that the total number of the 'closetoequilateral' triangles is finite.  Zak Seidov, Feb 23 2005
This is an infinite series; two sides are equal in length to the hypotenuse of almost 3060 triangles and the third side alternates between that length +/ 1.  Dan Sanders (dan(AT)ified.ca), Oct 22 2005
Heron's formula: a triangle with side lengths (x,y,z) has area A = sqrt(s*(sx)*(sy)*(sz)) where s = (x+y+z)/2. For this sequence we assume integer sidelengths x = y = z +/ 1. Then for A to also be an integer, x+y+z must be even, so we can assume z = 2k for some positive integer k. Now s = (x+y+z)/2 = 3k +/ 1 and A = sqrt((3*k +/ 1)*k*k*(k +/ 1)) = k*sqrt(3*k^2 +/ 4*k + 1). To determine when this is an integer, set 3*k^2 +/ 4*k + 1 = d^2. If we multiply both sides by 3, it is easier to complete the square: (3*k +/ 2)^2  1 = 3*d^2. Now we are looking for solutions to the Pell equation c^2  3*d^2 = 1 with c = 3*k +/ 2, for which there are infinitely many solutions: use the upper principal convergents of the continued fraction expansion of sqrt(3) (A001075/A001353).  Danny Rorabaugh, Oct 16 2015
