Question about decomposing an equilateral triangle into two triangles?

The height of the triangle is h where

h^2=(2n)^2-n^2=3n^2.

Also, m^2=h^2+1^2.

Therefore

m^2=3n^2+1

or

m^2-3n^2=1.

This is Pell's Equation (see Source). The smallest solution is trivial : (m,n) = (1,0). The smallest non-trivial solution (the fundamental solution) is (2,1). Then comes (7,4) as stated in the question.

All other solutions can be found from the coupled recurrence relations

m(k+1)=am(k)+nbn(k)

n(k+1)=bm(k)+an(k)

where (a,b)=(m0,n0) is the fundamental solution.

Here (a,b)=(2,1) so the next solutions are

m1=2*2+3*1*1=7

n1=1*2+2*1=4

m2=2*7+3*1*4=26

n2=1*7+2*4=15.

Excellent answer and thumbs up for Barry G! All solutions of the Pell's Equation m² - 3n² = 1, starting from (7, 4) exhaust the question.

Yet I decided to contribute a little bit, because I noticed some points of interest. Imagine not single Cevian as on Rita's picture above, but all 3, whose endpoints divide each side in ratio r = (n - 1)/(n + 1) as shown below. Trivially follows that the green triangle is also equilateral and the areas of the initial and the internal triangles are in ratio

(r² + r + 1) / (r - 1)²

http://en.wikipedia.org/wiki/Routh%27s_theorem

For r = (n - 1)/(n + 1) the above expression yields (3n² + 1)/4, what is integer for all n - odd, but if n is an odd solution of the cited Pell's Equation, then

3n² + 1 = m² with m - even (m and n have different parity in each solution), so we have in fact proven the following statement:

Given the conditions in the initial question, the ratio of areas of both equilateral triangles is also integer (and even a perfect square - m²/4) for every solution

(m, n) of m² - 3n² = 1 with m - even and n - odd.

First and second such non-trivial solutions are (26, 15) and (362, 209); for (2, 1) both triangles coincide of course.