First you have to prove the identity b²=a²+ac for such triangles. It can be proved using the similar triangles in the diagram or alternatively from the Sine and Cosine Rules.

Method 1 By construction DABX is an isosceles triangle. The angle of this triangle at B is 180^° - 2a, and hence the angles at A and X are a. Therefore DXAC and DABC are similar; hence

a b = BC AC = AC XC = b a+c ,

and thus b²=a²+ac.

Method 2 By the Cosine Rule:

b2 = a2 + c2 -2ac cos2a a2 = b2 + c2 -2bccos a

Subtracting the equations, using the double angle formula and rearranging them a bit we get :

2b2 - 2a2 = 2c(bcosa- acos2a+a).    (1)

By the Sine Rule:

a sina = b sin2a a sin a = b 2sinacosa cosa = b 2a .     (2)

Combining (1) and (2):

2b2 - 2a2 = 2c( b2 2a - 2ab2 4a2 + a) = 2c( b2 2a - b2 2a + a) b2 - a2 = ca b2 = a2 + ac.

To find integer solutions, we may assume that a, b and c have no common factors. Then a and a+c have no common factors (for if p divides a and a+c then it divides c, and hence also b). As a(a+c)=b², and as a and a+c are coprime, a and a+c must be perfect squares. Let a=u², a+c=v² so that

a=u2,     b=uv,     c=v2-u2.

Note that not every solution gives rise to a triangle (for all three triangle inequalities must be satisfied); for example, u=1 and v=4 gives (a,b,c) = (1,4,15) and this does not give a triangle. The following are a few examples of values which do give triangles:

Note: it can be proved that a, b and c give a triangle if and only if v > u > v/2; see Math. Gazette, Vol. 412, June 1976, p.130.