Suppose a = b.
Then we get one equation: a2 = ac + 1.
This reduces to a(a - c) = 1.
Therefore a = 1, a - c = 1, and a = -1, a - c = -1.
Thus we get (a, b, c) = (1, 1, 0) and (-1,-1, 0).
If a ≠ b,
subtracting the second relation from the rst we get
a2 - b2 = c(b - a).
This gives a + b = - c.
Substituting this in the rst equation, we get
a2 = b(- a - b) + 1.
Thus a2 + b2 + ab = 1.
Multiplication by 2 gives
(a + b)2 + a2 + b2 = 2:
Thus (a; b) = (1,-1), (-1,1), (1, 0), (-1, 0), (0, 1), (0,-1).
We get respectively c = 0, 0,-1, 1,-1, 1.
Thus we get the triples:
(a, b, c) = (1,1,0), (-1,-1, 0), (1,-1, 0), (-1, 1, 0), (1, 0,-1), (-1, 0, 1), (0, 1,-1), (0,-1, 1),