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Find all integers a, b, c such that a2 = bc + 1, b2 = ca + 1.

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Suppose a = b. Then we get one equation a2 = ac + 1. This reduces to a(a – c) = 1. Therefore 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 first we get

a2 - b2 = c(b - a).

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 = 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).

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