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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, 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),

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