Find all integers a, b, c such that a2 = bc + 4, b2 = ca + 4.

1.	Find all integers a, b, c such that a2 = bc + 4, b2 = ca + 4.
Answer» Suppose a = b. Then we get one equation: a² = ac + 4. This reduces to a(a - c) = 4. Therefore a = 1, a - c = 4, a = -1, a - c = -4, a = 4, a - c = 1, a = -4, a - c = -1, a = 2, a - c = 2, a = -2, a - c = -2. Thus we get (a, b, c) = (1, 1,-3), (-1,-1, 3), (4, 4, 3), (-4, -4,-3), (2, 2, 0), (-2, -2, 0). If a ≠ b, subtracting the second relation from the first we get a² - b²= c(b - a). This gives a + b = -c. Substituting this in the first equation, we get a² = b(-a - b) + 4. Thus a² + b² + ab = 4. Multiplication by 2 gives (a + b)² + a² + b² = 8. Thus (a, b) = (2, -2), (-2, 2), (2, 0), (-2, 0), (0, 2), (0, -2). We get respectively c = 0; 0, -2, 2, -2, 2. Thus we get the triples: (a, b, c) = (1, 1, -3), (-1,-1, 3), (4, 4, 3), (-4, -4, -3), (2, 2, 0),(-2,-2, 0), (2,-2, 0), (-2, 2, 0), (2, 0,-2), (-2, 0, 2), (0, 2,-2), (0,-2, 2).

Answer»

Suppose a = b.

Then we get one equation:

a² = ac + 4.

This reduces to a(a - c) = 4.

Therefore a = 1, a - c = 4,

a = -1, a - c = -4, a = 4, a - c = 1, a = -4, a - c = -1, a = 2, a - c = 2, a = -2, a - c = -2.

Thus we get

(a, b, c) = (1, 1,-3), (-1,-1, 3), (4, 4, 3), (-4, -4,-3), (2, 2, 0), (-2, -2, 0).

If a ≠ b,

subtracting the second relation from the first we get

a² - b²= c(b - a).

This gives a + b = -c.

Substituting this in the first equation, we get

a² = b(-a - b) + 4.

Thus a² + b² + ab = 4.

Multiplication by 2 gives

(a + b)² + a² + b² = 8.

Thus (a, b) = (2, -2), (-2, 2), (2, 0), (-2, 0), (0, 2), (0, -2).

We get respectively

c = 0; 0, -2, 2, -2, 2.

Thus we get the triples:

(a, b, c) = (1, 1, -3), (-1,-1, 3), (4, 4, 3), (-4, -4, -3), (2, 2, 0),(-2,-2, 0), (2,-2, 0), (-2, 2, 0), (2, 0,-2), (-2, 0, 2), (0, 2,-2), (0,-2, 2).

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