To solve the equations \( a^2 = bc + 1 \) and \( b^2 = ca + 1 \) for integers \( a, b, c \), we can follow these steps:
### Step 1: Write down the equations
We have two equations:
1. \( a^2 = bc + 1 \) (Equation 1)
2. \( b^2 = ca + 1 \) (Equation 2)
### Step 2: Assume \( a \neq 0 \) and \( b \neq 0 \)
To simplify our calculations, we will assume \( a \) and \( b \) are not equal to zero.
### Step 3: Multiply Equation 1 by \( a \)
Multiply both sides of Equation 1 by \( a \):
\[
a^3 = abc + a
\]
Let this be Equation 3.
### Step 4: Multiply Equation 2 by \( b \)
Multiply both sides of Equation 2 by \( b \):
\[
b^3 = abc + b
\]
Let this be Equation 4.
### Step 5: Subtract Equation 4 from Equation 3
Now, subtract Equation 4 from Equation 3:
\[
a^3 - b^3 = a - b
\]
The \( abc \) terms cancel out.
### Step 6: Factor the left-hand side
Using the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), we can rewrite the equation:
\[
(a - b)(a^2 + ab + b^2) = a - b
\]
### Step 7: Analyze the cases
1. **Case 1:** If \( a - b \neq 0 \), we can divide both sides by \( a - b \):
\[
a^2 + ab + b^2 = 1
\]
Let this be Equation 5.
2. **Case 2:** If \( a - b = 0 \), then \( a = b \).
### Step 8: Solve Equation 5
From Equation 5:
\[
a^2 + ab + b^2 = 1
\]
Substituting \( b = a \):
\[
a^2 + a^2 + a^2 = 1 \implies 3a^2 = 1 \implies a^2 = \frac{1}{3}
\]
This does not yield integer solutions.
### Step 9: Consider \( a = b \)
If \( a = b \), substitute back into the original equations:
From Equation 1:
\[
a^2 = ac + 1 \implies c = \frac{a^2 - 1}{a}
\]
This means \( c \) must also be an integer.
### Step 10: Analyze integer values of \( a \)
1. If \( a = 1 \):
\[
c = \frac{1^2 - 1}{1} = 0
\]
Thus, \( (a, b, c) = (1, 1, 0) \).
2. If \( a = -1 \):
\[
c = \frac{(-1)^2 - 1}{-1} = 0
\]
Thus, \( (a, b, c) = (-1, -1, 0) \).
### Step 11: Check other integer combinations
Now, check if there are other integer combinations for \( a \) and \( b \) that satisfy \( a^2 + ab + b^2 = 1 \). The only integers that satisfy this equation are \( (1, -1) \) and \( (-1, 1) \), both yielding \( c = 0 \).
### Conclusion
The integer solutions for \( (a, b, c) \) are:
1. \( (1, -1, 0) \)
2. \( (-1, 1, 0) \)
3. \( (1, 1, 0) \)
4. \( (-1, -1, 0) \)
