If the value of `ax^2+bx+c` is always an integer for integral values of `x` where a, b, c are rationals then a+b must be an integer.

Options

A. an integer.
B. a negative integer
C. a positive integer
D. none of these

Answer

To prove that if the polynomial \( ax^2 + bx + c \) takes integer values for all integral values of \( x \), then \( a + b \) must be an integer, we can follow these steps:

### Step 1: Evaluate the polynomial at specific integral values
We will evaluate the polynomial at \( x = -1, 0, \) and \( 1 \).

1. **At \( x = 0 \)**:
\[ f(0) = c \]
Since \( c \) is rational and \( f(0) \) must be an integer, we conclude that \( c \) is an integer.

2. **At \( x = -1 \)**:
\[ f(-1) = a(-1)^2 + b(-1) + c = a - b + c \]
Since \( f(-1) \) must also be an integer, we have \( a - b + c \) is an integer.

3. **At \( x = 1 \)**:
\[ f(1) = a(1)^2 + b(1) + c = a + b + c \]
Similarly, \( a + b + c \) is an integer.

### Step 2: Analyze the results
From the evaluations, we have:
- \( c \) is an integer.
- \( a - b + c \) is an integer.
- \( a + b + c \) is an integer.

### Step 3: Simplify the expressions
We can rearrange the equations:
1. From \( f(-1) \):
\[ a - b + c = k_1 \quad \text{(where \( k_1 \) is an integer)} \]
Rearranging gives:
\[ a - b = k_1 - c \]

2. From \( f(1) \):
\[ a + b + c = k_2 \quad \text{(where \( k_2 \) is an integer)} \]
Rearranging gives:
\[ a + b = k_2 - c \]

### Step 4: Add the two equations
Now, we add the two equations:
\[ (a - b) + (a + b) = (k_1 - c) + (k_2 - c) \]
This simplifies to:
\[ 2a = (k_1 + k_2) - 2c \]
Since \( k_1 \) and \( k_2 \) are integers and \( c \) is an integer, the right-hand side is an integer. Therefore, \( 2a \) is an integer, which implies that \( a \) is a rational number such that \( 2a \) is an integer.

### Step 5: Conclude about \( a + b \)
Now we can express \( b \) in terms of \( a \):
\[ b = (k_2 - c) - a \]
Substituting \( a \) from the previous step, we can see that \( a + b \) can be expressed as:
\[ a + b = (k_2 - c) - a + a = k_2 - c \]
Since \( k_2 \) and \( c \) are integers, \( a + b \) must also be an integer.

### Conclusion
Thus, we conclude that if \( ax^2 + bx + c \) is always an integer for integral values of \( x \), then \( a + b \) must be an integer.