For any integers a and b with HCF (a, b) = 1 , what is HCF (a +b, a - b)equal to?

Options

A. It is always 1
B. It is always 2
C. Either 1 or 2
D. None of the above

Answer

To solve the problem, we need to find the HCF (Highest Common Factor) of the expressions \(a + b\) and \(a - b\) given that \(HCF(a, b) = 1\).

### Step-by-Step Solution:

1. **Understanding HCF**:
The HCF of two numbers is the largest integer that divides both numbers without leaving a remainder. Since we know that \(HCF(a, b) = 1\), it means that \(a\) and \(b\) are coprime (they have no common factors other than 1).

**Hint**: Recall that coprime numbers have no common factors other than 1.

2. **Expressing the HCF**:
We need to find \(HCF(a + b, a - b)\). We can use the property of HCF that states:
\[ HCF(x, y) = HCF(x, y + kx) \]
for any integer \(k\). This means we can manipulate the expressions to find their HCF.

**Hint**: You can use properties of HCF to simplify the expressions.

3. **Applying the Property**:
Let's apply the property:
\[ HCF(a + b, a - b) = HCF(a + b, (a - b) + (a + b)) = HCF(a + b, 2a) \]
Now we need to find \(HCF(a + b, 2a)\).

**Hint**: Simplifying the HCF using addition or subtraction can help in finding the result.

4. **Finding HCF with Coprime Numbers**:
Since \(HCF(a, b) = 1\), we know that \(a\) and \(b\) do not share any common factors. Therefore, \(a + b\) and \(a\) will also not share any common factors with \(b\) unless \(b\) is even.

- If both \(a\) and \(b\) are odd, then \(a + b\) is even and \(a - b\) is even, hence \(HCF(a + b, a - b) = 2\).
- If one of them is even and the other is odd, then \(HCF(a + b, a - b) = 1\).

**Hint**: Consider the parity (odd/even nature) of \(a\) and \(b\) to determine the HCF.

5. **Conclusion**:
Therefore, \(HCF(a + b, a - b)\) can be either \(1\) or \(2\) depending on whether \(a\) and \(b\) are both odd or one is even and the other is odd.

**Final Answer**: \(HCF(a + b, a - b) = 1 \text{ or } 2\).