Two positive integers the difference of whose squares is 60 are ______ and ______ or ______ and _______

Answer

To solve the problem of finding two positive integers whose squares differ by 60, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables**:
Let the two positive integers be \( x \) and \( y \). According to the problem, we know that:
\[ x^2 - y^2 = 60 \]

2. **Use the Difference of Squares Formula**:
We can factor the left-hand side using the difference of squares:
\[ (x + y)(x - y) = 60 \]

3. **Identify Factor Pairs of 60**:
We need to find pairs of factors of 60 that can be expressed as \( (x + y) \) and \( (x - y) \). The factor pairs of 60 are:
- \( (1, 60) \)
- \( (2, 30) \)
- \( (3, 20) \)
- \( (4, 15) \)
- \( (5, 12) \)
- \( (6, 10) \)

4. **Consider Only Even Factor Pairs**:
Since both \( x + y \) and \( x - y \) must be even (as \( x \) and \( y \) are integers), we will only consider the even factor pairs:
- \( (2, 30) \)
- \( (6, 10) \)

5. **Solve for Each Pair**:
- **For the pair \( (2, 30) \)**:
\[ x + y = 30 \quad \text{and} \quad x - y = 2 \]
Adding these two equations:
\[ 2x = 32 \implies x = 16 \]
Substituting \( x = 16 \) into \( x + y = 30 \):
\[ 16 + y = 30 \implies y = 14 \]
So, one pair is \( (16, 14) \).

- **For the pair \( (6, 10) \)**:
\[ x + y = 10 \quad \text{and} \quad x - y = 6 \]
Adding these two equations:
\[ 2x = 16 \implies x = 8 \]
Substituting \( x = 8 \) into \( x + y = 10 \):
\[ 8 + y = 10 \implies y = 2 \]
So, the second pair is \( (8, 2) \).

6. **Final Answer**:
The two pairs of positive integers whose squares differ by 60 are:
\[ (16, 14) \quad \text{and} \quad (8, 2) \]