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 \( y \). We know from the problem statement that:
\[ x^2 - y^2 = 60 \]

2. **Factor the Difference of Squares:**
We can factor the left-hand side using the difference of squares formula:
\[ x^2 - y^2 = (x + y)(x - y) \]
Therefore, we can rewrite the equation as:
\[ (x + y)(x - y) = 60 \]

3. **Identify Factor Pairs of 60:**
Since \( x + y \) and \( x - y \) are both positive integers, we need to find pairs of factors of 60. The factor pairs of 60 are:
- (1, 60)
- (2, 30)
- (3, 20)
- (4, 15)
- (5, 12)
- (6, 10)

4. **Set Up Equations:**
For each factor pair \( (a, b) \), we can set:
\[ x + y = a \quad \text{and} \quad x - y = b \]
We can solve for \( x \) and \( y \) using these equations:
\[ x = \frac{(x + y) + (x - y)}{2} = \frac{a + b}{2} \]
\[ y = \frac{(x + y) - (x - y)}{2} = \frac{a - b}{2} \]

5. **Check Each Factor Pair:**
We will check each factor pair to ensure \( x \) and \( y \) are positive integers.

- For \( (2, 30) \):
\[ x + y = 30, \quad x - y = 2 \]
Adding these gives:
\[ 2x = 32 \quad \Rightarrow \quad x = 16 \]
Substituting back to find \( y \):
\[ y = 30 - 16 = 14 \]
So one solution is \( (16, 14) \).

- For \( (6, 10) \):
\[ x + y = 10, \quad x - y = 6 \]
Adding these gives:
\[ 2x = 16 \quad \Rightarrow \quad x = 8 \]
Substituting back to find \( y \):
\[ y = 10 - 8 = 2 \]
So another solution is \( (8, 2) \).

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