To solve the problem of finding how many pairs of natural numbers \( x \) and \( y \) satisfy the equation \( x^2 - y^2 = 60 \), we can follow these steps:
### Step 1: Use the difference of squares formula
The difference of squares can be factored as:
\[
x^2 - y^2 = (x + y)(x - y)
\]
So we can rewrite the equation as:
\[
(x + y)(x - y) = 60
\]
### Step 2: Factor 60
Next, we need to find all the pairs of factors of 60. The pairs of factors (including both positive and negative) are:
- \( (1, 60) \)
- \( (2, 30) \)
- \( (3, 20) \)
- \( (4, 15) \)
- \( (5, 12) \)
- \( (6, 10) \)
### Step 3: Set up equations for each factor pair
For each pair of factors \( (a, b) \) where \( a = x + y \) and \( b = x - y \), we can set up the following equations:
1. \( x + y = a \)
2. \( x - y = b \)
### Step 4: Solve for \( x \) and \( y \)
From the above equations, we can solve for \( x \) and \( y \):
\[
x = \frac{(x + y) + (x - y)}{2} = \frac{a + b}{2}
\]
\[
y = \frac{(x + y) - (x - y)}{2} = \frac{a - b}{2}
\]
### Step 5: Check each factor pair
Now we will check each factor pair to see if they yield natural numbers for \( x \) and \( y \):
1. **For \( (1, 60) \)**:
- \( x = \frac{1 + 60}{2} = 30.5 \) (not a natural number)
- \( y = \frac{1 - 60}{2} = -29.5 \) (not a natural number)
2. **For \( (2, 30) \)**:
- \( x = \frac{2 + 30}{2} = 16 \)
- \( y = \frac{2 - 30}{2} = -14 \) (not a natural number)
3. **For \( (3, 20) \)**:
- \( x = \frac{3 + 20}{2} = 11.5 \) (not a natural number)
- \( y = \frac{3 - 20}{2} = -8.5 \) (not a natural number)
4. **For \( (4, 15) \)**:
- \( x = \frac{4 + 15}{2} = 9.5 \) (not a natural number)
- \( y = \frac{4 - 15}{2} = -5.5 \) (not a natural number)
5. **For \( (5, 12) \)**:
- \( x = \frac{5 + 12}{2} = 8.5 \) (not a natural number)
- \( y = \frac{5 - 12}{2} = -3.5 \) (not a natural number)
6. **For \( (6, 10) \)**:
- \( x = \frac{6 + 10}{2} = 8 \)
- \( y = \frac{6 - 10}{2} = -2 \) (not a natural number)
### Step 6: Valid pairs
After checking all the pairs, we find valid pairs:
- For \( (30, 2) \):
- \( x = \frac{30 + 2}{2} = 16 \)
- \( y = \frac{30 - 2}{2} = 14 \)
- Pair: \( (16, 14) \)
- For \( (10, 6) \):
- \( x = \frac{10 + 6}{2} = 8 \)
- \( y = \frac{10 - 6}{2} = 2 \)
- Pair: \( (8, 2) \)
### Conclusion
Thus, the valid pairs of natural numbers \( (x, y) \) such that the difference of their squares is 60 are \( (16, 14) \) and \( (8, 2) \).
### Final Answer
There are **2 pairs** of natural numbers whose squares differ by 60.
