Explanation:
To solve this problem, we need to find pairs of natural numbers $(x, y)$ such that the difference of their squares equals 63. In other words, we want to find all pairs of natural numbers $(x, y)$ satisfying:
\[
x^2 – y^2 = 63
\]
Step-by-Step Solution:
1. Use the Difference of Squares Formula:
The expression $x^2 – y^2$ can be factored using the difference of squares formula:
\[
x^2 – y^2 = (x – y)(x + y)
\]
So, we have:
\[
(x – y)(x + y) = 63
\]
2. Find Factor Pairs of 63:
To satisfy the equation, we need to find pairs of natural numbers $(x – y)$ and $(x + y)$ such that their product is 63. The factor pairs of 63 are:
\[
(1, 63), \, (3, 21), \, (7, 9)
\]
3. Solve for Each Factor Pair:
For each factor pair $(a, b)$, we solve the equations:
\[
x – y = a \quad \text{and} \quad x + y = b
\]
We can solve for $x$ and $y$ by adding and subtracting these equations:
\[
x = \frac{a + b}{2}, \quad y = \frac{b – a}{2}
\]
Let’s check each factor pair:
– Factor pair $(1, 63)$:
\[
x – y = 1, \quad x + y = 63
\]
\[
x = \frac{1 + 63}{2} = 32, \quad y = \frac{63 – 1}{2} = 31
\]
Pair: $(x, y) = (32, 31)$
– Factor pair $(3, 21)$:
\[
x – y = 3, \quad x + y = 21
\]
\[
x = \frac{3 + 21}{2} = 12, \quad y = \frac{21 – 3}{2} = 9
\]
Pair: $(x, y) = (12, 9)$
– Factor pair $(7, 9)$:
\[
x – y = 7, \quad x + y = 9
\]
\[
x = \frac{7 + 9}{2} = 8, \quad y = \frac{9 – 7}{2} = 1
\]
Pair: $(x, y) = (8, 1)$
4. Count the Number of Pairs:
From the factor pairs, we found three valid pairs of natural numbers: $(32, 31)$, $(12, 9)$, and $(8, 1)$.
Conclusion:
There are 3 pairs of natural numbers such that the difference of their squares is 63.
The correct answer is (a) 3.