How many pairs of natural numbers are there such that the difference of whose squares is 63?

The problem asks for the number of pairs of natural numbers (a, b) such that the difference of their squares is 63. This can be written as: a^2 - b^2 = 63.

Using the difference of squares formula, a^2 - b^2 = (a - b)(a + b). So, we have (a - b)(a + b) = 63.

Let x = a - b and y = a + b. Then, x * y = 63.

Since a and b are natural numbers (positive integers), we can deduce some properties for x and y:

1. a^2 - b^2 = 63 > 0 implies a^2 > b^2, so a > b. This means x = a - b must be a positive integer.
2. Since a and b are positive, y = a + b must also be a positive integer.
3. From a > b, it follows that a + b > a - b, so y > x.
4. We can express a and b in terms of x and y: a = (x + y) / 2 b = (y - x) / 2 For a and b to be integers, (x + y) and (y - x) must both be even. This implies that x and y must have the same parity (both even or both odd).
5. Since their product x * y = 63 (an odd number), both x and y must be odd.

Now, we need to find pairs of factors (x, y) for 63 that satisfy these conditions (x * y = 63, x and y are odd, y > x): The factors of 63 are 1, 3, 7, 9, 21, 63.

Let's list the possible pairs (x, y) such that x * y = 63 and y > x:

1. x = 1, y = 63. (Both are odd, y > x) a = (1 + 63) / 2 = 64 / 2 = 32 b = (63 - 1) / 2 = 62 / 2 = 31 This gives the pair (32, 31). (Check: 32^2 - 31^2 = 1024 - 961 = 63)

2. x = 3, y = 21. (Both are odd, y > x) a = (3 + 21) / 2 = 24 / 2 = 12 b = (21 - 3) / 2 = 18 / 2 = 9 This gives the pair (12, 9). (Check: 12^2 - 9^2 = 144 - 81 = 63)

3. x = 7, y = 9. (Both are odd, y > x) a = (7 + 9) / 2 = 16 / 2 = 8 b = (9 - 7) / 2 = 2 / 2 = 1 This gives the pair (8, 1). (Check: 8^2 - 1^2 = 64 - 1 = 63)

These are all the possible pairs of factors (x, y) that satisfy the conditions. Each pair yields a unique pair of natural numbers (a, b). There are 3 such pairs of natural numbers.

The final answer is A) 3.