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

Correct Option

Let the two natural numbers be x and y. Without loss of generality, assume x > y. The problem states that the difference of their squares is 63, which can be expressed as:

x² - y² = 63

This equation can be factored using the difference of squares formula:

(x - y)(x + y) = 63

Let a = x - y and b = x + y. Since x and y are natural numbers (positive integers), a and b must also be positive integers. Additionally, since x > y, it follows that x + y > x - y, meaning b > a.

Furthermore, consider the sum and difference of a and b:

• a + b = (x - y) + (x + y) = 2x
• b - a = (x + y) - (x - y) = 2y

Since 2x and 2y are even, both a and b must have the same parity (either both even or both odd). As their product, a * b = 63, is an odd number, both a and b must be odd integers.

Now, we need to find pairs of factors (a, b) for 63 such that a * b = 63, a < b, and both a and b are odd. The factors of 63 are:

• 1 × 63
• 3 × 21
• 7 × 9

All these pairs satisfy the conditions that both factors are odd and the first factor is less than the second. We can now solve for x and y for each pair:

1. For (a, b) = (1, 63): x - y = 1 x + y = 63 Adding the equations yields 2x = 64, so x = 32. Subtracting the first from the second yields 2y = 62, so y = 31. This gives the pair (32, 31). (Check: 32² - 31² = 1024 - 961 = 63)
2. For (a, b) = (3, 21): x - y = 3 x + y = 21 Adding the equations yields 2x = 24, so x = 12. Subtracting the first from the second yields 2y = 18, so y = 9. This gives the pair (12, 9). (Check: 12² - 9² = 144 - 81 = 63)
3. For (a, b) = (7, 9): x - y = 7 x + y = 9 Adding the equations yields 2x = 16, so x = 8. Subtracting the first from the second yields 2y = 2, so y = 1. This gives the pair (8, 1). (Check: 8² - 1² = 64 - 1 = 63)

Thus, there are exactly 3 pairs of natural numbers whose squares' difference is 63: (32, 31), (12, 9), and (8, 1).

Incorrect Options:

Options 2, 3, and 4 are incorrect because the systematic factorization of 63 into pairs of odd factors (x-y, x+y) reveals precisely 3 unique pairs of natural numbers that satisfy the given condition, as demonstrated above.