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

Explanation :

Problem: Find pairs of numbers whose difference of squares is 63

Given:
– Two natural numbers x and y where x > y
– x² – y² = 63

Solution:
1. Factor the equation:
x² – y² = (x+y)(x-y) = 63

2. Find factors of 63:
63 = 1 × 63
63 = 3 × 21
63 = 7 × 9

3. For each factor pair (a,b):
x + y = larger number
x – y = smaller number

Solving:
x = (larger + smaller)/2
y = (larger – smaller)/2

4. Calculate pairs:
For 63,1:
x = (63 + 1)/2 = 32
y = (63 – 1)/2 = 31

For 21,3:
x = (21 + 3)/2 = 12
y = (21 – 3)/2 = 9

For 9,7:
x = (9 + 7)/2 = 8
y = (9 – 7)/2 = 1

Therefore, three pairs of numbers exist: (32,31), (12,9), and (8,1)