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

Natural numbers are integers starting with 1.

Let the two numbers be a and b then

a2 – b2 = 63 =>(a + b) (a – b) = 63. Now 63 = 9 × 7 = 21 × 3 = 63 × 1 So,

Case I: a + b = 9; a – b = 7. Here we will get a = 8 and b = 1

Case II: a + b = 21; a – b = 3. Here we will get a = 12 and b = 9.

Case III: a + b = 63; a – b = 1. Here we will get a = 32 and b = 31

Since 63 has only these 3 pairs of factors, so the correct option is (a).