Question

How many pairs of natural numbers is there the difference of whose squares are 63 .

Answer 1

Answer:

3 pairs . they are (32,31) , (12,9) & (8,1)

Step-by-step explanation:

(x2 - y2) = 63

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

1 x 63 = 63

3 x 21 = 63

7 x 9   = 63

Thus, the factors of 63 possibles are 1, 3, 7, 9, 21 and 63

from the above table , it implies that both the sum and difference of x & y are odd numbers

so either x or y , anyone one number have to be even and the other should be odd

x+y = 63      x+y =21    x+y = 9

x-y =   1        x-y =  3    x-y  = 7

______      ______    ______

2x = 64       2x = 24    2x = 16

x = 32          x = 12        x = 8

=>

y = 31            y = 9         y = 1     (on subst. x value in any of the eqn.s)

so 3 pairs exists, (32,31) , (12,9) & (8,1)