Question

Find two natural numbers which differ by 3 and whose squares have the sum 149

Answer 1

Let no.s be x and y.

x - y = 3

So, x = y + 3

x^2 + y^2 = 149

(y + 3)^2 + y^2 = 149

y^2 + 6y + 9 + y^2 = 149

2y^2 + 6y = 140

2y^2 + 6y - 140 = 0

Dividing by 2

y^2 + 3y -70 = 0

y^2 + 10y - 7y -70 = 0

y(y + 10) - 7(y - 10) =

(y - 7)(y - 10) = 0

y = 7 or y = 10

So,

x = 10 or x = 13

Hope this helps! :)