Question

two natural numbers differ by 4 and the sum of whose squares is 58.find the numbers

Answer 1

1) put x=7 in (3)
y=21/7=3
2)put x= -3 in (3)
y= 21/-3= -7
therefore
required
numbers are
(x=7,y=3) or (x=-3, y=-7)

Answer 2

Let the numbers be x and y.
y - x = 4... (1)
x^2 + y^2 = 58... (2)
From (1) y = 4 + x
Use above in (2)
(4+x)^2 + x^2 = 58
16+ x^2 + 8x + x^2 = 58
2x^2 + 8x + 16 = 58(Divide by 2 on both sides)
x^2 + 4x + 8 = 29
x^2 + 4x + 8 - 8 = 29 - 8(Transposing)
x^2 + 4x - 21 = 0
x^2+7x-3x-21 = 0
x(x+7)-3(x+7) = 0
(x-3)(x+7) = 0
x = 3,x = - 7 ; -7 is integer.
When x = 3 ; y = 7 (substituting in (1)).
Hence the numbers are 3 and 7.