Question

A natural number is greater than the other by 5. The sum of their squares is 73. Find those numbers.

Answer 1

Let the natural numbers be x and (x + 5).
By the given condition, we get:
x² + (x + 5)² = 73
$\Rightarrow$ x² + x² + 10x + 25 = 73
$\Rightarrow$ 2x² + 10x – 48 = 0
$\Rightarrow$ 2(x² + 5x – 24) = 0
$\Rightarrow$ x² + 5x – 24 = 0
On splitting the middle term 5x as 8x – 3x, we get:
x² + 8x – 3x – 24 = 0
$\Rightarrow$ x(x + 8) – 3(x + 8) = 0
$\Rightarrow$ (x + 8)(x – 3) = 0
$\Rightarrow$ x + 8 = 0 or x – 3 = 0
$\Rightarrow$ x = –8 or x = 3
Since x is a natural number, it cannot be negative.
Therefore, x = 3 and (x + 5) = 8
Thus, the two required natural numbers are 3 and 8.