Question

Prove algebraically that the difference between the squares of any two consecutive
odd numbers is always a multiple of 8

Answer 1

Answer:

See below ↓

Step-by-step explanation:

Let the odd integers be :

• 2x + 1
• 2x + 3

Solving

• (2x + 3)² - (2x + 1)²
• 4x² + 12x + 9 - 4x² - 4x - 1
• 8x + 8 = 8(x + 1)
• On substituting any value for 'x', the value will be a multiple of x.

Examples

• 8(1) + 8 = 16 = 8 x 2
• 8(2) + 8 = 24 = 8 x 3
• 8(3) + 8 = 32 = 8 x 4

Answer 2

Answer:

Let one odd number be (2n + 1)

Therefore, its consecutive odd number is (2n + 3)

Difference between the two squares of these numbers:

$\begin{aligned}\sf (2n+3)^2-(2n+1)^2 & = \sf (4n^2+12n+9)-(4n^2+4n+1)\\ & = \sf4n^2+12n+9-4n^2-4n-1\\ & = \sf 8n+8\\ & = \sf8(n+1)\end{aligned}$

Thus proving that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.