Answer:
See below ↓
Step-by-step explanation:
Let the odd integers be :
Solving
Examples
Answer:
Let one odd number be (2n + 1)
Therefore, its consecutive odd number is (2n + 3)
Difference between the two squares of these numbers:
[tex]\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}[/tex]
Thus proving that the difference between the squares of any two consecutive odd numbers is always a multiple of 8.
