The difference of squares of two consecutive odd numbers is always
divisible by 8
divisible by 3
divisible by 16
None of the above
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CISF-AC-EXE – 2018
The difference of their squares is:
(2n + 3)^2 – (2n + 1)^2
Using the algebraic identity a^2 – b^2 = (a – b)(a + b):
= [(2n + 3) – (2n + 1)] * [(2n + 3) + (2n + 1)]
= [2n + 3 – 2n – 1] * [2n + 3 + 2n + 1]
= [2] * [4n + 4]
= 2 * 4(n + 1)
= 8(n + 1)
Since n is an integer, (n + 1) is also an integer. Therefore, the expression 8(n + 1) is always a multiple of 8.
Consecutive odd numbers 1 and 3: 3^2 – 1^2 = 9 – 1 = 8. 8 is divisible by 8.
Consecutive odd numbers 3 and 5: 5^2 – 3^2 = 25 – 9 = 16. 16 is divisible by 8.
Consecutive odd numbers 5 and 7: 7^2 – 5^2 = 49 – 25 = 24. 24 is divisible by 8.
The difference is always in the form 8 * (an integer), so it is always divisible by 8.