N square is divisible by 8 if n is
Asked by admin @ in Math viewed by 315 People
Show that n2-n is divisible by 8 if n is an odd positive integer
Answered by admin @
Hello Dear.
There is the little Printing mistake in the Question. Correct Question would be like this ⇒
Show that (n² - 1) is divisible by 8 if n is an odd positive integer.
Solutions ⇒
To Prove ⇒ (n² - n) is divisible by 8.
Given ⇒ n is an odd positive integer.
Proof ⇒
We know that an Odd Positive Integer n is always written in the form of (4k + 1) or (4q + 3) or (4q + 5) and so on.
If n = 4k + 1
Then, n² - 1 = (4k + 1)² - 1
= 16 k² + 1 + 8k - 1
[∵ (a + b)² = a² + b² + 2ab]
= 16k² + 8k
= 8k(2k + 1)
Hence, it is divisible by 8.
If n = 4k + 3
Then, n² - 1 = (4k + 3)² - 1
= 16k² + 9 + 24k - 1
= 16k² + 24k + 8
= 8(2k² + 3k + 1)
Hence, it is also divisible by 8.
If n = 4k + 5
Then, n² - 1 = (4k + 5) - 1
= 16k² + 25 + 40k - 1
= 16k² + 40k - 24
= 8(2k² + 5k - 3)
Hence, it is also divisible by 8.
Now, For any value of n, n² - 1 is always be divisible by 8.
Hence Proved.
Hope it helps.
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