Question

Show that n^3-n is divisible by 8 if n is an odd positive integers

Answer 1

Answer:

Any odd positive number is in the form of (4p+1)or(4p+3) for some integer P.

letn=4p+3n2−1=(4p+1)2−1=16p2+8p+1−1=8p(2p+1)⇒n2−1isdivisibleby8n2−1=(4p+3)2−1=16p2+24p+9−1=16p2+24p+8=8(2p2+3p+1)⇒n2−1isdivisibleby8​

Therefore, n2−1 is divisible by 8 if n is an odd positive integer.

Answer 2

Answer:

Step-by-step explanation:

we see that

$n^{3} - n\\n(n^{2} - 1)$

we know that $x^{2} - y^{2} = (x+y)(x-y)$

therefore

(n-1)(n)(n+1)

given that n is odd, let it be of the form 2k+1, substituting

(2k+1-1)(2k+1)(2k+1+!)

(2k)(2k+1)(2k+2)

taking 2 common from last bracket

2(2k)(2k+1)(k+1)

4k(2k+1)(k+1)

it is always divisible by 4 for know, so we need either k or k+1 to be even to make it divisible by  8

now k can either be odd or even

is

K --> odd, then k+1 will be even only as successor of an odd number is always even

therefore = it will be divisible  by 8

k--->even then the k with 4 will be even making it divisible  by 8

therefore it will always be divisible by 8

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