Question

prove that n^2-n is divisible by 2 for positive integer n​

Answer 1

Here is the answer to your question.
 Any positive integer is of the form 2q or 2q + 1, where q is some integer
When n = 2q, 

n²+n=(2q)²+2q
      =4q²+2q
      = 2q(2q+1)   
which is divisible by 2
 
when n=2q+1
n²-n=(2q+1)²+(2q+1)
      = 4q²+4q+1+2q+1
      =4q²+6q +2
     =  2(2q²+3q+1)
which is divisible by 2
hence n²+n is divisible by 2 for every positive integer n

Answer 2

Answer:-

We can write that $\it n^2 - n = n(n - 1)$

We know that every integer is either in form of $\it 2q + 1$ (odd) or $\it 2q$ (even) if q ∊ Z.

Case 1: in any odd integer,

if $\it n = 2q + 1$,

$\it \therefore \dfrac{ n(n - 1) }{2} \\ \it = \dfrac{(2q + 1)(2q + 1 - 1)}{2} \\ \it = \dfrac{(2q)(2q + 1)}{2} \\ \it = q(2q + 1)$.

Thus, it is divisible by 2.

Case 2: in any even integer,

if $\it = 2q$,

$\it \therefore \dfrac{(2q)(2q -1)}{2} \\ \it = q(2q - 1)$.

Thus, it is also divisible by 2.

By observing, we can conclude that $\it n^2 - n$ is divisible by 2 for any integer n.