Question

show that n^2-1 is divisible by 2 for any positive integer n.
please thanks my answers​

Answer 1

__________________________

$\huge{\textbf{\textsf{{☆ AN}}{\purple{SW}}{\pink{ER ☆} \: {{}{:}}}}}$

CORRECTED QUESTION :

Show that n² - n is divisible by 2 for every positive integer n.

ANSWER :

$\large \sf \fbox{even \: integer : n = 2q }$

$\large \sf \fbox{odd \: integer : n = 2q + 1}$

Here, n = 2q or n = 2q + 1 indicate the set of even and odd positive integers respectively where q ∈ Z.

CASE 1 :

➳ When n is an even integer :

\$When \: n \: = \: 2q \\ \\ In \: this \: case \: we \: have \\ \\ {n}^{2} − n = (2q)^{2} − 2q \\ \\ \: =4{q}^{2} − 2q \\ \\ = 2q(2q − 1) \\ \\ {n}^{2} − n = 2r,where r = q(2q − 1) \\ \\ {n}^{2} − n \: is \: divisible \: by \: 2$

CASE 2 :

➳ When n is an odd integer :

\$When \: n = 2q + 1 \: \\ \\ In \: this \: case \\ \\ {n}^{2} −n=(2q + 1)^{2} − (2q + 1) \\ \\ =(2q + 1)(2q + 1 − 1) \\ \\ = 2q(2q + 1) \\ \\ {n}^{2} − n = 2r,where r = q(2q + 1) \\ \\ {n}^{2} − n \: is \: divisible \: by \: 2$

☆ Hence, n² - n is divisible by 2 for every integer n.

__________________________

Hope it helps you !