Question

Show that (n^2-n) is divisible by 3 where n is an positive number

Answer 1

we know that any positive integer is in the form of 3q, 3q+1, 3q+2
so there are following case

case I. when n= 3q
in this case n^2-n= (3q)^2 - 3q = 9q^2 -3q
= 3q(3q-1)
so n^2-n= 2m where m=3q-1
so now we can say that n^2 - n is divisible by 2

case II.
when n= 3q+1
then
n^2-n= (3q+1)^2 - (3q+1) = (3q+1)(3q+1 - 1)= 3q (3q+1)
now
n^2-n=3m where m= q(3q+1)
so n^2-n is divisible by 3

case III.
when n=3q+2
then
n^2-n=(3q+2)^2-(3q+2)
= (3q+2)(3q+2-1)
= (3q+2)(3q+1)
=9q^2 +3q+6q+2
= 9q^2+9q+2
= 3q(3q+3)+2
so we can say that

n^2-n= 3m where m= q(3q+3)+2
so n^2-n is divisible by 3

in the above all case we see that all are divisible by 3
here it has been proove mark it brainly