Question

If m=n^2-n where n is an integer then m^2-2m is divisible by

Answer 1

GIVEN :

If $m=n^2-n$ where n is an integer

TO SHOW :

The expression $m^2-2m$ is divisible by 24

SOLUTION :

Given that $m=n^2-n$ where n is an integer

Substitute $m=n^2-n$ in the expression $m^2-2m$ we get,

$(n^2-n)^2-2(n^2-n)$

$=(n^2)^2-2n^2.n+n^2-2n^2+2n$

$=n^4-2n^3-n^2+2n$

$=n(n^3-2n^2-n+2)$

$=n(n^2(n-2)-1(n-2))$

$=n((n^2-1)(n-2))$

By using the algebraic identity:

$a^2-b^2=(a+b)(a-b)$

$=n(n-1)(n+1)(n-2)$

=(n-2)(n-1)n(n+1)

The above expression n(n-1)(n+1)(n-2) represents the product of 4 consecutive numbers.

∴ the expression (n-2)(n-1)n(n+1) is divisible by 24 for n>2.

Hence showed.