$12$
divides $n^4  n^2$
for all positive integers $n.$

How about factorizing the expression.

How can one show divisibility by
$12$
by showing divisibility by smaller numbers? 
Would splitting the problem into cases help prove divisibility by
$4?$

… such as odd and even?

Factorise to get
$n^4n^2$
$=n^2(n^21)$
$=n^2(n+1)(n1)$
and we’ll show that it’s divisible by both 3 and 4. As$n1, n,n+1$
are three consecutive integers, 3 must divide the expression. To prove that 4 divides the expression, we will consider two cases. If$n$
is even then$n^2$
is divisible by 4. If$n$
is odd then both$n+1$
and$n1$
are even, hence their product is divisible by 4.