Show that 4n^3 + 6n^2 + 4n + 1 is a composite integer for each n E N.?

Question

n E N = > n is an element of the set of natural numbers

Answer 1

The following algebra uses binomial expansion as well as difference of two squares:
4n^3+6n^2+4n+1 =
4n^3+6n^2+4n+1+n^4-n^4 =
n^4+4n^3+6n^2+4n+1-n^4 =
(n+1)^4-n^4 =
[ (n+1)^2-n^2 ] * [ (n+1)^2+n^2 ] = (2n+1)(2n^2+2n+1).

So 4n^3+6n^2+4n+1 is always divisible by 2n+1. For any natural number n, 1 < 2n+1 < 4n^3+6n^2+4n+1, so 4n^3+6n^2+4n+1 is composite.

Answer 2

4n^3 + 6n^2 + 4n + 1 = (2n+1)(2n^2 + 2n + 1). Since n is a natural number, both factors are bigger than 1, so it is composite.