Prove $n^2+4n+3$ is not prime for $n \in \mathbb{Z}^{+}$. I am trying to write a proof for this theorem:

For every positive integer $n$, $n^2+4n+3$ is not a prime.

Proof: Let $n \in \mathbb{Z}^{+}$. Note that $$n^2+4n+3=(n+1)(n+3)>1\text{,}$$ 
and $n+1 >1$ and $n+3 >1$.
Let $a = n+1$ and $b = n+3$. Then we have $$\dfrac{(n+1)(n+3)}{a}>\dfrac{1}{a}$$ and $$\dfrac{(n+1)(n+3)}{b}>\dfrac{1}{b}\text{.}$$ 
Therefore, $n^2+4n+3$ is not prime. $\square$
I don't think my proof is right and miss many things. Can anyone give me a hit or show me how to write a better proof for this question? 

One easy way of going about this is simply to consider the parity of $n$:
$n$ is even: We have that
$$
(2\ell)^2+4(2\ell)+3=4\ell^2+8\ell+3=\underbrace{(2\ell+1)(2\ell+3)}_{\text{composite}}.
$$
Thus, $n^2+4n+3$ is not a prime when $n$ is even.
$n$ is odd: We have that
$$
(2\ell+1)^2+4(2\ell+1)+3=\underbrace{2(2\ell^2+6\ell+4)}_{\text{composite}}.
$$
Thus, $n^2+4n+3$ is not a prime when $n$ is odd.
Consequently, when $n$ is even or odd, we have that $n^2+4n+3$ is not a prime. $\Box$ 
