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? 

I don't know what your definition of prime is.  According to Wikipedia, a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.  You have already factored the expression as $(n+3)(n+1)$.  It should suffice to note that since $n>0, n+3>n+1>1$.  Since your expression has at least $2$ positive factors, neither of which is $1$, it follows that $(n+3)(n+1)$ is not prime.
