Why can $5^{2n+3} + 3^{n+3} \cdot 2^{n} \quad \forall n \in \mathbb{N}$ never be a prime number? It seems to be true for the first thousand $n$ but I really can't think of a way to  prove this statement. 
Any kind of help will be appreciated!

Oops...I think I found a proof.
We can notice that for every $n$ our sum seems to be divisible by $19$.
So we have to prove that
$$5^{2n+3} + 3^{n+3} \cdot 2^{n} \equiv 0 \pmod{19}\quad \forall n \in \mathbb{N}$$
We can prove this using induction.
Base-case:
    \begin{align*}
  &19|152&
 \end{align*}
Inductive step:
    \begin{align*}
  0 &\equiv 5^{2(n+1)+3} + 3^{n+4} \cdot 2^{n+1} &\pmod{19} \\
  0 &\equiv 25\cdot 5^{2n+3} + 6 \cdot 3^{n+3} \cdot 2^{n} \\
  0 &\equiv 6\cdot 5^{2n+3} + 6 \cdot 3^{n+3} \cdot 2^{n} \\
  0 &\equiv 6\cdot ( 5^{2n+3} + 3^{n+3} \cdot 2^{n} )\\
  0 &\overset{\text{IH}}{\equiv} 6 \cdot 0 \equiv 0 \\&&\square
 \end{align*}
