If $p>5$ is prime and $2p+1$ is prime, then $4p+1$ is composite. I'm a relatively new discrete mathematics student and would like to receive feedback on whether my proof for this problem is valid or not.
Proof:
Suppose that $p\in\mathbb{Z}$ and $p>5$ and $p$ is prime and $2p+1$ is prime. By the quotient-remainder theorem, $p$ can be written in one of the forms

$3q$ or $3q+1$ or $3q+2$

for some $q\in\mathbb{Z}$. In fact, since $p$ is prime and $p\neq3$ and $3\mid 3q$, $p$ must have one of the forms

$3q+1$ or $3q+2$.

Case 1 ($p=3q+1$ for some $q\in\mathbb{Z}$): Since $p=3q+1$,

$2p+1=2(3q+1)+1=6q+3=3(2q+1)$.

Let $m=2q+1$. Then $m\in\mathbb{Z}\because 2,q,1\in\mathbb{Z}$ and sums and products of integers are integers. Thus, substituting,

$2p+1=3m$ where $m\in\mathbb{Z}$.

$3\mid (2p+1)\because 3\mid 3m$. Thus $2p+1$ is prime and $2p+1\neq 3$ and $3\mid (2p+1)$, which is a contradiction. Hence this case negates one of the premises.
Case 2 ($p=3q+2$ for some $q\in\mathbb{Z}$: Since $p=3q+2$,

$4p+1=4(3q+2)+1=12q+9=3(4q+3)$.

Let $m=4q+3$. Then $m\in\mathbb{Z}\because 4,q,3\in\mathbb{Z}$ and sums and products of integers are integers. Thus, substituting,

$4p+1=3m$ where $m\in\mathbb{Z}$.

$3\mid (4p+1)\because 3\mid 3m$. Hence $4p+1$ is composite by definition of composite.
Cases 1 and 2 show that

$\forall p\in\mathbb{Z}, p>5$ and $p$ is prime and $2p+1$ is prime $\rightarrow 4p+1$ is composite.

QED.
 A: Everything looks okay to me. Some critiques:
Overall, it's a little overly formal for my tastes and I feel like you overused substitutions a lot. For example, I doubt anyone would have doubted your results if you simply stopped at $2p+1 = 3(2q+1)$ and said this implies that $3|2p+1$, contradicting that $2p+1$ is prime since $p>5 \implies 2p+1 > 11 > 3$ and thus $2p+1$ would have to be composite.
That said I feel like this is something that might be a remnant of your class and how it's managed and also might be influenced by how relatively new to this sort of thing you claim to be, so I can't really fault you for that. I fell into the same habits too. Odds are you'll be writing less clunky proofs with time and practice. If nothing else, your proof does show you're thinking through things well and thoroughly, which is always a good thing!
One could also argue that you should prove that sums/products of integers are indeed integers, if you want to be super formal. This can be done once you understand how the integers are constructed in a formal manner, but given how relatively new you are to this topic it might be a bit above your level and would be something you could just assume holds for the time being. Though if you are familiar with this construction you might find it neat to look at some proofs: here's the closure of multiplication and here is the closure of addition. I'll stress again, though, this is probably something you can assume to be true at your level, so don't sweat it too much.
A: I think this is how the problem came into being
Observe that:one of $4p, 4p+1, 4p+2$ must be divisible by $3$ as they are three consecutive integers
If prime $p>5>3,3\nmid4p$ as $3\mid4p\iff3\mid p\implies p=3$
Similarly as $4p+2=2(2p+1)$ and 
$2p+1>3,3\nmid(2p+1)\implies3\nmid2(2p+1)$
So, $3$ must divide $4p+1$
