1-to-1 correspondence between twin primes and $n^2-1$ I am trying to establish the one-to-one correspondence of twin primes to integers $n$ where $n^2-1$ has 4 divisors.  It is clear to me that this is the case, since 
$$n^2-1=(n+1)(n-1)$$
where the RHS would represent the twin primes.  I think I can set my bijection up using the function
$$f(n-1,n+1)=(n-1)(n+1)$$
HOwever, I'm having difficulties starting this argument off.  I need to show injectivity and surjectivity.  Thus, 
$$f(n_1-1,n_1+1)=f(n_2-1,n_2+1)$$
$$(n_1-1)(n_1+1)=(n_2-1)(n_2+1)$$
$$n_1^2-1=n_2^2-1$$
$$n_1^2=n_2^2$$
$$\sqrt{n_1^2}=\sqrt{n_2^2}$$
$$n_1=n_2$$
I took a liberty here since I'm pretty sure since dealing with twin primes and primes in general the assumption would be we are working in the positive integer realm (or at least non-negative)
Now for surjectivity, I take an integer from the set of numbers of the form $n^2-1$  Then
$$n^2-1=(n-1)(n+1)=(6k-1)(6k+1)=f(6k-1,6k+1).$$
I chose $6k-1, 6k+1$ since a previous exercise had me show there twin primes come in the form.  To finalize the surjection, I have to tie in the fact that there are twin primes (3,5) that do not fall in this form.  So how do I complete my argument, if it is correct to begin with?

$\begin{align}\text{Let}\ T & = \{ (p, q) \mid p, q \in \mathbb{P},\ q - p = 2 \}\\
V & = \{n^2 - 1 \mid n \in \mathbb{N},\ \text{d}(n^2-1) = 4\}
\end{align}$
Theorem [Eleven-Eleven]
$T \cong V$
Proof
Let $f:T \to V$, $(p, q) \mapsto pq$.


*

*$f(p_1, q_1) = f(p_2, q_2) \Rightarrow\\
p_1q_1  = p_2q_2 \Rightarrow\\
p_1 = p_2,\ q_1 = q_2\ [\text{unique prime factorization of $p_1q_1 = p_2q_2$, with $p_1 < q_1$, $p_2 < q_2$}]$
$\therefore$ $f$ is an injection.

*Let $n^2 - 1 \in V$.
$1$, $n - 1$, $n + 1$, $n^2 - 1$ are distinct factors of $n^2 - 1$, and $d(n^2 - 1) = 4$ $\Rightarrow$
$(n - 1), (n + 1) \in \mathbb{P} \Rightarrow$
$(n - 1, n + 1) \in T,\ f(n - 1, n + 1) = n^2 - 1$
$\therefore$ $f$ is a surjection.
1, 2 $\Rightarrow$ $f$ is a bijection.
$\therefore T \cong V$
