I want to share with my attempt to prove the infinity of twin primes, and i want to have you opinion:
Every odd composite $N \geq9$ can be written as: $p^2_{n}+2p_{n}c=N$
With $c \in Z^+$ and $p \in P$ such as $p_{n} \geq 3$
Let $p_{n}$ the $n-th$ prime number and $p_{n+1}$ the next prime, when two primes generate the same odd composite accordingly with the above equation, we get
$$p^2_{n}+2p_{n}c_{1}= p^2_{n+1}+2p_{n+1}c_{2}=N$$
By rearranging we get \begin{equation} p_{n+1}-p_{n}= 2 \frac{p_{n}c_{1}-p_{n+1}c_{2}}{p_{n+1}+p_{n}} \end{equation} Where $g_{n}=2 \frac{p_{n}c_{1}-p_{n+1}c_{2}}{p_{n+1}+p_{n}}$ denotes the twine prime gap
$$\lim_{n\to\infty} p_{n+1}-p_{n}= \lim_{n\to\infty} 2 \frac{p_{n}c_{1}-p_{n+1}c_{2}}{p_{n+1}+p_{n}}$$
For any $\epsilon > 0$ and sufficiently large $n$ we have $c_{1}=c_{2}+ \epsilon$, thus $$\frac{p_{n}c_{1}-p_{n+1}c_{2}}{p_{n+1}+p_{n}}= c_{2} \frac{p_{n}-p_{n+1}}{p_{n+1}+p_{n}}+\epsilon \frac{p_{n}}{p_{n+1}+p_{n}} $$
For any $\alpha >0$ and large $n$ we have $p_{n+1}<(1+\epsilon)p_{n}$ , by prime number theorem. This implies $$\lim_{n\to\infty} c_{2} \frac{p_{n}-p_{n+1}}{p_{n+1}+p_{n}}=0$$
However, according to the prime number theorem, we have $$ \lim_{n \to \infty} \frac{p_n}{p_{n+1}} = \lim_{n \to \infty} \frac{p_n}{p_{n+1}} \cdot \frac{n \ln(n)}{n \ln(n)} \cdot \frac{(n+1)\ln(n+1)}{(n+1)\ln(n+1)} $$ $$ = \lim_{n \to \infty} \frac{p_n}{n\ln(n)} \cdot \frac{(n+1)\ln(n+1)}{p_{n+1}} \cdot \frac{n\ln(n)}{(n+1)\ln(n+1)} $$ $$ = \lim_{n \to \infty} \frac{n\ln(n)}{(n+1)\ln(n+1)} $$ $$ = \left[\lim_{n \to \infty} \frac{(n+1)\ln(n+1)}{n\ln(n)}\right]^{-1}$$ $$ = \left[\lim_{n \to \infty}1 + \frac{\ln[(n+1)/n]}{\ln(n+1)} + \frac{\frac{\ln(n+1)}{\ln(n)}}{n }\right]^{-1} = 1^{-1} = 1. $$ it follows that $$ \lim_{n \to \infty} \frac{p_n}{p_n + p_{n+1}} = \lim_{n \to \infty} \frac{1}{1 + \frac{p_{n+1}}{p_n}} = \frac{1}{1+1} = \frac 12. $$ Then
$$\lim_{x\to\infty} inf (p_{n+1}-(p_{n}))=\epsilon$$ And since $\epsilon \in [2, \infty[$, we conclude \ $$\lim_{x\to\infty} inf (p_{n+1}-(p_{n}))=2$$ Proof of the lower bound of $\epsilon$.
Let $p_{n}$ be a prime number, $p_{n+1}$ the next prime and ($c_{1};c_{2} \in Z^*{+} $)\ When 2 primes generate the same odd composite, we have the following equation \
$$p^2_{n}+2p_{n}c_{1}= p^2_{n+1}+2p_{n+1}c_{2}$$
Since $p_{n}<p_{n+1}$, then $c_{1}>c_{2}$. We put ($c_{1}=c_{2}+ \epsilon$), the minimum value of $\epsilon$ occurs when $p_{n+1}=p_{n}+2$, after changing these values, the above equation becomes \ $$p^2_{n}+2p_{n}c_{1}= (p_{n}+2)^2+2(p_{n}+2)(c_{1} - \epsilon)$$ It implies when solving for $\epsilon$ \ $$\epsilon=\frac{4p_{n}+4+4c_{1}}{2p_{n}+4}$$ \ And since $c_{1}>0$, then its minimum value is $1$, we conclude \ $$\lim_{n\to\infty} \inf (\frac{4p_{n}+4+4c_{1}}{2p_{n}+4})=2$$
The first error is
In fact, for an odd composite to satisfy your equation for both $p_n$ and $p_{n+1}$ then it must equal $qp_np_{n+1}$ for some odd $q\geq 3$ (since it has to be divisible by both primes but greater than $p_{n+1}^2$ and odd).
Thus $c_1=\frac{qp_{n+1}-p_n}{2}$ and $c_2=\frac{qp_{n}-p_{n+1}}{2}$, giving $c_1-c_2\geq p_{n+1}-p_n$. So certainly the statement above is not true unless you restrict it both by requiring $\epsilon\geq 2$ and that you are looking for some sufficiently large $n$ (not every sufficiently large $n$). And with those restrictions it's simply a restatement of the twin prime conjecture, so you can't assume it.