lets fixing my proof here.... lets try to solve the goldbach conjectire

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\title{Towards a Proof of Goldbach's Conjecture}
\author{Muhammad Alfa Ridzi}
\date{}
\maketitle

\section*{Introduction}

Goldbach's conjecture, proposed in 1742, states that every even integer greater than 2 can be expressed as the sum of two prime numbers. Despite significant numerical evidence supporting the conjecture, a complete proof remains elusive. In this attempt, we combine computational verification, results from analytic number theory, and elementary number theory principles to strengthen the argument for the validity of Goldbach's conjecture.

\textbf{Goldbach's Conjecture:}
Every even integer 
2
𝑛
>
2
2n>2 can be written as the sum of two prime numbers 
𝑝
p and 
𝑞
q, such that:

𝑝
+
𝑞
=
2
𝑛
.
p+q=2n.
\section{Preliminary Steps: Verifying Small Cases}

We begin by verifying the conjecture for small even numbers. Specifically, we manually check that for even integers up to a reasonably large value (say 100), each number can indeed be written as the sum of two primes. These verifications lay the foundation for generalizing the conjecture to larger numbers.

\begin{itemize}
\item 
2
𝑛
=
4
2n=4, 
4
=
2
+
2
4=2+2
\item 
2
𝑛
=
6
2n=6, 
6
=
3
+
3
6=3+3
\item 
2
𝑛
=
8
2n=8, 
8
=
3
+
5
8=3+5
\item 
2
𝑛
=
10
2n=10, 
10
=
3
+
7
10=3+7
\item 
2
𝑛
=
12
2n=12, 
12
=
5
+
7
12=5+7
\item 
…
…
\item 
2
𝑛
=
100
2n=100, 
100
=
47
+
53
100=47+53
\end{itemize}

Thus, the conjecture holds for all even numbers up to 100. Numerical verifications have extended this range to much larger values. Now, we turn to a more general approach for larger even numbers.

\section{Vinogradov's Theorem for Large Numbers}

For large numbers, we invoke Vinogradov's theorem, which asserts that every sufficiently large odd integer can be expressed as the sum of three primes. This gives us a strategy to handle even numbers greater than a certain threshold.

\begin{theorem}[Vinogradov's Theorem]
Every sufficiently large odd integer can be expressed as the sum of three primes.
\end{theorem}

\noindent Since 
2
𝑛
2n is an even integer, consider the number 
2
𝑛
−
2
2n−2, which is odd. By Vinogradov's theorem, we know:

2
𝑛
−
2
=
𝑝
1
+
𝑝
2
+
𝑝
3
,
2n−2=p 
1
​
 +p 
2
​
 +p 
3
​
 ,
where 
𝑝
1
,
𝑝
2
,
𝑝
3
p 
1
​
 ,p 
2
​
 ,p 
3
​
  are primes. Adding 2 to both sides, we have:

2
𝑛
=
(
𝑝
1
+
𝑝
2
)
+
(
𝑝
3
+
2
)
.
2n=(p 
1
​
 +p 
2
​
 )+(p 
3
​
 +2).
Thus, for large 
2
𝑛
2n, we have a way to express 
2
𝑛
2n as the sum of two primes, as 
𝑝
3
+
2
p 
3
​
 +2 is prime if 
𝑝
3
=
3
p 
3
​
 =3.

\section{Application of the Prime Number Theorem}

To refine the approach, we use the Prime Number Theorem, which gives an approximation for the number of primes less than or equal to 
𝑥
x:

𝜋
(
𝑥
)
∼
𝑥
log
⁡
(
𝑥
)
,
π(x)∼ 
log(x)
x
​
 ,
where 
𝜋
(
𝑥
)
π(x) is the prime-counting function.

For a large even number 
2
𝑛
2n, the number of primes less than or equal to 
𝑛
n is approximately:

𝜋
(
𝑛
)
≈
𝑛
log
⁡
(
𝑛
)
.
π(n)≈ 
log(n)
n
​
 .
This approximation gives us the number of primes available to form sums. The same holds for 
2
𝑛
2n:

𝜋
(
2
𝑛
)
≈
2
𝑛
log
⁡
(
2
𝑛
)
.
π(2n)≈ 
log(2n)
2n
​
 .
Given this, the number of prime pairs 
(
𝑝
,
𝑞
)
(p,q) such that 
𝑝
+
𝑞
=
2
𝑛
p+q=2n increases as 
𝑛
n increases, and the asymptotic density of primes guarantees the existence of such pairs for large 
2
𝑛
2n.

\section{Intermediate Cases: Pigeonhole Principle}

For intermediate values of 
2
𝑛
2n, we apply the pigeonhole principle. Consider two sets:
\begin{itemize}
\item 
𝑆
1
=
{
1
,
2
,
…
,
𝑛
}
S 
1
​
 ={1,2,…,n}
\item 
𝑆
2
=
{
𝑛
+
1
,
𝑛
+
2
,
…
,
2
𝑛
}
S 
2
​
 ={n+1,n+2,…,2n}
\end{itemize}

We count the number of primes in each set using the prime-counting function. Let 
𝑝
(
𝑛
)
p(n) represent the number of primes less than or equal to 
𝑛
n, and 
𝑝
(
2
𝑛
)
p(2n) represent the number of primes less than or equal to 
2
𝑛
2n. The number of primes in 
𝑆
2
S 
2
​
  is 
𝑝
(
2
𝑛
)
−
𝑝
(
𝑛
)
p(2n)−p(n).

Applying the pigeonhole principle, we ensure that there are enough primes in both sets to form at least one pair 
(
𝑝
1
,
𝑝
2
)
(p 
1
​
 ,p 
2
​
 ) such that:

𝑝
1
+
𝑝
2
=
2
𝑛
.
p 
1
​
 +p 
2
​
 =2n.
Thus, for intermediate values of 
𝑛
n, Goldbach's conjecture holds by this principle.

\section{Conclusion}

By combining the verification of small cases, the application of Vinogradov's theorem for large numbers, and the use of the prime number theorem and pigeonhole principle for intermediate values, we have shown strong evidence in favor of Goldbach's conjecture. While this does not yet constitute a complete proof, the reasoning outlined above provides significant support for the conjecture's validity across all even numbers.

\end{document}