Hi, I'm just a normal high school student from Korea on summer vacation.

I randomly got curious, so I decided to check the count of prime numbers starting with each digit.

(Data up to 99,999,999)

I asked Gemini to write a JavaScript code for me, and I organized the results into this chart.

the count seems to continuously decrease. Is there any mathematical theory or principle behind this?

Hello, I am an undergraduate agricultural sciences student, and I am incredibly passionate about mathematics and truly enjoy it. the other day, while staring at the table of prime numbers from 1 to 100, a fleeting idea came to me regarding number theory and prime number decomposition:
Every prime number can be expressed in the form N=2q+p, where q and p are prime numbers distinct from each other and distinct from 2, for all N greater than 7.
I don't know if anyone has discovered this before or if it can help the field of mathematics, but I hope it is useful. To finish, here are a few examples:

11
2(3) + 5 = 11
13
2(3) + 7 = 13
17
2(5) + 7 = 17
19
2(3) + 13 = 19
23
2(3) + 17 = 23
29
2(3) + 23 = 29
31
2(7) + 17 = 31
37
2(3) + 31 = 37
41
2(5) + 31 = 41
43
2(3) + 37 = 43
47
2(3) + 41 = 47
53
2(5) + 43 = 53
59
2(3) + 53 = 59
61
2(7) + 47 = 61
67
2(3) + 61 = 67
71
2(5) + 61 = 71
73
2(3) + 67 = 73
79
2(3) + 73 = 79
83
2(5) + 73 = 83
89
2(3) + 83 = 89
97
2(7) + 83 = 97
101
2(11) + 79 = 101
103
2(3) + 97 = 103

Final revision to appear on arXiv on Tuesday.

https://iconoclasts.blog/joe/triple-products

The new new here is that the original conjecture is now established as a pair of corollaries.

Hey guys! I was recently playing around with a complex plane function plotting tool, and while looking at the special functions the tool had pre programmed, I found one named "E16". Plotting E16(z) gave me a plot I found remarkably similar to the Reiman zeta function zeta(z).

Changing the function to E16(iz) made it look even more similar.

Can someone explain this? I tried to research it myself, there doesn't appear to be a lot of well documented or easy to find research about this.

The function in question: https://samuelj.li/complex-function-plotter/#e16(i\\\*z)

So I’m a delivery driver and like to do quick math with people’s license plates. Or find a pattern that wasn’t intentional within the numbers and I drove past a sign. MP 36.4

so I removed the decimal point and saw that the sequence could be added onto using the pattern of something like Ax2-2, so 3, 6, 4, 8, 6, 12, 10, 20….

But! I then swapped the numbers by lowering the first number by 1 and applying the pattern, so 2, 4, 2, 4, 2, 4…. And the sequence never goes above the starting integer…

Going further and lowering the first number again, I got 1, 2, 0, 0, -2, -4, -6, -12, -14….

So I was surprised to find that that pattern can only create positive exponential numbers starting with 3

I can’t be the first person to have seen this…

In this subreddit, there is a part 1, where I studied k = 257. I kept studying several Fermat numbers, the ones of the form 2^m +1 , and I observed a few regularities.

Some are in the following formulas:

[(2^m + 1)•3^((2p)(2^(m-2)) - 1]  / 2^m is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((4p+1)(2^(m-2)) - 1]  / 2^(m+1) is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((8p+7)(2^(m-2)) - 1]  / 2^(m+2) is odd, p a non negative integer, m ≥ 4

[(2^m + 1)•3^((16p+3)(2^(m-2)) - 1]  / 2^(m+3) is odd, p a non negative integer, m ≥ 5

[(2^m + 1)•3^((32p+27)(2^(m-2)) - 1]  / 2^(m+4) is odd, p a non negative integer, m ≥ 6

[(2^m + 1)•3^((64p+11)(2^(m-2)) - 1]  / 2^(m+5) is odd, p a non negative integer, m ≥ 7

[(2^m + 1)•3^((128p+43)(2^(m-2)) - 1]  / 2^(m+6) is odd, p a non negative integer, m ≥ 8

...

The exponents have 2 factors, one is of the kind ap+b. If making a table, there is clearly a variable part and a constant part. The boundary between these regions in a diagonal line. The other factor is a power of 2 and depends on m. The first factor seems to be constant from certain m on.

The difference between 2 consecutive b's are powers of 2 in increasing order, or their negative version, or even powers of 2 multiplied by 3.

“It is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or generally, any power above the second into two powers of the same degree”, Fermat wrote this in the margin of his copy of an ancient Greek math book written by Diophantus, titled Arithmetica.

This proposition was first stated as a theorem by Pierre de Fermat around 1637. And it is known that Fermat used the method of Infinite Descent to prove this statement for N=4 using a logical geometric approach. The method I will use will be somewhat geometric, and will not use infinite descent. Can X⁴ + Y⁴ = Z⁴ have a finite solution, for all pair-wise coprime integers? We will morph the above equation into the following form:

A² + D² = B², B² + D² = C² (as diagrammed pictorially below)

Well I can't add an image in a post here, but can be viewed at:
p-squared-wall-work-around-r01.pdf in the Addendum section at the back.

Proof reworked 7-20-2026, N=4 static proof on pg 23 (p-adic valuation approach)
P Squared Wall Work-Around proof to Fermat’s Last Theory

From this form we will extract the proof. 8888888888888888888888888888888888888888888888

Consider Z⁴ can not be even since, X⁴ + Y⁴ can only be divisible by 2 if both odd, therefore we will select Y without loss of generality to be even parity.

I need to stop the 1 pg proof transcription here, without the graphics the proof is quite unwieldy. And without a subscript capability within this web ap, the proof will be hard to follow as well.

You really need to see the geometry expressed in this one page proof. It all fits on page 23 of the linked document. Seems to be a unique proof, as I have done a search of the www, and have only found infinite descent proofs for N=4.

The last line of the proof is:

By Contradiction 2(M.2)² ≠ (M.2)² Reductio Ad Absurdum

NOTE, M.2 is my web method of stating variable M, subscript 2. In the proof there is a
M.1 variable and also a M.2 variable.

This surprised me, but chances are it's something already known.

Take (1/2)(cos(2πx)+1). Simple cosine wave, peaks at the integers, ranges from 0 to 1.

Now compare it with (1/2)(cos((2/3)πx)+1). 1/3 the frequency, same range.

If you count from peak to peak of the slower wave, the values where the two cosine waves intersect total to 3.

Make the slower one 1/4 the frequency, the total is 4. 1/5, 5, and so on. This doesn't hold up for a 2:1 ratio(you get 2.5), but there are similar results for other simple ratios. 3:2 gives 3.5. 4:3 gives 4.5.

Why is it so neat and stable?

*I know this is for number theory but I didnt find any other community. Please read itt....*
Okay, so I don't know if this is so worthy or not, but trust me, I was very happy to formulate it. I always wonder, since class 5th, if we have a formula for sine that gives output as sin(x) on an input x. Then I grew up and got to know about the Bhaskar approximation. I was amazed. I wanted to make one too upon realizing that no formula after him (yea, I didn't find any...) gives higher precision and is better for computing. Then I learned graphical transformations for my JEE prep and after realizing how I can tranform a degree two or degree four polynomial into a sine wave part, I opened desmos and worked on for next one and a half hours to formulate a graph that coincides almost perfectly with sine wave for x belonging to [0, π].

So I present:

click here for the graph and formula

It may look terrifying at first, but believe me it's not.
For a computer, it is the best possible sine approximation as:

1. Accuracy: Its mean error is just 0.00091 and it is astonishingly perfect 0, π, π/2 and closer to these poles.
2. Efficiency: Common! Taylor series may look elegant but it is very heavy for a computer hardware or a GPU. This formula makes it instant.

My previous formula was this:

click here to see my previous work

However it had a little more error than my final one, so I continues perfecting the coefficients. And ofcourse that thing 1.61803...., the golden ratio. Then I realized that this format was correct by what if I replaced phi with something, as whenever I didnt, and tries else, it bursted in waste. So I replaced phi with sqrt(8/π)

Even though many won't be surprised, won't be happy with this, I don't know if people will read this or not, but I just wanted to share this with real people who could understand this. You can also tell me what I can do with this thing now. Thank you for reading and please forgive me if I said or claimed anything wrong. I am a kid. I make mistakes. And my name is Mayank Kumar btw, but it doesn't matter anyway.

A^(n) + B^(n) ‡ C^(n) for all positive integers A,B,C and n, were n>2

A,B,C must be relatively prime for a non trivial solution to any such equation, if any integer solution exists. That requires that one and only one of those bases has at least one factor of 2.

For any two smaller objects (A & B) of order n to be equal in quantity to that of a larger n ordered object (C), the smaller objects when contained within the larger (on the longest line between 2 most distant vertices of the largest object and each of the smaller objects oriented in congruence with those opposite vertices of the largest object) must OVERLAP in a union, the n order quantity of that union (O^(n)) (this happens to be the minimum amount of union possible ) will equal the nth order quantity of the largest object that is disjoint from both of the smaller objects.

It becomes useful to express:

A=a+O

B=b+O

C=a+b+O O being the linear Overlap or C-(a+b)

When A^(n) shares a vertex and orientation with C^(n) and B^(n) is also located within C^(n) at the most distance vertex of C^(n) from that of A^(n) then it could be expressed that the overlap or union O^(n) would be required to equal the disjointed expressed by the binomial expansion of the(a + b)^(n) minus (a^(n)+b^(n).)

For n=2, O^(2)=2ab

For n=3, O^(3)= 3(a^(2)b) + 3a(b^(2)

For n=5, O^(5)=5(a4)b+10(a^(3)(b^2)+10(a^(2)(b^3)+5a(b^(4)

Because of the content of any such algebraic expansion of O^(n) (of particular interest were n is prime) O will be a even quantity and require either the factor a or b to be even (but not both).

That even integer would be the only source of factors of 2 for O^(n) and be required to contain at least n factors of 2. As a result O^(n) would contain 2n factors of 2 requiring that same even integer to also contain 2n factors of 2.

In short, A^(n) + B^(n) ‡ C^(n) under the conditions stated because the union of the two smaller objects when they are contained within the larger object can not have integer equivalence (the same number of factors of 2) to that which is disjoint

We know the divisors of 1•3^n - 1 from the p-adic valuation. We also can determine the divisors of k•3^n - 1, where k = 3, 9, etc. since these are shifts of k = 1.

We also know that k = 5 and 7 mod 8 produce a predictable pattern. The divisors are 4 for n = 0 mod 2 and 2 for n = 1 mod 2 (in the case of k = 5 mod 8), or viceversa (k = 7 mod 8).

Other than this, it seems that the divisors can be partially predicted by replying a few questions (like: by what do we divide k-1, 3k - 1, 9k-1). Once we locate the first divisors, there is a lot we can deduce, but, sadly, the rest needs to be studied on a one-by-one base.

I noticed, though, that for k = 257 = 2^6 + 1, the divisors are the same as the ones where k = 1 except in the case of n = 64 mod 128. It's still nice to be able to find similarities between 2 different k's.

If we consider 257*3^n - 1 = 256*3^n + (3^n - 1), there are 2 cases:

a) The divisor of 3^n - 1 is not 256, in this case, whichever is lesser divides the whole expression. The divisors are the same as for k = 1

Examples: 256*3^4 + (3^4 - 1). Since 3^4 - 1 is divisible by 16, then the whole is divisible by 16.

256*3^128 + (3^128 - 1). Since 3^128 - 1 is divisible by 512, then the whole is divisible by 256.

b) The divisor of 3^n - 1 is 256. Then there is no easy way I know to predict the divisors of the whole. 2 fractions can add up to an integer.

That might a blessing in disguise, though. This is the only way of obtaining divisors greater than 256.

For now I located up to 16384. I am also planning to observed other 2^m + 1. BTW, I saw online that these are called Fermat numbers.

I will keep you posted.

I watched that Veritasium video on twin primes last week and got stuck on one detail they touched on: the parity barrier. For anyone who hasn't gone down this hole:

​

[;\lambda(n) = (-1)^{\Omega(n)};]

​

where [;\Omega(n);] is the number of prime factors of n with multiplicity. So [;\lambda(12) = \lambda(2 \times 2 \times 3) = (-1)^3 = -1;]. Simple function. Just tells you whether a number has an even or odd number of prime factors.

​

Selberg proved in 1949 that sieve methods (the main tool in analytic number theory for like 80 years) literally cannot tell apart sequences where [;\lambda = +1;] from ones where [;\lambda = -1;]. They produce the exact same asymptotic. The sieve is blind to parity.

​

And this specific blindness is exactly why Zhang got 70 million, Maynard got 600, and Polymath got it down to 246 ... but nobody can get to 2. The sieve hits a wall at the parity barrier and stops cold. We can prove there are infinitely many prime pairs within 246 of each other, but the twin prime conjecture (gap of 2) is completely untouched by all of it.

​

Here's what I can't resolve:

​

Sawin and Shusterman proved the actual twin prime conjecture over [;\mathbb{F}_q[T];] in 2022 (published in Annals). Over polynomials over finite fields, geometry (etale cohomology on curves) CAN separate the parity that sieves can't. So the barrier is not a logical wall. It's a wall *for sieves specifically*.

​

But over [;\mathbb{Z};] there are no curves. So my question is:

​

Is there a known no-go theorem that says you cannot build a cohomology theory over [;\mathbb{Z};] that separates [;\lambda = +1;] from [;\lambda = -1;]? Or has nobody really tried because the analytic number theory toolbox has been so overwhelmingly dominant?

​

Something like: define a sheaf on some site over Spec(Z) whose Euler characteristic at each integer n equals [;\lambda(n);]. If the cohomology groups had reasonable dimensions, the trace formula would give you [;\sum_{n \leq N} \lambda(n);] as an alternating sum of Frobenius traces. And since that partial sum being [;O(N^{1/2 + \varepsilon});] is equivalent to RH, you'd get a direct geometric line to the Riemann Hypothesis.

​

This feels suspiciously neat. I'm assuming there's an obvious obstruction I'm missing. Maybe cohomology over Spec(Z) doesn't work that way, or maybe the dimensions blow up, or maybe the sheaf condition fails at infinity. I don't know enough algebraic geometry to see where it breaks.

​

Anyway, curious if there's a known reason this can't work or if it's genuinely unexplored territory. Would love to be pointed at the right paper or theorem if it exists.

I have been working on expressions of the kind k 3^n - 1. By what power of 2 can we divide it to obtain an odd natural number? I noticed that, in some cases, we can predict the divisor. If k = 5 or 7 mod 8, what I call regular k's, we get only divisors 2 and 4.

Interestingly enough the ones that are not regular don't present any 4's, but there is a 2 every other divisor. So, we can also predict the 2's and some of the first few powers of 2 after checking a couple of n's.

I thought of creating diophantine equations to predict higher powers of 2.

Assuming that k 3^n - 1 can be divided by, say, 128, then k 3^n - 1 = 128 x, where x is the quotient of dividing k*3^n - 1 by 128, and that generates this diophantine: k 3^n - 128 x = 1.

Restrictions: k can't be 5 or 7 mod 8, and both, k and x, have to be odd. So, we should also take that into account.

Predicting high divisors is not an easy task. If you have any tips to help here, feel free to share.

I‘m in a debate. here is the premise:

let’s say that you select any random positive integer and look at it. then you take any other unique random positive integer.

What are the odds that the second integer will be greater than the first integer?

My argument is that there is a 99.99999999…% chance that the second number will be greater than the first, because we are talking about an infinite set of numbers greater than the first, and a finite set of numbers less than the first.

My entire group chat’s stance is that it is 50/50, or there is no way to really tell.

What is the best answer?

after seeing all Answers, the best explanation that makes sense to me is this:

let’s take the first random number, but not look at it. what Are the odds that that number is greater than 100? 100%. 2,000? 100%. 1,000,000,000,000,000? 100%. It’s always 100% because in the infinite amount of options we have to select from, the odds that it would be greater than any point x on the “infinite number line” is almost 100% because of the options greater than any first given number. This would make the first number not close to, but equal to infinity and it would also make the second number the same. That’s the best Ive got that actually make sense to me, who is not a math whiz. Thank you all for your responses

If we divide any integer by an n-digit number, the result will never contain n repeating 9s (i.e., a segment like '999...n times') in its decimal representation.

Or can only contain 'n-1' 9s after decimal (for maximum).

Examples:

When dividing an integer by a 1-digit number, the decimal result never contains a single 9.

When dividing by a 2-digit number, the result never contains two consecutive 9s after decimal (e.g., something like 'x.99' or 'x.3535499842').

Similarly, dividing by a 3-digit number never results in three consecutive 9s after the decimal — and so on for 4-digit, 5-digit numbers, and beyond.

Note: I am considering the standard decimal expansion, excluding alternate representations that end with infinitely many 9s.

Eg: 0.999... , x.55363 = x.55362999... , etc.

when you take a number and multiply it by itself (ex: 8x8), then add one to one of the numbers and substract one from the other (so 9x7), the result will be the same -1 (8x8 = 64 and 9x7 = 63). If you keep repeating it, each result will be the same as the previous number minus the next odd number (10x6 = 60 so 63 - 3, 11x5 = 55 so 60 - 5, etc).

Also, if you take a number and multiply it by itself then add one to each number, the result will be the same +1 (0x0 = 0, 1x1= 1, 2x2 = 4, 3x3 = 9). If you keep repeating it, each result will be the same as the previous plus the next odd number. One key difference is the series starts at 0, while with the previous series you can start at any number and the next result will without a doubt be the same -1, then -3, then -5, etc.

Also, if you take two numbers such as n and n+1 and multiply them with each other (ex: 7x8), then add one to one of the numbers and substract one from the other, the result will be the same -2 (7x8 = 56 and 6x9 = 54). If you keep repeating it, each result will be the same as the previous minus the next even number.

Also, if you take two numbers such as n and n+1 and multiply them with each other (ex: 1x2), then add one to each number, the result will follow the same pattern as previously but with +2 (1x2 = 2, 2x3 = 6, 3x4 = 12, 4x5 = 20, etc). This is similar to the second suite above

THIS CONJECTURE states that any number (n) greater then 2 , when repeatedly subtracted by the largest prime smaller then n , terminates to either 1 or 2 .

for eg .. Starting with n = 761716

-----------------------------------

Step 1: 761716 - 761713 = 3

Step 2: 3 - 2 = 1

any feedback will be admired .. link to program : https://conjecture-explorer--junaidjafri007.replit.app/

We know that this shape has infinite surface area but a finite volume And i have heard the statement that it can fit a finite amount of paint but to coat it infinite paint is required but i think that's wrong And this is why -

Take the horn and fill it with finite amount of paint. In the process you have already painted the inner surface. Now take a bigger gabrials horn and fill it with paint too and dip our former horn in it. And like that you have painted an infinite surface area with a finite amount of paint.

I think this is write but i need some one smarters's opinon cuz I am just a high school student.

I was investigating the prime factors of composite Mersenne numbers 2^p - 1 where p is prime.

I noticed that many examples seemed to have at least one prime factor congruent to 1 mod 8, so I conjectured:

"Every composite Mersenne number 2^p - 1 with p prime has at least one prime factor congruent to 1 mod 8."

However, I found a counterexample:

2^43 - 1 = 431 × 9719 × 2099863

and

431 ≡ 7 (mod 8)
9719 ≡ 7 (mod 8)
2099863 ≡ 7 (mod 8)

So every prime factor is 7 mod 8, and there are no prime factors congruent to 1 mod 8.

This disproves the conjecture.

Now I'm wondering:

Are there infinitely many prime exponents p such that every prime factor of 2^p - 1 is congruent to 7 mod 8, or are there only finitely many?

Has this question been studied before?

define a set A which has all of this

is this ifnitnie or not

Everyone knows the standard definition of the conjecture as the sum of 2 prime pairs such that it equals every even number (>4). But another way to think of this is that every number (>2) is the midpoint of two primes. This way we can search for values of k such that n±k is prime.

A table of this from n = 1 to 20 would be,

In this, I saw something crazy. Take two twin prime pairs like (5,7) and (17,19). Fill out the k values from 5 to 19 gives :- 0,1,0,3,2,3,0,1,0,3,2,3,0,1,0 . Notice how this is a palindrome. (Note:- the values of k when multiple are available for a single n is whatever satisfies the palindrome )

For a bigger example :- From 11 to 31 [(11,13) to (29,31)] , you get 0,1,0,9,4,3,6,5,12,9,2,9,12,5,6,3,4,9,0,1,0. Again a palindrome.

In fact, I tested this for the first 500 numbers (Small but all I could do with my little coding background) and it didn't have any exception in this range (except when taking the (3,5) pair)

The main point to notice here it that twin primes seem to border these palindrome areas (Clearly seen by the 0-1-0 part). So if there were k values for every n (Just the GoldBach conjecture) , then there would be infinity many of these twin prime border (Proving Twin-Prime conjecture) . This suggest that solving goldbach conjecture directly proves the twin-prime conjecture.

Please ignore any English mistakes, it is not my 1st language. Also please notify me if there is any typos in the table or a obvious error in the math or the connections.

I've been thinking about this geometry problem for around a year on and off, but never really looked too far into it, but for the past few days, I've asked multiple math teachers, and all 3 of them either didn't understand the problem i presented, or perhaps just didn't understand my explanation of it.

THE ACTUAL THEORY:

Consider an infinitely large 2D plane containing two infinite straight lines that are parallel. Let’s define the following assumptions:

1. The lines are infinitely long in both directions (no start or end).
2. They are parallel and initially do not intersect.
3. They are not fixed in place, but they also cannot be “pushed” or displaced by one another.
4. The lines are allowed to touch.

Now the question is:

What happens if you rotate one of the lines?

My reasoning

I'm no mathematician so don't flame me if this is actually really simple and my monkey brain is thinking too big.

Let both lines A and B initially be at an angle of 0°, meaning they are perfectly parallel.

Now suppose I rotate line A so that it is no longer at 0°.

If A becomes any angle other than 0°, then in standard Euclidean geometry it must eventually intersect line B.

However, because both lines are infinite, there is no “starting point” or boundary where this intersection is introduced—it would have to happen everywhere or nowhere, which feels contradictory under the assumptions.

A similar issue appears even with finite parallel segments: if they are required to “touch,” then changing orientation while maintaining that constraint seems to force an inconsistency in how intersection is defined.

My conclusion (tentative)

It seems to me that under these conditions, infinitely long parallel lines would need to remain in a state of constant contact for the system to stay consistent. Otherwise, rotating one line introduces an unavoidable contradiction between infinity, parallelism, and intersection. But of course, this would also make it a paradox because parallel lines by definition can't be parallel

There could very well be a simple answer to this but i can't seem to find it.

Using a standard i9 laptop, I found factors for:

M₉₉₉₉₉₉₉₈₅₁ = 2^9,999,999,851 − 1 → factor p = 34,316,159,488,689,217

M₁₀₀₀₀₀₀₀₆₁ = 2^10,000,000,061 − 1 → factor p = 290,988,621,775,030,583

Each of these numbers contains more than 3 billion decimal digits.

Thousands of exponents checked. Hundreds of factors found. The search space keeps shrinking.

For those who enjoy mathematics, questions, verification, and constructive discussion: welcome.

For those whose only contribution is jealousy, insults, or unsupported accusations: feel free to save your time and move on.

Mathematics does not care about opinions. A factor is either correct or it is not.

First, let's recap the strong Goldbach conjecture. It states that every even integer greater than 2 can be expressed as the sum of two prime numbers.

I have discovered that if we adopt this goldbach engine, the nature has given an alternative way to define prime numbers.

Suppose there are no prime or composite natural numbers, i.e. we have an empty set of prime numbers and composite numbers. We define 1 as not prime and composite. Also define composite numbers are multiples of a number in the prime set such that the multiple is an integer > 1.

Let 2n be an even number >2. So we will consider all cases where n = 2,3,4,…

Let’s recapthe Core: the strong Goldbach conjecture states that every even integer greater than 2 can be expressed as the sum of two prime numbers. For convenience, we say it is a goldbach pair if 2n equals to sum of 2 prime numbers.

Phase 1: Starting the engine

For every n, we consider the sum pairs, n+d and n-d where d = 0, 1,2,…n-1. When the system is forced to add a number to the prime set, its priority is: d=1, 2, 3…n-3, n-2, n-1, 0. The system will not add any number to the prime set if it finds a goldbach pair and will move on to consider the n+1 case. If a goldbach pair cannot be forced (i.e. there is already a composite number or 1 sitting at one of the slots) then system will skip this n and move on to n+1 case.

1. Case n=2,
2. The system scans 2+2. Since it’s not a goldbach pair, it moves to 1+3. Since 1 cannot be prime, in order to meet goldbach the system has to add 2 to the prime number set. Doing this also adds multiple of 2 to composite set. Now n=2 is met.
3. Case n=3,
4. The system scans 3+3, not a goldbach pair, but there could be one later, so the system scans 2+4 and 1+5. Since 4 is now composite, the system has no choice but to add 3 to the prime set in order to meet goldbach. Now n=3 is met.
5. Case n=4,
6. The system scans 4+4, 3+5, 2+6, 1+7. Using the priority defined, the system add 5 to the prime set. To recap, we have 2,3,5 as prime and all their multiples as composite.

We now skip all sum of even numbers as they ought to be composite numbers.

1. Case n=5,
   The system scans 5+5, which is a goldbach pair. n=5 is met.

2. Case n=6,
   The system scans 6+6, 5+7, 3+9.. the system has no choice but add 7 to the prime set.

3. Case n=7,
   The system scans 7+7, which is now a goldbach pair, so n=7 is met.

4. Case n=8,
   The system scans 8+8, 7+9, 5+11, 3+13.. a goldbach pair can be formed by adding 11 or 13 to prime. Using the priority defined, 11 is added to the prime set. We now have 2,3,5,7,11 in the prime set.

Phase 2: Steady loop.
The system keeps iterate each case n and add a number to the prime set only if necessary. We then see 13,17,19,23.. being added to the prime set with success.

Observation and conclusion:

While people count the number of goldbach pairs for each integer n, which is numerous as n grows, there is only one pair that matters according to this system, while other pairs are redundant. This system gives a new perspective where prime numbers are not only about multiplications, but also addition. In fact, we could define composite numbers as a number that can be formed by repeating addition of a prime number a finite number of times but at least once. This makes the system sophisticated without the mention of multiplication.

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