Goldbach's Conjecture Proof

This proof does not demonstrate Goldbach’s Conjecture.\color{#D61F06}{\text{This proof does not demonstrate Goldbach's Conjecture.}}
 Can you figure out what this proof actually shows? \color{#D61F06}{\text{ Can you figure out what this proof actually shows?}}

Goldbach's Conjecture says that all even integers greater than 22 can be written as a sum of 22 prime numbers (may or maynot be distinct).

2n=p1+p2 \therefore 2n = p_1 + p_2

PROOF

CASE 1\textbf{CASE 1} - p1p_1 & p2p_2 are not equal to 22

We know that p1+p2p_1 + p_2 is always divisible by 22. It can easily be proven -

For all pip_i not equal to 22,

p1=1(mod2)p_1 = 1 \pmod{2}

p2=1(mod2)p_2 = 1 \pmod{2}

p1+p2=0(mod2)\therefore p_1 + p_2 = 0 \pmod{2}

Let p2p1p_2 \geq p_1, so let p2=p1+ap_2 = p_1 + a for some integer aa.

2n=p1+p1+a\therefore 2n = p_1 + p_1 + a

n=2p1+a2\therefore n = \dfrac{2p_1 + a}{2}

We know 2(p1+p2)2 \mid (p_1+ p_2) and 22p1    2a2 \mid 2p_1 \implies 2 \mid a

Thus, let a=2xa = 2x

n=2(p1+x)2\therefore n = \dfrac{2(p_1 + x)}{2}

n=p1+x\therefore n = p_1 + x

Taking (modp1)\pmod{p_1} on both sides,

n(modp1)=(p1+x)(modp1)\therefore n \pmod{p_1} = (p_1 + x) \pmod{p_1 }

For n=p1n=p_1,

x\therefore x would be 00 .

For np1n\neq p1,

np1+xx(modp1)n \equiv p_1+x\equiv x \pmod{p_1}

Now (n(modp1)n \pmod{p_1}) can be any number less than p1p_1 and we can assume that (x(modp1))(x\pmod{p_1}) can also be any number less than p1p_1 keeping in mind that there are infinite primes and there is no sequence in the gap between any prime.

To keep in mind- Also, xx cannot be equal to p1p_1 because then, p2p_2 will have a prime factor.

One more fact is that for nn being a multiple of p1p_1, it cannot be proven.

Hence, it is proven for p1p_1 and p2p_2 not equal to 22.


CASE 2\textbf{CASE 2}- p1p_1 or/and p2p_2 equal to 22

2n=p1+p22n = p_1 + p_2

where one or both maybe 2. It is obvious that for p1p_1 being 22 and p2p_2 any other prime, it will be an odd integer.

So, the only possibility is that p1p_1 and p2p_2 are equal to 22.

So, 2n=2+2    n=22n = 2 + 2 \implies n=2 (See, 44 can be written as sum of 2 primes as 2+22+2 )

Hence, proved that both the primes have to be 22 simultaneously, or both have to be 2\neq 2.


Thank You for reading it. If you want to know anything, you can e-mail me. Please tell if my proof is correct or not!!

Note by Kartik Sharma
5 years, 2 months ago

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I think you've misread the Goldbach Conjecture? If any two prime numbers >2>2 are added, the result will always be even. Goldbach's Conjecture proposes that for any integer n>2n>2, there exists 22 prime numbers that add up to 2n2n. It' s trivial to show that at least one of them can be a prime number, but to show that both can be is a bit harder.

Michael Mendrin - 5 years, 2 months ago

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A 'bit'?

Mursalin Habib - 5 years, 2 months ago

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Yes, a bit.

Michael Mendrin - 5 years, 2 months ago

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That's what was bugging me! @Kartik Sharma

Sharky Kesa - 5 years, 2 months ago

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If only I could format the LaTeX for this.

Sharky Kesa - 5 years, 2 months ago

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I will do it, I can do that as a moderator. @Sharky Kesa .

@Kartik Sharma, this is not the proof of Goldbach Conjecture, this doesn't prove the existence of 2 primes that sum up to give the number, for any even integer. What you proved is, if sum of 2 primes is 2n2n, then either both primes are 22 or none of them is 22. Isn't this quite obvious ? If you want sum to be even , then both the numbers have to be odd, or both even. BTW, if it was this easy, why would it be set as a challenge in mathematics ?

Aditya Raut - 5 years, 2 months ago

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Thanks, @Krishna Ar for letting me know about this conjecture! And you too @Sharky Kesa for giving me some reviews! You can share your view on it here!

Kartik Sharma - 5 years, 2 months ago

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Dear Kartik, Goldbach Conjecture is NOTNOT that

\bullet If an even integer can be written as sum of 2 prime numbers, then both of them have to be odd primes or both have to be equal to 22     ×\implies \Huge{\color{#D61F06}{\times}}


Goldbach Conjecture is that

\bullet Every even integer greater than 22 can be written as sum of 22 primes.     \implies \Huge{\color{#20A900}{\surd}}


Please note this. I have edited the note, do not re-edit it to remove the changes I have made.

Aditya Raut - 5 years, 2 months ago

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Hey, I think I have proved it only. The Case 2 should not have been there, it is misleading my 'proof'. Why do you think I haven't proved it?

I took the equation as 2n = p1 + p2 which is correct, right?

And I proved it equal, but it is wrong, why? Which step is wrong?

Kartik Sharma - 5 years, 2 months ago

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@Kartik Sharma

Now n(modp1) n \pmod{p_1} can be any \ldots \ldots between any prime.

Could you explain what you did in that paragraph?

Siddhartha Srivastava - 5 years, 2 months ago

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@Siddhartha Srivastava Well, I have just assumed one thing. It says that xmodp1x mod {p}_{1} may vary as we don't know for sure what x is with respect to any prime p1{p}_{1} and it is quite obvious that nmodp1n mod {p}_{1} can also vary.

Kartik Sharma - 5 years, 2 months ago

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@Kartik Sharma Nope. Can't understand a bit. You need to define a lot of things first. What is p1p_1? What is p2 p_2 ? What is n n ? Are they fixed?

How about this. Instead of proving it for all even integers, prove there exist two primes whose sum is 434674.

Siddhartha Srivastava - 5 years, 2 months ago

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@Kartik Sharma Well, your proof just deals with that p1p_1 and p2p_2 will be odd primes, (started with that assumption only) , but nowhere in the proof you can assume what is to be proved. And btw, your proof also leads to that 1818 can be written as 3+153+15 or 9+99+9, which is not wanted (we want primes)

Your proof doesn't prove that there will be 2 primes for every even positive integer. Because you assumed only that 2n=p1+p22n=p_1+p_2 , and then you proved that p1p2(mod2)p_1\equiv p_2 \pmod{2}. That's it.

Aditya Raut - 5 years, 2 months ago

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@Aditya Raut I still didn't understand what's wrong with it... Well, 18 can be written as 3 + 15 is not shown by my proof. What are you saying? I have already said p1+x{p}_{1} + x should lead to another primeprime p2{p}_{2} and as that gap may vary, I have taken x as any number.

I have never proven the latter. Well, I am not convinced by your review. Actually, you are neglecting many parts of my proof- mainly the one Siddhartha Srivastva has asked.

Kartik Sharma - 5 years, 2 months ago

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@Kartik Sharma Typo, 13+5, not 3+5.

Sharky Kesa - 5 years, 2 months ago

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@Sharky Kesa Well, actually 3 + 15 another typo!

Kartik Sharma - 5 years, 2 months ago

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