# Goldbach's Conjecture Proof

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

Goldbach's Conjecture says that all even integers greater than $2$ can be written as a sum of $2$ prime numbers (may or maynot be distinct).

$\therefore 2n = p_1 + p_2$

PROOF

$\textbf{CASE 1}$ - $p_1$ & $p_2$ are not equal to $2$

We know that $p_1 + p_2$ is always divisible by $2$. It can easily be proven -

For all $p_i$ not equal to $2$,

$p_1 = 1 \pmod{2}$

$p_2 = 1 \pmod{2}$

$\therefore p_1 + p_2 = 0 \pmod{2}$

Let $p_2 \geq p_1$, so let $p_2 = p_1 + a$ for some integer $a$.

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

$\therefore n = \dfrac{2p_1 + a}{2}$

We know $2 \mid (p_1+ p_2)$ and $2 \mid 2p_1 \implies 2 \mid a$

Thus, let $a = 2x$

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

$\therefore n = p_1 + x$

Taking $\pmod{p_1}$ on both sides,

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

For $n=p_1$,

$\therefore x$ would be $0$ .

For $n\neq p1$,

$n \equiv p_1+x\equiv x \pmod{p_1}$

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

To keep in mind- Also, $x$ cannot be equal to $p_1$ because then, $p_2$ will have a prime factor.

One more fact is that for $n$ being a multiple of $p_1$, it cannot be proven.

Hence, it is proven for $p_1$ and $p_2$ not equal to $2$.

$\textbf{CASE 2}$- $p_1$ or/and $p_2$ equal to $2$

$2n = p_1 + p_2$

where one or both maybe 2. It is obvious that for $p_1$ being $2$ and $p_2$ any other prime, it will be an odd integer.

So, the only possibility is that $p_1$ and $p_2$ are equal to $2$.

So, $2n = 2 + 2 \implies n=2$ (See, $4$ can be written as sum of 2 primes as $2+2$ )

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

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
6 years, 10 months ago

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I think you've misread the Goldbach Conjecture? If any two prime numbers $>2$ are added, the result will always be even. Goldbach's Conjecture proposes that for any integer $n>2$, there exists $2$ prime numbers that add up to $2n$. 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.

- 6 years, 10 months ago

A 'bit'?

- 6 years, 10 months ago

Yes, a bit.

- 6 years, 10 months ago

That's what was bugging me! @Kartik Sharma

- 6 years, 10 months ago

If only I could format the LaTeX for this.

- 6 years, 10 months ago

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 $2n$, then either both primes are $2$ or none of them is $2$. 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 ?

- 6 years, 10 months ago

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!

- 6 years, 10 months ago

Dear Kartik, Goldbach Conjecture is $NOT$ 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 $2$ $\implies \Huge{\color{#D61F06}{\times}}$

Goldbach Conjecture is that

$\bullet$ Every even integer greater than $2$ can be written as sum of $2$ 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.

- 6 years, 10 months ago

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?

- 6 years, 10 months ago

Now $n \pmod{p_1}$ can be any $\ldots \ldots$ between any prime.

Could you explain what you did in that paragraph?

- 6 years, 10 months ago

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

- 6 years, 10 months ago

Nope. Can't understand a bit. You need to define a lot of things first. What is $p_1$? What is $p_2$? What is $n$? Are they fixed?

- 6 years, 10 months ago

Well, your proof just deals with that $p_1$ and $p_2$ 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 $18$ can be written as $3+15$ or $9+9$, 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 $2n=p_1+p_2$ , and then you proved that $p_1\equiv p_2 \pmod{2}$. That's it.

- 6 years, 10 months ago

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 ${p}_{1} + x$ should lead to another $prime$ ${p}_{2}$ 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.

- 6 years, 10 months ago

Typo, 13+5, not 3+5.

- 6 years, 10 months ago

Well, actually 3 + 15 another typo!

- 6 years, 10 months ago