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# Goldbach's Conjecture Proof

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

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
2 years, 6 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. · 2 years, 6 months ago

A 'bit'? · 2 years, 6 months ago

Yes, a bit. · 2 years, 6 months ago

That's what was bugging me! @Kartik Sharma · 2 years, 6 months ago

If only I could format the LaTeX for this. · 2 years, 6 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 ? · 2 years, 6 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! · 2 years, 6 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 $$2$$ $$\implies \Huge{\color{Red}{\times}}$$

Goldbach Conjecture is that

$$\bullet$$ Every even integer greater than $$2$$ can be written as sum of $$2$$ primes. $$\implies \Huge{\color{Green}{\surd}}$$

Please note this. I have edited the note, do not re-edit it to remove the changes I have made. · 2 years, 6 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? · 2 years, 6 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. · 2 years, 6 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. · 2 years, 6 months ago

Typo, 13+5, not 3+5. · 2 years, 6 months ago

Well, actually 3 + 15 another typo! · 2 years, 6 months ago

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

Could you explain what you did in that paragraph? · 2 years, 6 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. · 2 years, 6 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?

How about this. Instead of proving it for all even integers, prove there exist two primes whose sum is 434674. · 2 years, 6 months ago