\(\color{Red}{\text{This proof does not demonstrate Goldbach's Conjecture.}}\)

\( \color{Red}{\text{ Can you figure out what this proof actually shows?}}\)

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!!

## Comments

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TopNewestI 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. – Michael Mendrin · 3 years ago

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– Mursalin Habib · 3 years ago

A 'bit'?Log in to reply

– Michael Mendrin · 3 years ago

Yes, a bit.Log in to reply

@Kartik Sharma – Sharky Kesa · 3 years ago

That's what was bugging me!Log in to reply

If only I could format the LaTeX for this. – Sharky Kesa · 3 years ago

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@Sharky Kesa .

I will do it, I can do that as a moderator.@Kartik Sharma, this is

notthe 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 ? – Aditya Raut · 3 years agoLog in to reply

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 · 3 years ago

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\(\bullet\) 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-editit to remove the changes I have made. – Aditya Raut · 3 years agoLog in to reply

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 · 3 years ago

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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. – Aditya Raut · 3 years ago

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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 · 3 years ago

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– Sharky Kesa · 3 years ago

Typo, 13+5, not 3+5.Log in to reply

– Kartik Sharma · 3 years ago

Well, actually 3 + 15 another typo!Log in to reply

Could you explain what you did in that paragraph? – Siddhartha Srivastava · 3 years ago

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– Kartik Sharma · 3 years 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.Log in to reply

How about this. Instead of proving it for all even integers, prove there exist two primes whose sum is 434674. – Siddhartha Srivastava · 3 years ago

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