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hypsin_0

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Registered: 2025-07-05

Posts: 146

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Try to prove it!

We've known that for any positive integer n,there is always exists a prime number p such that n≤p≤2n.
So,try to prove:
If there is a minimum even number 2m that it is greater than 2,and it can't be the sum of two prime number,then 4m can be the sum of 3 or 4 prime number.
(I am weak in grammar,so it's really hard for me to write such a long sentence.Perhaps my sentences is hard to understand with some mistakes.Sorry:)

Last edited by hypsin_0 (2026-04-05 21:08:23)

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hypsin_0

Member

From: Earth

Registered: 2025-07-05

Posts: 146

Website

Re: Try to prove it!

Nobody?Ok,I'll give the proof.
We've known that there is always exists a prime number such that m≤p≤2m,so m+m≤p+p≤4m.2m can't be the sum of two prime numbers.so2m≠p+p.also,if 4m=p+p,p must equal to 2m.2m is greater than2,so it's impossible.so 2m<2p<4m.Let 4m=p+p+n,n must be an even number.n=4m-2p<4m-2m=2m.Noticed:2m is the minimum even number that can't the sum of two prime numbers.So n=2 or n can be the sum of two prime numbers.Sm 4m = p+p+n can be the sum of 3 or 4 prime numbers.

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hypsin_0

Member

From: Earth

Registered: 2025-07-05

Posts: 146

Website

Re: Try to prove it!

I've try to use it to prove something that you must know
But I failed.
How silly I am to use sth from an eight grade's math book to prove world-famous question!

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