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Goldbach's conjecture proof By Wadï Mami

Mami, Wadï

Abstract

Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states that every even natural number greater than 2 is the sum of two prime numbers. ---------------------------------------------------------------------------------------------- A prime number must be an odd number The sum of 2 odd numbers is an even number Then The sum of two prime numbers is an even number (A). Erdös Theorem : For every integer n > 1, it exists always a prime number between n and 2n (Source : Le Beau livre des Maths De Pythagore à la 57 dimension DUNOD edition, author Clifford A.Pickover) By récurrence of Erdös Theorem mentioned above and (A) There is always k even number which is the sum of two prime numbers p and q. (B) p for n n <= p < =2n (i) q for n/2 n / 2<= q <= n (j) (i) + (j) n + n /2 <= p+q <= 3n ie 3n/2 <= p+q <= 3n wich implies k= p+q is an even number because of (A) and n integer > 1 and k is between lower limit n+ n/2 and upper limit 3n We can state then every even natural number greater than 2 is the sum of 2 prime numbers (what needed to be demonstrated) Goldbach’s conjecture proven.

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Goldbach’s conjecture proven By me Wadï Mami Email : [email protected] / [email protected] Date : 30/10/2025 Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states that every even natural number greater than 2 is the sum of two prime numbers. ---------------------------------------------------------------------------------------------- A prime number must be an odd number The sum of 2 odd numbers is an even number Then The sum of two prime numbers is an even number (A). Erdös Theorem : For every integer n > 1, it exists always a prime number between n and 2n (Source : Le Beau livre des Maths De Pythagore à la 57 dimension DUNOD edition, author Clifford A.Pickover) By récurrence of Erdös Theorem mentioned above and (A) There is always k even number which is the sum of two prime numbers p and q. (B) p for n n <= p < =2n (i) q for n/2 n / 2<= q <= n (j) (i) + (j) n + n /2 <= p+q <= 3n ie 3n/2 <= p+q <= 3n wich implies k= p+q is an even number because of (A) and n integer > 1 and k is between lower limit 3n/2 and upper limit 3n f(x) = 3x/2 and g(x) = 3x the two f(x) and g(x) are linear function the area between them that intersect for x>0 contains points solutions which are sum of p and q prime numbers and their abscissa are x integer > 1. (B) We can state as we have (A) and (B) : every even natural number greater than 2 is the sum of 2 prime numbers. (what needed to be demonstrated) Goldbach’s conjecture proven. -- Minds, like parachutes, function best when open. ,,, (o o) / --------oOO--(_)--OOo--------------------\ | Wadï Mami didipostman | Github : https://www.github.com/didipostman