# Proof for Goldbach's Conjecture with Fundamental theorem of Arithmetic [closed]

The Goldbach Conjecture states that "Every even integer greater than 2 can be written as the sum of two primes".

Fundamental theorem of Arithmetic states that "Every integer can be expressed in terms of powers of prime factors"

So, in order to prove the Goldbach Conjecture, couldn't we have used the fundamental theorem of arithmetic (Taking multiplication as repeated addition and power as repeated multiplication)?

For example, the number 24=11+13.

Taking 24=(2^3)*(3)=[(2+2)+(2+2)+(2+2)]+[(2+2)+(2+2)+(2+2)]+[(2+2)+(2+2)+(2+2)] Now if we could take one prime number out (since as per the fundamental theorem, we could take one prime number) and there is at least a probability of 1 to get another prime number to form a pair of prime numbers. [5,19],[11,13].

(It seems every even number N if split by a prime number A, the other part B need not be a prime number but there exists a prime number C which could add up A+C=N)

## closed as unclear what you're asking by Lord_Farin, Lord Shark the Unknown, Jean-Claude Arbaut, KReiser, CesareoJan 11 at 11:22

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• "in terms of" is a bit vague. $210=2\cdot 3\cdot 5\cdot 7$, but how does that help us find two primes that add up to $210$? – Thomas Andrews Apr 13 '17 at 22:46
• Goldbach's conjecture would not be still open if such trivial approaches would lead to a proof. – Peter Apr 13 '17 at 23:08
• "Couldn't we have used...": other than by typing "goldbach $2003948561694$" on wolframalpha, how would your idea work in this instance? – user228113 Apr 13 '17 at 23:32

The Goldbach Conjecture states that every even number is the sum of two primes. There don't exist two primes that sum to give $123$, since $121$ and $125$ are both composite and the only way an odd number is the sum of two primes is if one of them is $\pm 2$.
So let's consider even numbers. Even numbers are of the form $2k$. If $k$ is prime, then yes we have $2k=k+k$. However, if $k$ is composite this stops producing the right kind of sum. $66=2\cdot 3\cdot 11=33+33$, but $33$ isn't prime. $66=7+59$, but there doesn't seem to be an obvious way to find that from the prime factorization alone.
• @usir0 I don't know what that means. But yes, $k+k=2k$ – Stella Biderman Apr 22 '17 at 15:22
• @usir0 1) You're wrong. If $f(x)=x$ and $g(x)=x$ and $h(x)=2x$ then $f+g=h$, 2) there's no meaningful difference between variables and polynomials, and 3) if we were to pretend that there's a difference, then I'd still be right because I'm talking about variables. – Stella Biderman Apr 22 '17 at 15:45