My idea regarding the Goldbach Conjecture doesn't quite work.

 

Nevertheless, I thought I would publish it anyway.  Here is the write-up, maybe someone can see a way to fix it:

Proof Attempt Regarding the Goldbach Conjecture

by Philip White

April 9, 2026

(In this write-up, != means "is not equal to.")

The Goldbach conjecture states that every positive even number can be written as the sum of two positive prime numbers.

We can re-write this conjecture in the following way:  For every positive integer Q > 2, there exists a natural number v s.t. 0 < v < Q and both Q-v and Q+v are positive prime numbers.

We can re-write the concept of primality using a logically equivalent condition:  We say that an integer L is prime if, for any integer J >= L, for all prime numbers p_i, written as p_1, p_2, ... , p_k, from 2 up to the square root of J inclusive, sqrt(J), we have that for all integers i in 1, 2, ... , k, it is the case that L (mod p_i) != 0.

Now, since by the Chinese Remainder Theorem, every integer Q is equal to exactly 1 value for each prime modulus p_i above--that is, Q (mod p_i) = B--we can state that the Goldbach conjecture is logically equivalent to the following statement:

For every positive integer Q > 2, letting J = 2*Q, there exists an integer v such that for all primes p_i in the list of primes from 2 up to and including sqrt(2*Q), Q-v != 0 (mod p_i) and Q+v != 0 (mod p_i) .

Importantly, to satisfy this condition, we must select an integer v such that for each prime modulus p_i, a particular value A and its additive inverse in the set of integers intersected with [0, p_i) is avoided.  That is, given an integer Q, for each prime modulus p_i in the list of primes up to J = 2*Q, there exists a value A such that our value v satisfies the condition that v != +/- A (mod p_i).

There are many values less than product(p_i, i, 1, k) that satisfy this condition.  We can select, using the Chinese remainder theorem, any integer v such that 0 < v < product(p_i, i, 1, k) such that v (mod p_i) != +/- A, where A is equal to Q (mod p_i), for all p_i.  (Please verify for yourself that this process works even when p_i = 2 or 3.)  This value generated from the Chinese remainder theorem may or may not work as a satisfactory value for the conjecture, in the sense that if the value is too large--larger than Q--it will yield a negative number for one of the two primes.

That is why this proof idea doesn't work.

It would seem that it is "highly likely" that the Goldbach conjecture is true, because the product of the moduli is probably not all that much larger than 2*Q.  There are tons of values that could work, so if it were random, you would say, the odds of not having a value that works that falls under the value of Q are "very low."  Unfortunately, this is not a random sample.  At least this proof idea should present some evidence that the Goldbach conjecture is true...if there were some notion of "the randomness of number theory," the proof could perhaps be made to work, based on a statistical analysis.