ON GOLDBACH'S CONJECTURE

An equivalent form of the Goldbach Conjecture is stated using manipulation of characteristic equations and simple logical arguments that lead to an equation which restates the conjecture. A new form of the number of unordered partitions of an even number into two primes is presented.

For n>0, is clearly 0 if n is an integer and otherwise.
Therefore, is 1 if n is an integer and if n is not an integer. So by the property of the floor function, End Proof It is well known that for an integer n. So, .
By Wilson's theorem, for a natural number n >1, So by the proof of the previous theorem, . This is clearly the characteristic equation of the primes.
Theorem 2: where is the number of unordered partitions of into two primes. Proof: For an even number , I will use in this example, the numbers from to may be written in order and then one may write the numbers backwards, offset by , directly above as follows, 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 It is clear to see that each column will be equal to , in this example. Where the top and bottom both have primes, this is a solution for of the Goldbach Conjecture. To avoid counting a solution twice and noting that 1 is not in the solution set, it is clear to see that it is only necessary to count solutions between and inclusive. The primes may be counted backwards using the characteristic equation of the primes at the value and ranging from to , the primes may be counted forward by using the characteristic equation of the primes at the value and ranging from to . This will include any possible solution. A characteristic equation can only be or , so multiplying these two characteristic equations together ensures that both must be one for the product to be , otherwise it will be . In this case, multiplying these two characteristic equations together and summing will count only primes which sum to and thus, The form of the number of partitions of into two primes, which I will denote as ,

End Proof
Theorem 3: = Where is the number of unordered partitions of into primes, is the number of unordered partitions of into composites, and is the prime counting function.

Proof:
As I have shown, Expanding the sum, The first of these sums is clearly . In the second, the that was added to the floor function has been omitted, so the sum has been subtracted from . The sum with added to the floor function would have been , so this is simply the number of composites less than or equal to n minus 1 which is . With this and noting that the sum is now negative in its form, .
The second sum in the theorem, using the same approach as before would have been if was still added to the floor function, because the characteristic equation would have been counting primes in this interval. So this sum is the negative of .
The third sum in the theorem, is still counting partitions of , but now it is clearly counting composites. So this sum is equal to the number of unordered partitions of into two composites, which I will denote as . This sum is positive because the characteristic equations are multiplied together.
Adding the sums together therefore gives, = End Proof From this list it is clear that every partition is an odd number paired with an even number. Any even number is composite. Therefore, a pair will be an unordered partition of into two composites if the odd number in the pair is not prime. Excluding of course , which will always be paired with 2 and excluding 1, because it is not in the solution set of composites. There are exactly unordered partitions total, because the partitions range from to . Ignoring the pair that includes , this gives possible solutions. Subtracting from this will exclude any odd primes in these remaining possible solutions and will also leave out leaving only partitions with a composite odd term. Since we can count all odd composite solutions within the partitions, and these will always be paired with even numbers , this count includes all possible solutions. Therefore, End Proof Recall the equation proven in theorem 3, which was We can now replace in this equation with . The equation therefore becomes, Now, is just the characteristic equation of the primes. So this may be simplified further as,

Where
is the characteristic function of the primes which is if is a prime and otherwise.
From this, the Goldbach Conjecture can be restated as I will show in the following theorem.
Theorem 5: The Goldbach Conjecture is equivalent to proving that for composite Proof: By Proof of theorem 4, the number of unordered partitions of into primes has the form, .
If is prime, a solution will always exist in the form . So, it suffices to prove there is a solution for all composite But, for composite , So for composite Now, if the Goldbach Conjecture is true, than the left side of this equation must be greater than and so the right side must be as well. This means of course that we cannot have Therefore, we must have for composite End Proof By proof of theorem 4, this of course implies that we must have for composite . By conjecture, this statement is true for all CONCLUSION It can now be seen that proof of even a close approximation to the number of unordered partitions of into composites may be a key factor in proving that the number of unordered partitions of into primes cannot be for From this it is clear that studying is just as important as studying They will hereby forever be connected by their relationship in the form, This can only add to the complexity of such a simple and beautifully impossible question. Is every even number the sum of two primes?