Goldbach’s Conjecture Proof

Here’s a proof sketch of Goldbach’s conjecture, which states that any even number can be formed as the sum of two prime numbers.

We start by envisioning any even number as the sum of any column within two rows. E.g. 14 is:

 2  3  4 5 6 7
12 11 10 9 7 7

The bolded “columns” are those ones that satisfy the conjecture, i.e. are composed of two primes.

Conceptually, the conjecture merely says that it’s impossible to “eliminate” every number below the primes in the top row via division by one of the primes in the top row. For example, the 3 eliminates the 9 from consideration (which is why the 5 doesn’t count despite being prime itself), but nothing eliminates the 11 or the 7 on the bottom row.

To start, we recognize that the pattern of “hits” created by the primes in the top row is unique within the scope of any composite prime series. For instance, the prime series 2, 4, 6… and 3, 6, 9… form the composite prime series 2, 3, 4, 6… By bolding the numbers in this series, and italicizing the rest, we can see this as a visible pattern that is symmetrical.

0123456

Lemma 1: This pattern (hit, miss, hit, hit, hit, miss, hit) can be repeated if and only if we add the product of the primes. In this case, 2⋅3=6. And indeed, if we add 6 we get:

6 7 8 9 10 11 12

And this is true for any multiple of 6. Suppose we add 12 instead:

12 13 14 15 16 17 18

Now let’s try some number that’s not a multiple of 6, such as 8.

8 9 10 11 12 13 14

We don’t hit the middle number, which in this case is 11.

Now we can extend Lemma 1 in two ways to make Lemma 1b.

  1. It is true even if we replace the “skip” with “wildcard”. I.e. we could say that the miss on 11 is disqualifying in that last sequence, but the hit on 9 isn’t, and it will still be true.
  2. It is true even if we confine our search to either half of the symmetry.

In other words, even if we were only searching for a hit-wildcard-hit-hit, we’d only find it starting on 6a + 2.

Let’s do another quick example for funsies, using 2, 3, and 5. Their product is 30, so we can add 30 to the original sequence and see the congruence.

 0  1  2  3  4  5
30 31 32 33 34 35

32 is divisible by 2 (but not 3 or 5). 33 is divisible by 3 (but not 2 or 5). And 35 is divisible by 5 (but not 2 or 3). We can also operate in reverse, to see that 30 was the end of a symmetry as well.

0 1 2 3 4 5 ... 25 26 27 28 29 30

28 is divisible by 2 (but not 3 or 5). 27 is divisible by 3 (but not 2 or 5). 25 is divisible by 5 (but not 2 or 3). And here is the crucial part! This symmetry exists if and only if the final number is a multiple of the primes that form the series. I.e. 30 in our example with 2, 3, and 5.

Why does this matter? Because, if we go back to our original problem, we’re trying to prove that it’s impossible to create a number where the primes in the top row (less than or equal to half of that number) are paired with numbers on the bottom row that are divisible by the primes in the first row. Now we see that the only way for those bottom numbers to all be divisible is if they are at the end of the composite prime sequence. I.e. we have to find an even number, 2k, that equals the product of all primes less than or equal to k. And (Lemma X), that’s impossible.

For instance, 14 is not equal to 2⋅3⋅5⋅7=210.

QED

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