I know, I know. Primes are the cool topic. But maybe we’ll understand a bit more of what they are by understanding a bit more of what they aren’t.

Here’s something I first noticed back in the late 90s. Make a grid of the different ways you can add to numbers to get a specific number. We’ll start with 2, since we aren’t interested in the “degenerate” case of n + 0 = n.

2 = 1+1

3 = 1+2

4 = 1+3 = 2+2

5 = 1+4 = 2+3

6 = 1+5 = 2+4 = 3+3

7 = 1+6 = 2+5 = 3+4

8 = 1+7 = 2+6 = 3+5 = 4+4

Let’s stop here for a moment. Let us now take each pair of numbers in each row and instead of adding them, let’s multiply them.

2: 1

3: 2

4: 3; 4

5: 4; 6

6: 5; 8; 9

7: 6; 10; 12

8: 7; 12; 15; 16

Note that in this small sample, the same number occurs after the colon in two consecutive rows only twice. 4 appears in the rows headed by 4 and 5. 12 appears in the rows headed by 7 and 8. If you add 4 and 5, you get 9, a composite number. If you add 7 and 8, you get 15, a composite number. If you add any of the numbers that head any of the OTHER pairs of consecutive rows, you get, respectively, 5, 7, 11 and 13, all primes.

What gives?

Let’s look at some other examples. We know that 25 and 27 are composite. So if we take the numbers that add up to 12 and multiply them and take the numbers that add up to 13 and multiply them, we should find one product in common on the two lists. Similarly for 13 and 14. So the set of numbers after 13 should intersect once with those after 12 and once with those after 14. Cutting to the chase, note that:

12 = 6+6 and 6*6 = 36

13 = 4+9 and 4*9 = 36

13 = 5+8 and 5*8 = 40

14 = 4+10 and 4*10 = 40

Conversely, we know that 29 and 31 are prime. So no pair of numbers that sum to 15 will, when multiplied, match up with any pair of numbers that sum to 14 or 16 when those are multiplied, if the pattern holds firm. And it does.

14: 13; 24; 33; 40; 45; 48; 49

15: 14; 26; 36; 44; 50; 54; 56

16: 15; 28; 39; 48; 55; 60; 63; 64

So then, will this pattern hold up in all cases? That is, will we find one of these matches between the numbers in the sequences headed by consecutive numbers n and n+1 if and only if 2n+1 is an odd composite number and not a prime? The answer is yes, and it is quite easy to show.

BBA Theorem 1

Let us first consider whether ALL odd composites will exhibit this phenomenon. Every odd composite 2n+1 can by definition be expressed as the product of  two odd numbers 2x+1 and 2y+1, where x ≥ 1 and y ≥ 1. Hence, 2n+1 = (2x+1)(2y+1) = 4xy+2x+2y+1.

Therefore, n = 2xy+x+y and n+1 = 2xy+x+y+1.

Therefore, n = (xy+y)+(xy+y) and (xy+x)(xy+y) = (x^2)(y^2)+(x^2)(y)+(x)(y^2)+xy

Meanwhile, n+1 = (xy)+(xy+x+y+1) and (xy)(xy+x+y+1) = (x^2)(y^2)+(x^2)(y)+(x)(y^2)+xy

QED. All odd composites will exhibit this property.

Note, by the way, that these equivalent products are also equivalent to the product of the two pronic numbers x^2+x and y^2+y.

It’s just a little more complicated to show that ONLY odd composites, and no primes, will exhibit this phenomenon.

The phenomenon can be slightly recast as follows: For a given pair of consecutive numbers n and n+1, which of course add up to odd number 2n+1, are there an i and a j such that (n-i)(i) = (n+1-j)(j) and, if so, what does it tell us about the odd number 2n+1? As it turns out, it tells is that 2n+1 is composite.

First note that we are talking about natural numbers here, that is, n (and therefore n+1) and i and j are all natural numbers, hence (n-i)(i) and (n+1-j)(j) are also natural numbers. Also, we are not interested in trivial cases in which one of  i or n-i and one of j or n+1-j equal zero. With all this understood, we have:

(n-i)(i) = (n+1-j)(j)

ni-i^2 = nj+j-j^2

n(i-j) = i^2-j^2 +j

n= i+j+(j/i-j)

Since n is a natural number, j must be divisible by i-j

Let j = xy and i-j = x

Hence i = j+x = xy+x and j/i-j = y

Hence n = i+j+(j/i-j) = xy+x+xy+y = 2xy+x+y  and 2n+1 = 4xy+2x+2y+1 = (2x+1)(2y+1), ie a composite number as long as x≥1 and y≥1. But they must be. Recall j≥1 and j=xy.

QED

This is the basic proof.

In future installments, I will show that the phenomenon adduced here appears to be far, far more general. But first, I will play around with some algebra to tidy up the proof to preempt some questions that some may have about it at this point. But this is a good stopping point for now!