It makes sense that the factors of 2 would be important here.Quote:
What I now realize is that tremendous size and lying close together is not the key at all, is not what makes two primes behave a certain way in QR. The key is just as simple, however. The key is how many factors of 2 are involved in (p-1))(q-1), from the pure Eisensteinian perspective.
With two 4n+3 primes, p-1 and q-1 each have only a single factor of 2 to contribute. That is to say after dividing the total number of interior lattice points by 4, we come to an odd number, which of course cannot be divided evenly, so the two numbers have to have opposite characters when WAXY is divided once more by the diagonal. If only one more factor of 2 is available (which it always will be, as long as both primes are not 4n+3) to deal with this further division performed by the diagonal, then the diagonal will be dividing (apportioning) an even number of points in WAXY. If the power of 2 in the multiplication is only 23, then WAXY is forced to produce negative exponents for both primes upon the further division by the diagonal. But if the power of 2 is 24 or greater the exponents must always both be positive.
It appears that the nature of the exponents (and thereby reciprocity) depends only on the power of 2 in (p-1)(q-1), nothing else, in terms of Eisenstein's representation. The essence, the causal mechanism is none other than highly evenness. This was my original insight when I first started thinking about Brocard's problem and switched to QR, and before I actually understood what I was talking about. It has taken me this long to understand that my own insight was hitting the nail on the head squarely, to intend a pun.
With twin primes we are always guaranteed at least 23. What we simply need in order to always be even, i.e. in one another's residue set, is a 4n+1 number wherein n itself carries at least one factor of 2. From knowing nothing more we can always state the character of both primes with respect to each other in QR.
The idea of two unequal primes acting as units makes sense.Quote:
QR does not work when p=q. Imagine an 11X11 square. φ is 110, not 100. This means you cannot even get four squares (not rectangles in this case) all with equal lattice points.
But what about a 17X17 square where there are plenty of 2's to go around? This case will provide four equal squares all right. But it is a dead end, a non-sequitur, because no number between 1 and 16 inclusive will ever square out to 17 (mod 17), and so forth for all primes.
A visibly cogent fact is that the line p=q on graphing paper is a 45 degree angle and is our diagonal, and goes through all the points (1,1), (2,2), (3,3),...(17,17). The method does not work on squares. It only works on rectangles. The diagonal hits eight lattice points in WAXY. 256 interior lattice points divided by four is 64 for our quadrant square, but eight of these cannot count because they hit lattice points, bringing WAXY down to 56 servicable points, and each small triangle to 28, indeed equal, but meaningless except perhaps for why it is meaningless. Only on rectangles where p≠q are there no lattice points on the diagonal. P and q respond identically but meaninglessly when p=q because they do not kick against one another rationing out squares under the other as modulus. At the moment I do not know how to subtract those eight extra points in the context of something meaningful, I just know eight would have to be subtracted in this particular case to somehow fictionally redirect the apparently nonsensical. This is all about finding the logic of why it is illogical for squares themselves.
Only on rectangles where p≠q are there no lattice points on the diagonal.
So where p=q, it would have to look like:
[(p-1)(q-1)]-[(p-1)/2].
This is
p2-2p+1-(p-1)/2, is equivalent to
2p2-4p+2-p-1=2p2-5p+1, which means nothing to me but the sense of the nonsense.
A modulus is about division and remainders, and division is about ratios, and QR is about two unequal primes acting as units for the other under the operation of squaring, spitting out squares as remainders. Pitcher and catcher. Then switch places while the other acts as divider and see which numbers its overlap spits out as squares.
I think I remember reading something like that in Derbyshire's text. I think ultimately the ratio of the number of primes in the two sets are supposed to converge to 1 implying they have the same number, but initially the 4n + 3 set has more. I couldn't find the page to reference it.Quote:
An interesting note:
In Prime Obsession Derbyshire states that 4n+3 primes consistently out number 4n+1 primes. There may be one brief interlude where 4n+1 primes hold the lead, but then it reverts back to a 4n+3 lead, supposedly for good. If they will always hold the lead is probably unproven. I cannot remember, or if he said.

