Maybe you need to study one of the other QR proofs to help stimulate new ideas?
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Maybe you need to study one of the other QR proofs to help stimulate new ideas?
The two sets of quadratic residues for p and q are: {1 4 2} {1 4 9 5 3}.
The first set actually represents the height of the rectangle, though p in the final result is represented by the lower half of WAXY, and q the upper half, and q represents the width.
(1, 1) is the first pair, and we will take those to be coordinates. That one is easy to assign a point. (1, 4) is next, and we take those to be coodinates, too, and so on and so forth.
* * * *
(1, 1) (1, 4) (1, 9) (1, 5) (1, 3)
(4, 1) (4, 4) (4, 9) (4, 5) (4, 3)
* * * *
(2, 1) (2, 4) (2, 9) (2, 5) (2, 3)
Only those pairs with an asterik have unique coordinates in WAXY, the other pairs simply reduce to duplicates. See a pattern? Remember, the first coordinate is reduced (mod 3), and the second is reduced (mod 5). And do not forget!! our first coordinates above are vertical coordinates, and the second coordinates are the horizontal. That is the opposite of what we are all used to from algebra where the horizontal x-coordinate is always the first element in the ordered pair and the vertical y-coordinate is the second element.
If I reversed the order of the digits in the ordered pairs, we still get eight ordered pairs with asteriks, as long as I reverse the moduli too.
Were eight asteriks a coincidence above, or will there always be exactly the same number of pairs with natural coordinates as there are lattice points in one of the triangles of WAXY?
Only more grinding will know.
This is a new idea, lad. It takes quite a lot of labor to investigate one idea that is already deep in, as you can see from my last post, which dealt with only one pair of primes, 7 and 11. I am so scattered around I cannot do it all in one day. But an amateur is having fun.Quote:
Originally Posted by YesNo
I like how you paired the residues. I hadn't thought of them in that way before.Quote:
Originally Posted by desiresjab
I hadn't either. Last night I did more numerical experiments. Those pairs that work directly as coordinates within WAXY without being reduced, and those that do not, always seem to partition the two halves indentically as the diagonal does, but I can no longer conjecture the larger half will always be the one expressible as coordinates. For 5 and 11, WAY and YAX contained 6 and 4 lattice points respectively, but only four expressions for lattice points out of the ten total. There goes that one. The expression went with neither the larger half nor the one on top. At least we know the one on top will always have as many or more points than the bottom, since it wil always get the point (1, 1). Just the lean of the diagonal because of our convention of always making the short side the vertical dimension gets that done. Because we can remember a few conventions and have wiki-pejia access with Eisenstein's rectangle, we can communicate more easily. Tug lines are good in deep water.Quote:
Originally Posted by YesNo
More might come of investigating residue pairs further. It could also be a deadend that dispalys a lot of not unlikely connections without helping to get deeper into the process. Both are true of any investigation, however, and since I do not see any other method that might possibly lead ahead right now, I will roll it around for a while, as usual, without doing any more work.
If I can figure out what the number of natural expressions represents, that might be used as another means of going deeper into the machine.
I suppose the thing to look at now is the distribution of the "natural points" within WAXY, besides (1, 1), which is always predictable, and see if they tell me anything new or are simple reflections of ideas I already know.
One thing jumps out--the pairs that make it will be 1-heavy, with both coordinates small. The 1 column and the 1 row should be more full than the others.
The residue pairs seem like an interesting approach. I don't know enough about the topic to know if anyone else has looked into this. It may lead to some other unexpected results.
There are many experts right now who could tell us exactly where pairing residues leads. I never even pretend I might out think them or dream up an approach no one else has tried. The smartest people in math are simply too smart. Questions we have to dig in for some numerical on, they solve accurately in their visions. It is probable that if this leads anywhere we will end up in the territory of other maths we know nothing about, speaking in our usual language.Quote:
Originally Posted by YesNo
All the "unnatural pairs" pairs we generate with our multiplication are in reality duplicates of the natural pairs. I feel strongly that the natural pairs have some other quality not possessed by the pairs that have no representation at all. I could almost put it into words right now, but they would be only words with no mathematical help yet. It is a matter of plugging both sets in somewhere and seeing the difference in their behavior.
Now wouldn't it be a shade of wonderful if it was something as simple as they are divided according to those which are residues two ways and those which are quadratic residues only one way? Now I just have to decide how I am going to decode it. But how does that work if the division is something like 6 and 4, as it is with primes 5 and 11? I am learning not to box myself in with conjectures.
For now we can feel confident assuming we know the pairs will always partition out the same, through the coordinates or through the diagonal division, but are unable to yet say why or how. Euler might be giggling at us right now.
* * * * *
People might wonder how we can spend so much time on quadratic reciprocity without being even dumber than we admit to, when college juniors and seniors take classes in number theory where it is presented and pass their tests. It was just one more class in a long string of difficult classes for college students majoring in math or science.
Do they really understand it? They pass the tests, I believe we would pass them, too. Furthermore, we would amaze many, probably including the professor, with how thoroughly we understood many aspects of the theory.
I took calculus and differential equations, and I did better than merely pass those classes. What you learn in such classes is how to operate the formulae and which ones to use when. Normal college classes are not about in depth looks at particular problems, but learning each new language and how to operate in it. None of them go away with a greater understanding than we have right now.
As a graduate student in math they might confront this problem full on in a class, once they have had both group theory and abstract algebra. Those proofs are from the eagle's perspective, many levels above our ground level and subterranean approaches. Their view tells them something like the crankshaft will not turn, at any rate, if the carbeurator is disconnected. It does not take them down in the engine where the numbers are. At best, they learn how certain classes or fields behave, which is where we may end up yet.
The 38 lectures I watched on abstract algebra did not prove QR. A few times it seemed like they were getting close to the same ideas, though. They did prove quite a few other propositions, however, including Fermat's little theorem. The thing that knocks you off your feet is how brief the proofs are. A few sweeps of the chalk and they are done. That is how high they are soaring.
* * * *
(1, 1) (1, 4) (1, 9) (1, 5) (1, 3)
(4, 1) (4, 4) (4, 9) (4, 5) (4, 3)
* * * *
(2, 1) (2, 4) (2, 9) (2, 5) (2, 3)
Here are the quadratic pairs for 7 and 11. On this word processor, the asteriks may not line up where I want them. They didn't. The point is, if you take the moduli back to three and five, those fifteen points reduce back to only eight points. Only eight points are represented above. There are more lattice points in WAXY, that have no representation at all. Let us track down the other seven.
(1, 2) (2, 2) (3, 3) (3, 2) (4, 2) (3, 1) (3, 4)
the point (4, 2) is not actually in WAXY. It is too large. Reduced all the way it is merely the point (1, 2)
Wait again. (3, 5) is a point reperesented nowhere. (4, 2) can be reduced by the smaller moduli, (3, 5), cannot, except to (0, 0), which mamma don't allow none of around here.
(3, 5) must replace (4, 2) in the list of pairs above. Here it is:
(1, 2) (2, 2) (3, 3) (3, 2) (3, 5) (3, 1) (3, 4)
We have their names, now what can we investigate or interpret?
What can they not do that the others can?
* * * *
(1, 1) (1, 4) (1, 9) (1, 5) (1, 3)
(4, 1) (4, 4) (4, 9) (4, 5) (4, 3)
* * * *
(2, 1) (2, 4) (2, 9) (2, 5) (2, 3)
Only eight distinct points are represented above, if you use the reduced moduli. All are valid, they just happen to lie outside of WAXY on the coodinate system, though of course they are within the larger ABCD, to keep things in perspective. Magically, they all reduce back to natural pairs already listed within the matrix. There are more lattice points in WAXY. Let us track down the other seven.
(1, 2) (2, 2) (3, 3) (3, 2) (3, 5) (3, 1) (3, 4)
Here are the two residue sets again. {1 4 2} {1 4 9 5 3}.
We intuitively understand why the point (1, 1) is represented. Can we understand why a point like (1, 2) above is not? We want to understand it in a better way than just that the two sets when combined in the order shown, cannot produce that pair. What do we need to try?
For (1, 2)
1 is a residue of 7, but 2 is not a residue of 11. That is a start. Keep going.
For (2, 2)
2 is a residue of 7, but 2 is not a residue of 11.
For (3, 3)
3 is not a residue of 7, but 3 is a residue of 11
For (3, 2)
3 is not a residue of 7, and 2 is not a residue of 11
For (3, 5)
3 is not a residue of 7, but 5 is a residue of 11
For (3, 1)
3 is not a residue of 7, but 1 is a residue of 11
For (3, 4)
3 is not a residue of 7, but 4 is a residue of 11
Remember, these are pairs which are coordinates, but were not generated in our combinatorial multiplication. There is at least one-way rejection, for the entire list. But for the matrix of pairs above with fifteen pairs, both never reject, all the time. Even the pairs that lie outside of WAXY are valid, they are just not inside WAXY, which is true for some of the quadratic residue pairs generated in the multiplication. Since some of the valid pairs generated in the combinatorial multiplication lie outside of WAXY, we can only expect some of the points inside WAXY to be valid residue pairs. In fact, eight of them are, and seven of them are not. This accords exactly with the split of the diagonal, and the split of the natural expressions.
Those lattice points within WAXY that aren't two-way accepting, fill the second column and fifth rows, exclusively, forming a right-heavy bar on the capital T of its shape. This puts 5 of the 7 points in WAY, but 2 of them in YAX. Those seven are the same ones we said earlier had no “Natural expression.” Natural expressions accompany pairs of two-way acceptance only. We wonder on the side if only one is usual.
For primes 7 and 11 and the rectangle WAXY, eight of the lattice point coordinates (the same ones with natural expressions) do indeed have two-way acceptance pairs as coordinates, and seven have at least one rejection. A perhaps complexifying aspect here is that one of those seven pairs (3, 2) is two-way rejecting.
When you represent the pairs as coordinates when Eisenstein's rectangle ABCD is on coordinates, then eight of the fifteen pairs of coordinates within WAXY are coordinates where both elements are residues of their respective prime in the ordered pair. Those are exactly the same pairs that have natural expressions that were generated in the combinatorial multiplication. Six pairs of coordinates have only one quadratic residue, and one pair (3, 2), has none, i.e. 3 is not a residue of 7 and 2 is not a residue of 11. Graph-wise, (3, 2) is where the leg of the T and its bar intersect and occupy the same lattice point.
Somehow, the diagonal and the value of the number of natural expressions, manage to cut WAXY correctly in terms of two numbers to be used as exponents, innocent lattice points, and actual coordinates. The diagonal does not cut all the naturals to one side, but gets the numerical partition correct. The T marks the exact positions. That coordinates even apply is great!
7 and 11, being a pair of 4n+3 primes, are never mutal hosts to one another, but some of the pairs generated are, in a sense. We generated the pairs the old fashioned way—through a multiplication process more basic than the one taught in grade school, and found out that 15 is the number of distinct pairs generated from two sets of 3 and 5 members, respectively. Eight of these pairs are mutally accepting pairs, and seven of them are not.
Now it's back to the think tank.
As you mentioned math classes are more to show students how to use the language and solve some problems. Generalizing does take one away from the details.
I was looking at the Sierpinski problem recently since that is the one that I have a computer working on for PrimeGrid. I am trying to see if I can get Python and MySQL to help generate covering congruences for some numbers that are unknown whether they are Sierpinski numbers or not.
A Sierpinski number is a number of the form k 2n + 1 where k is odd. A Sierpinski number is a k such that for all n none of the numbers are prime.
Paul Erdos made some discoveries concerning covering numbers. The Sierpinski gasket is a fractal object.
5 and 17 are a strange pair. They are both 4n+1. Phi/4 is 16, so there are plenty of factors of two left. Yet when the diagonal divides them it partitions them to 9 and 7. Since they are mutually rejective with that many factors of two left, that is all the diagonal could do.
Sometimes the diagonal is forced to parttion an odd number into two odd ones, as in 7 and 11, when there are only two factors of two (the minimum), or break an even one into two odds of different value, such as for 5 and 11, where there are only three factors of 2. But this is the first time I have seen a highly even number partitioned unevenly. The diagonal always gets it right. I have not had time to check the "natural" pairs for for 5 and 17 yet to see what they say. There are only 16 of them, so it will not be difficult.
I did check those sixteen pairs of coordinates, and any conjecture crashes between the number of naturally expressible coordinates and the partitioning of lattice points. Those coordinates are merely the ones found strictly within WAXY. Only four coordinates out of sixteen are of this type for the two primes 5 and 17, and we know that number does not correspond to either value of the partition.Quote:
Originally Posted by desiresjab
With certainty, we know that the diagonal for two 4n+3 primes will "thread," through an odd number of lattice points to partition them into one or other of WAY or YAX in unequal odd numbers.
With certainty, we know that a 4n+3 prime and a 4n+1 prime of lowest evenness for its kind (only divisible by 2 twice), will "thread" through an even number of points, partitioning them into odd halves which may be either equal or unequal, as far as we know.
With certainty, we know the two situations above were forced by a limited number of factors of 2.
In the Eisenstein rectangle lattice point graph for the two primes 5 and 17, the "lean" of the diagonal steals away three lattice points on the bottom row, but two of them get made up somewhere.
Whether or not the values of WAY and YAX can ever differ by more than two, becomes an interesting question in itself.
You might try constructing proofs of the certain items if for no other reason than to get a foundation for future results. There is a theory of lattice points that I am unfamiliar with that might be a place to start.
I thought you found examples where the WAY and YAX differed by more than two lattice points. Perhaps not.
I'm working on a problem at the moment and trying to get Python to generate some examples or a solution. The claim is that the sum over n starting with 1 of (n-1)n is never prime. So the goal is to find a prime in that sequence of integers or find a covering congruence to show that the sequence has no primes.
I'll be happy if I can get a script to generate composites of this form up to n = 1000. I think I have a workable algorithm for that, but I don't even have a way to show that a covering set of primes actually covers all of the numbers in the sequence.
I may have stated incorrectly once I had an example. I believe I have no examples of WAY and YAX with a difference of more than two lattice points. These problems get huge to generate by hand with relatively small primes. I have no mathematical software to assist.Quote:
Originally Posted by YesNo
Your problem sounds interesting, and either has an echo or a false echo of Fermat.
It is a snap to show that (p-1)(q-1)/4 is equivalent to Euler's totient function. Pretend that these two primes are really, really, really huge. We can always tell their types, but ascertaining whether one is a quadratic residue of the other may be next to impossible by hand. What can we do?
We also pretend we have a computer capable of multiplying (p-1)(q-1), in fact we will need one. Trusty division by four is our next step. Now we have something to look at. We can tell the species of this quotient, too.
If the quotient is already an odd number, we know the diagonal will produce an even and an odd number. We must have been dealing with two 4n+3 primes.
If the quotient is a 4n+3 number, at least the possibility if not the certainty of (1)(1) is preserved, though I see no way yet to determine if it is that or (-1)(-1) for very huge numbers.