Printable View
Interestingly enough, suppose one intended a proof of quadratic reciprocity derived from the totient function. This would only work for odd primes. We have to use the formula (p-1)(q-1), where one or the other of the expressions is (2-1). A diagonal of the Eisenstein diagram of these dimensions will not make numerical sense.
The totient function is only a shortcut for odd primes, but what a shortcut it is.
If you can use a spreadsheet you can check some of these for low numbers. I use Google sheets since it is convenient, in a cloud storage and free. Here's a link to a Google sheet I made some time ago about your conjecture: https://docs.google.com/spreadsheets...it?usp=sharingQuote:
Originally Posted by desiresjab
It looks like 5 and 23 have a difference of 4. (Of course I might have constructed the sheet incorrectly.)
There are four tabs on the spreadsheet. On the Configuration tab there are entries for "Prime A" and "Prime B". Those are the only values to change. On the Lattice Points tab is a graph of how the lattice points look. The fractions represent deviations from 0 or the diagonal. The Twin Primes tab is a list of tests for twin primes. It is not fully filled out. The References tab are places I looked for lattice point information. The sheet is limited to numbers under 100.
You should be able to copy the spreadsheet from the link and modify that copy should you want to use something like this. You may need a Google account to set up Google drive if you don't have this already.
I am planning on using Fermat's Little Theorem to simplify the calculations in Python. Basically I would be using ap-1 = 1 (mod p) for prime p. I have this set up on a Google sheet, but these easily get past the max size of integers on a spreadsheet. So I have to use something like Python. That's a free tool you also might find useful. I am still feeling my way around it, but I have programmed in many different languages for decades.Quote:
Originally Posted by desiresjab
I did it twice by hand, trying to make sure. My rough graphs by hand are not ultimate arbiters, but it appears you may be right. Yet several lattice points are really hard to judge by eye. If we judge there to be 9 and 13 points respectively, and are wrong on one lattice point, that the brings the totals to a respectable 10 and 12, except we know that is wrong--neither one of these is the quadratic residue of the other, by the application of easy properties.
However, if we judge there to be 8 and 14 points respectively, being off by one point transfer would make both values negative again, and fulfill that needed condition.
Since you are using software, I tend to go with your results as more definitive in the judging-by-eye department.
I tried it by hand initially, but then I realized I was making too many mistakes. I think the spreadsheet is correct, but I'm not sure. One of the problems with software is that it has many more pieces to check than a proof.
That is why we are still down in the engine room, looking for the fundamental mechanical principle of mere numbers everything relies on and quadratic reciprocity expresses. Sometimes investigations turn out to be superflous to solving the problem, but add to one's knowledge base. I would say that is the rule rather than the exception. We know now that the principle is not related to the number of lattice points expressible in WAXY through primitive combinatorial multiplication to generate coordinates. That was an important thing to get out of the way, once the idea came up.Quote:
Originally Posted by YesNo
Have the masters really captured everything there is know of QR from the ground level view? I cannot allay the suspicion that the mechaical principle is visible through all the gears, wires and steaming valves, if one stands exactly in the right place and bends down just so with a crane of the neck and peers through the complexity at the cause of it all.
Time was I was sure I had it, and up to a point I did. Once the combined totals of factors of 2 in (p-1) and (q-1) reach 24, however, the outcome of the diagonal split of this even number of lattice points in WAXY, cannot be predicted, though one has other rules, properties and laws to consult to usually clear up what the mutual reciprocity is, which is what one is usualy after.
We were looking at the number of points in WAXY and their coordinate names as a side issue. I no longer know if it is relevant.
The answer I am looking for, and the way I am looking for it, make a fantastically difficult yet solvable problem, I suspect. This is what I wanted. I see no need to move on. I keep learning more. The engine room is fine.
I've started using pari/gp for calculations needing multi-precision arithmetic. It is also an algebra package primarily for number theory. It allows you to work with matrices and you could probably implement the WAXY pattern for integers larger than what the Google sheet allowed.
The gp part is a calculator and the pari part is a C library. I am more interested in the calculator. There are free C compilers available which could use the libpari library, but I'll give Python a chance first although I suspect pari might be faster. Here is the location of pari/gp: http://pari.math.u-bordeaux.fr/download.html
Since Moffat's gravity theory made a prediction about gravity waves from the big bang which is different from what the Newton-Einstein theory would predict, I was looking at LIGO which showed the existence of gravitational waves last year. I am not sure what Moffat's prediction is, but at https://losc.ligo.org/about/ there is a tutorial about LIGO's recent findings with an interactive Jupyter notebook allowing you to play around with the data.
There are other alternative gravity theories. Moffat discusses them in "Reinventing Gravity". One of the benefits of a modified gravity is that dark matter would not be necessary and black hole singularities could be eliminated.
The inability to find dark matter and the observations of the rotational speed of galaxies are evidence that the Newton-Einstein theory of gravity is incorrect and needs modification. The observed movements of galaxies falsifies the Einstein gravitation theory without dark matter.
But all of these are theories or models of the universe. Even Moffat's theory is only a map. It is not the reality. Sometimes it is hard to keep the map and reality separate since the only way we can make sense out of reality is through a map.
When Moffat discussed other alternative graitational theories and the problems in them, he wrote this regarding "quantum gravity", an attempt to combine gravitational theory with quantum theory: (John W. Moffat, "Reinventing Gravity", Harper Collins, 2008, page 142)
Some theorists simply claim that since gravity is observed and quantum mechanical effects are also observed that qualifies as enough experimental evidence that a quantum gravity theory is necessary.
The implication is that quantum gravity is not necessary. The problem with getting quantum gravity to work is to take any theory that works on the quantum level and getting predictable results that match what is observed about gravitation on the cosmic level.
If one doesn't need quantum gravity then there is no need for a "graviton", a quantum particle of force associated with gravitation like the photon is associated with electromagnetic radiation which so far has not been detected.
What results if any to report of Moffat's attempt at a time machine? I know he proposes a coil of lasers to produce warpage, where only light is involved in this bending instead of massive objects. He expects to find particles (marked somehow, I assume) when he performs his early experiments that he has already sent back in time to himself, which is mind bending. A single particle is what he is trying to send back or ahead, for now.
Whoops! Excuse me, please. I got names mixed up. The guy I was thinking of is Ron Mallett.
I remember reading something by Ron Mallett regarding time travel some years ago, but I don't think time travel is possible. That would violate the second law of thermodynamics where we can only go from low entropy to high entropy, from past to future. By the way, I haven't run into any time travelers.
John Moffat has an interesting thing about time at t = 0 ("big bang") which he does not consider to be a singularity. He assumes there are two universes one going into the past and the other into the future. We don't know which one we are in. It helps avoid the singularity at t = 0. Other singularities such as black holes are also eliminated. And he doesn't need dark matter or a multiverse in which the anthropic principle can get us to where we are now. That is, it makes predictions which could be falsified or verified if LISA becomes operational and we can view gravitational waves from the origin of the universe.
But it is only a model. Its main goal is to fit the observations of acceleration in the galaxies that cannot be explained by Einstein's general relativity without assuming there is more matter in the universe than we can observe. It is useful or not if it can make accurate predictions.
By narrowing our look at Quadratic Reciprocity to twin primes only, we are able to initially highlight those two instances that interest us most, that is, where the larger twin is a quadratic residue of the smaller (and therefore the smaller is a residue of the larger, as well). We want to know what to expect of a set of twins at a mere glance.
Only 8n+1 and 8n+7 primes have 2 (the difference of any two twins) as a quadratic residue. In the case of 19 and 17, 19 is an 8n+3 prime which is a quadratic residue of 17, the 8n+1 prime. In the case of 73 and 71, 73 is a 9n+1 prime and 71 is an 8n+7 primes, whose difference is 2. Those should be the only two types of cases where the larger twin is a quadratic residue of the smaller.
Quadratic residues of 17 are 1, 4, 9, 16, 8, 2, 15, 13. Another way of expressing this group is: -9, -4, -2, -1, 1, 2, 4, 9,
The latter expression illustrates how the quadratic residues are grouped symmetrically around zero. Of course, we could always substitute 19 in for 2, to make it even more clear that 19 is a quadratic residue of 17. We can easily see, that is, when 6x6 is divided by 17, it leaves a remainder of 2, making 2, and thereby 19, a quadratic residue of 17.
Now let's make a list of the quadratic residues of 19.
1, 4, 9, 16, 6, 17, 11, 7, 5, It also looks like this:
-14 -13 -8, -3, -2, 1, 4, 9, 16.
The groupings are asymmetrical, for 19 is an 8n+3 prime, and of course therefore a 4n+3 prime.
We see that 17 is a quadratic residue of 19, as well. What we also see is that in either of these cases (an 8n+3 and an 8n+1, or a 9n+1 and an 8n+7), the two primes will be quadratice residues of each other. We further can note that in the case of the Legendre symbols for these two numbers, they will always be positive, so one is always multiplying two positive Legendre symbols together.
Of course the above cannot be the case in general for twin primes, but only for the two cases we looked at.
What happens for other twin prime combinations? Well, what has to happen? First of all, we think we can guarantee that all other combinations of twins will generate two negative Legendre symbols to multiply together to acheive positive 1. All we have to do is try a few.
11 and 13 are 8n+3 and 8n+5. Quadratic residues of 11 are:
1, 4, 9, 5, 3. An identical expression is: -8, -6, 1, 4, 9.
Quadratic residues of 13 are:
1, 4, 9, 3, 12, 10. An identical expression is: -10, -3, -1, 1, 4, 9.
Notice that neither group contains its twin in its residue set. Both Legendres will be negative, producing a positive upon multiplication.
The remaining case is 8n+5 with 8n+7. The twins 29 and 31 will fit this bill.
The residues of 29 are:
1, 4, 9, 16, 25, 7, 20, 6, 23, 13, 5, 28, 24, 22.
The residues of 31 are:
1, 4, 9, 16, 25, 5, 18, 2, 19, 7, 28, 20, 14, 10, 8.
Notice that once again, neither is in the other's quadratic residue set. The Legendres will be negative by themselves, producing a positive product.
The only case we did not explore was 9n+1 and 8n+7. Unfortunately the the smallest pair of twins with this form are 71 and 73, for which I do not care to calculate the sets. But I can guarantee they are quadratic residues of one another.
It appears that if primes, and in particular twin primes, are equally distributed by type, then half the pairs will be residues of one another and half will not be. In any case, for twin primes the combined Legendre symbols will always create positive reciprocity, whether it attains it through (1)(1) or (-1)(-1).
I conjecture for the moment that the latter [(-1)(-1)] occurs when only three factors of 2 are involved between the minus ones of the two twins.
This conjecture feels hopeful. If it were true, it would enable us to immediately "see" the characters of the separate Legendres involved. Everything seems pinned on the 8n+5 number, since it plays a part in both cases. If its minus one contains more than two factors of 2, I am saying it will never cut Eisenstein's lower triangle into odd halves.
Perhaps this is obvious, but I have to find a way to prove it or demonsrtate its truth or falsity clearly. The method would be to find any 8n+5 prime involved in a twinship, whose minus1 has more than two factors of two yet still divides Eisenstein's triangle into odd halves. For instance, 39+41 instead of 40+40, for eighty lattice points et al. This seems like an interesting question to pursue. We need to make a list of 8n+5 primes to see if any are both super even (more than two factors of 2) and involved in a twinship. It does not matter if our 8n+5 is the larger or the smaller of the twins.
5, 13, 29, 37, 53, 61.....
(101, 103)
It turns out that most of these primes are involved in a twinship. This is no way to proceed.
Wait. I have seen it. When 4 is added to any 8n number, it reduces its evenness by one factor of 2, the case with all 8n+5-1 numbers. Thank God for Gauss. This is exactly the condition we need for the conjecture to be true, and we see that the conjecture is indeed true.
A diagonal through WAXY in Eisensteins' rectangle in Wikipedia will always cut WAXY into two equal but odd numbers of lattice points whenever the diagram is for twin primes either of which is an 8n+5 number. The exact principles used here apply anytime one looks at any pair of primes of opposite type. This concludes the investigation.
* * * * *
It is now clear that 8n+5 numbers will always have two and only two factors of 2. This becomes clear when we tie these numbers to the ruler sequence, which expresses the degree of evenness of each consecutive even number, in other words, the number of factors of 2 it contains.
1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 4...
Now I understand what Gauss was doing when he proceeded on from 4n numbers to investigate 8n numbers, and why.
At this time I am prepared to guarantee how a diagonal will apportion the lattice points of WAXY when the diagram is for twin primes, according to the types of primes involved. I am not quite prepared to guarantee how the diagonal will apportion lattice points for any two primes at all, based simply on their ruler function positions. But Halelujah anyway! I have learned much this time.
What I need to know now is this: When he have two super even (more than two factors of 2 apiece) 4n+1-1 numbers, I believe they will not always apportion the lattice points into two even sets just because they are super even. Only Eisenstein diagrams for rectangles with very low eccentricity (like twin primes) can guarantee what the cut will be. Eccentric rectangles may not always produce the result of two equal sets of lattice points. Which has me wondering if hugely eccentric rectangles can ever produce equal sets. Somehow, I suppose they can. But I really have no clue whether they can ever produce two unequal sets with an odd number of elements.
It would sure be nice if the ruler function ruled the whole law. For all I know, it does. I certainly hope it does. What that would mean is this: We could determine the cut, and thereby the Legendre symbols for any two primes at a glance. Let us pray the ruler function rules, which I doubt.
A look back into my own papers reveals the answer quickly. Both (5, 13) and (5, 17) arrive at their positive 1 through a multiplication of (-1)(-1), yet their minus ones all are highly even and 16 is super even, suggesting that past a certain level eccentricity plays a greater part than evennes in deciding the cut. This is now clear. As p/q varies from 1/1, a square, the distance of p/q from 1 is its eccentricity. I believe this ratio of p/q is the key factor in bringing the rest of the law under the reign of understanding. I can imagine a graph, two graphs intersecting, one for evenness, one for eccentricity. They have to cross somewhere. That is where one takes over dominance from the other.
To put it in an even smaller nutshell: The minus ones of the four types of 8n+z primes all have their own evenness which is perfectly predictable. Only the minus ones of 8n+1 type primes are ever higher than two factors of 2. Just as the ruler function shows, all the action that changes is in the 8n slots, making 8n+1 primes the only primes whose minus one can be super loaded with factors of 2. How super evenness and eccentricity of the Eisenstein rectangle interact, is now the question.
I will now make a conjecture which I myself may be able to quickly dash.
Any two 4n+1 primes far enough out the number line, one of which is an 8n+1 prime, will make the Eisenstein cut into two identical and even sets. Even separated by four instead of two, if they are far enough out the number line the eccentricity of their rectangle will approach zero, and should force the apportioning of the diagonal into two even and equal sets. It only takes one counter example to dash the conjecture to smithereens.
Must two highly even primes far out the number line really always be quadratic residues of one another merely because they are relatively close together? Hmmm. This sounds very suspicious. But we shall see.