Cosmology

[desiresjab]

Let me sum up, try to anticipate any questions, and move these results back into a discussion of general cosmology.

The whole giant digression involving QR took place because I wanted to take the proposition that God could not make a universe where 2 is not the successor of 1 to a higher level, on a road to what might even include all of mathematics, but at minimum enlarging the statement to God could make no universe where the statements of mathematics would be false, from at least the founding axiom through to somewhere beyond the law of quadratic reciprocity. Anything theoretically true here would be theoretically true in any other universe as well, and vice versa.

No such universe is imaginable, might be a less religiously provocative way of stating it. Now that I can state it, there is only to wrap up the discussion of QR and return to cosmology, where the above will be one of the postulates of my personal philsophy within cosmology--mathematical cosmology, I suppose.

A gem that comes out of this is that we have the capability to understand any universe, and any universe of any description, no matter how different from our own, would have the capability of understanding our universe. That capability, consisting of mathematics and its growing extensions, would remain invariant across universes, while retaining its elastic variability and variety.

* * * * *

Eisenstein's P by Q rectangle must be viewed as a scaling object, just as if we had two gears intermeshing with radii corresponding to the lengths of the two 4n+3 primes, it expresses their ratio. I think his rectangle must be the simplest scaling object for this problem.

As I suspected a while back, the problem is a paucity of 2's when both primes are 4n+3. Once the rectangle is divided into four quadrants, there is no 2 left over, in other words, each quadrant contains an odd number of lattice points. That paucity of 2's forces the diagonal to cut the smaller rectangle of WAYX into two unequal halves of different parity.

The dimensions of the interior rectangle within ABCD containing only interior lattice points is

(p-1)(q-1).

This is none other than Euler's phi function Φ(p), the measure of how many numbers less than p are prime to it. There are sixty lattice points in the interior of ABCD. Points on the perimeter are not prime to one or other prime, so cannot be included. As usual, (mod 0) not allowed.

Once odd primes have the extra freedom of at least one more factor of 2, the problem is resolved, and the two primes are forced to act together, forced to the same parity because their smaller rectangle contains an even number of lattice points.

At the beginning, Eisenstein calculates the number of even lattice points in ABCD. He knows if they are aysmmetrical in the two halves of ABCD, so will the odd lattice points be, to make up for the discrepancy. The number of even lattice points in ABCD is of course thirty. His initial additive method was good enough for the parity of p, (-1¹⁷), but not good enough to obtain q's parity.

(P-1/2)(q-1)/2) is indeed the total number of points in WAYX, 1/4 of the total points in ABCD, and 1/2 the number of even points.

-1⁽⁵⁾⁽³⁾ is -1^(60/4) or -1^Φ(p)/4 after all. So Eisenstein's exponents are correct not only in their parity but they are also the "appropriate" exponents in that they are a factorization of the number of lattice points in WAYX. More importantly, (p-1/2) and (q-1/2) are the number of quadratic residues for each prime, we already know.

Now, that is everything about Eisentein's rectangle. Does it really prove quadratic reciprocity?

Yes, here is why I think so. Since (p-1/2) and (q-1/2) are a simple count of the number of quadratic residues for each prime, and their product is used as an exponent to count -1 back and forth from negative to positive, both are represented, and their product forms the dimensions for the rectangle of inner points of WAYX, and the exponentiation's result can only be negative when both (p-1/2) and (q-1/2) are odd.

For me that quite settles the issue, not only for 4n+3 prime pairs but for any prime pair, excluding 2, which I can recite the forumla for but have not yet reasoned out. My focus has been the 4n+3 primes, knowledge of which I hoped would illuminate the triggering mechanism for other prime pairs as well, which has happened for me.

4n+3 primes have to be the same thing in any universe. The only fundamental difference between 4n+1 and 4n+3 numbers is the degree of evenness when you subtract 1 from them.

* * * * *

Now that we all know this fundamental fact of numbers and how it constrains universes, we can proceed with broader cosmology again.

[YesNo]

A gem that comes out of this is that we have the capability to understand any universe, and any universe of any description, no matter how different from our own, would have the capability of understanding our universe. That capability, consisting of mathematics and its growing extensions, would remain invariant across universes, while retaining its elastic variability and variety.

I agree that given the initial axioms, the mathematical results are invariant across all possible universes. That there exist other universes can be assumed based on knowing that our universe is not eternal. In particular, the big bang shows it had a beginning.

Edit: Regarding the "barely even" number, those having only one factor of 2, there is a concept called "singly even" or "oddly even" that matches that: https://en.wikipedia.org/wiki/Singly_and_doubly_even

[desiresjab]

I agree that given the initial axioms, the mathematical results are invariant across all possible universes. That there exist other universes can be assumed based on knowing that our universe is not eternal. In particular, the big bang shows it had a beginning.

Edit: Regarding the "barely even" number, those having only one factor of 2, there is a concept called "singly even" or "oddly even" that matches that: https://en.wikipedia.org/wiki/Singly_and_doubly_even

I like highly even and barely even better. I modeled it after Ramanujan's idea of highly composite numbers, which assigns an index to each number to rank how composite it is. With the ruler function I can calculate and assign any even number its degree of evenness, without calculating those that precede it.

A few last details to focus up.

For the 7x11 rectangle, Eisenstein's exponents of 5 and 3 give a correct factorization of the lattice points in WAXY, but for the 7x3 rectangle where there are three points in WAXY, the exponent is 3, which does not factor the lattice points of WAXY, but sums them. Both the additive and the multiplicative coincidences may have been just that. I am not worried about that, it is minor and will sort itself out.

I saw the mechanism behind the behavior of 4n+3 primes from ground level. Mission accomplished. But not quite. For of slightly more consternation is that I understand the proof but not why it proves QR. True, WAXY is a quadrant, but the last bit of "seeing" has not clicked into place as to how this proves whether or not p or q are in each other's quadratic residue sets.

Can I not simply say Φ/4, where Φ is Euler's totient function, divided by four will always give the correct number of points in WAXY, and be done with it? I believe I can. This has to work for both species.

I understand everything about this proof except why it proves what it proves.

[desiresjab]

In the meantime, I see clearly that Φ/4, where Φ is Euler's totient function, will always give the correct number of lattice points in WAXY. This may work for 4n as well as 4n+2 numbers. I find this connection with the totient function highly intriguing.

[Φ(pq)]/4, in other symbols. Φ(pq) is what the RSA encryption system is based on. It is extremely hard to find Φ(pq) unless you know what p and q are. But you only know what pxq is, which means you have to find its factors, and that is almost impossible for huge, "barely composite" numbers with today's computing technology and math techniques.

[desiresjab]

Here is a very interesting near hit for the rectangle 11x7.

Φ(77)=60, Φ(60)=16, Φ(16)=8.

If Φ(60) had been 15 instead of 16...? Well, that is how research begins, I guess. But what if there is a pattern in the descending chain of Φ's anyway that only more investigation will ferret out? It should be easy to look at other examples, but my brain is shutting down right now. More later.

[desiresjab]

Why, looky here.

Φ(7x3)=12→ Φ(12)=4→ Φ(4)=2.

The reduction by a chain of Φ's instead of by factors of 2, worked out exactly as before, probably close enough to warrant even further investigation.

The second Φ again equaled one more than the total number of lattices in WAXY; the third Φ equaled the greatest number of lattices in either of the triangles. Uh-oh, now we have to go on.

Φ(19x23)=396

Φ(396)=120

Φ(120)=32.

Here everything goes quite amiss and we see we are on a deadend using the chain of Φ's. It was merely another coincidence. But Φ/4 is not a coincidence. That baby is real and will get you the right parity and the correct number of points in WAXY, which in this case is 99.

Φ/4 rules, Φ of Φ of Φ does not. Now we know for sure. I do not think 120 is one more than 99, and I do not think 32=49. That case closed.

[desiresjab]

It seems wonderful, it is wonderful. Why do Φ and these squares interact at all?

What do Φ(pq)=(p-1)(q-1)=a, or for that matter pq=z itself, and rt=d, the distance formula, have to do with each other? They are in some kind of equivalence class with V=IR, the voltage formula, and the equations for uncountable (literally) other phenomena. In abstract algebra they are called isomorphisms. That means they are just re-lablings of each other. They have the same group characteristics. A popular way of saying it is: they are the same under the hood. Their matrices look identical except for the difference in labels. Isomorphism. Of all those creepy isms in abstract algebra they are the easiest to see clearly. Just when you think you see automorphism or homomorphism clearly, they add some more bugaboo to it or manage to make them unclear by other means.

(p-1)(q-1) just happens to belong to this equivalence class of functions, linear relationships known as directly proportional to. Something like that. One cannot help noticing the similarity of the equations for work done by two workers doing a job and resistance from two resistors in parallel in electronics, as another example. The phenomena of the world around us express certain classes of functions replicated endlessly with only different labels under the hood. A few basic classes dominate much of the action, it seems to me at this moment.

I guess I will continue to loiter around the QR lobby until I see why Eisenstein has proved it. I see everything else about his beautiful proof, it has illuminated the operative principle of both 4n+3 primes and 4n+1 primes in QR by rectangular illustration, it provides exact numbers, I might as well hang around to see the reason it does what it purports to do.

I said I would be happy if I could see the general mechanism, and I have, but I guess I lied, for I am not satisfied now until I can see what makes Eisenstein's rectangle a proof of QR. All of it is right on the page and obvious the way math always is, a grand tautology, so eventually it will pop out clearly, the way the mechanism did, after my staring at it ignorantly forever. At the next moment of revelation I am bound to see more clearly something I have already stated, if past is precedent.

[YesNo]

But Φ/4 is not a coincidence. That baby is real and will get you the right parity and the correct number of points in WAXY, which in this case is 99.

If p and q are distinct primes then Φ(pq) = Φ(p)Φ(q) = (p-1)(q-1). All we need is to divide by 4 to get the number of lattice points. So I agree Φ/4 will always equal the number of lattice points in Eisenstein's rectangle.

[YesNo]

said I would be happy if I could see the general mechanism, and I have, but I guess I lied, for I am not satisfied now until I can see what makes Eisenstein's rectangle a proof of QR. All of it is right on the page and obvious the way math always is, a grand tautology, so eventually it will pop out clearly, the way the mechanism did, after my staring at it ignorantly forever. At the next moment of revelation I am bound to see more clearly something I have already stated, if past is precedent.

It is good you are not satisfied. Otherwise you would stop looking.

One of the problems with both Gauss's and Eisenstein's proofs is that they do not directly show why (p-1)(q-1)/4 should be the exponent. The direct proofs are in the Gauss Lemma and the Eisenstein Lemma. Those exponents in those lemmas are different from (p-1)(q-1)/4. But they want the exponent to be (p-1)(q-1)/4 because that is computationally easier to work with and so they transform their results so that the parity is preserved, which is all I think they are interested in.

At least that is how I see it at the moment.

[desiresjab]

It is good you are not satisfied. Otherwise you would stop looking.

One of the problems with both Gauss's and Eisenstein's proofs is that they do not directly show why (p-1)(q-1)/4 should be the exponent. The direct proofs are in the Gauss Lemma and the Eisenstein Lemma. Those exponents in those lemmas are different from (p-1)(q-1)/4. But they want the exponent to be (p-1)(q-1)/4 because that is computationally easier to work with and so they transform their results so that the parity is preserved, which is all I think they are interested in.

At least that is how I see it at the moment.

I feel that 3 & 5 must be the natural exponents. The only way to get lower is to have 1 & 3, like the 7x3 rectangle. Also there is the connection of Φ to them.

I do not believe the sum of the two exponents has much bearing. The exponents using Eisenstein's algorithm are 1 & 3 for the 3x7 rectangle. Those exponents do the job, Φ/4 does part of the job even faster, but it does not distinguish between p and q, as the exponents do.

Something just became slightly more focused in my head that is still blurred. If (p-1) & (q-1) were the same number and we multiplied them together we would be squaring, in which case we would merely add their exponents, wouldn't we? This is why adding exponents cannot work here. But the symmetries between squaring and what we are doing is intriguing. One feels that box might contain some secrets.

Several things we know for sure: Two sure ways to the correct result are Eisenstein's algorithm and Φ/4. It just happens that the intermediary step Φ/2 is the number of quadratic residues of each prime, and the final step the dimensions of the rectangle inside WAXY the lattice points sit on.

I am beginning to see that QR is as centrally connected as its reputation says it is. These connections are only the tip of the iceberg, QR reaches into everything. No doubt there are proofs that exploit the connection to Φ, just as there are trigonometric proofs, proofs that exploit the Pythagorean theorem, proofs that exploit Fermat's little theorem, and all kinds of proofs from group theory and abstract algebra, including at least one vector proof.

It will fall into place, but not without more staring and effort. I am hoping for a few weeks or less.

Something about 5x3 is really nagging me.

[YesNo]

Something just became slightly more focused in my head that is still blurred. If (p-1) & (q-1) were the same number and we multiplied them together we would be squaring, in which case we would merely add their exponents, wouldn't we? This is why adding exponents cannot work here. But the symmetries between squaring and what we are doing is intriguing. One feels that box might contain some secrets.

From one direction, we really should be adding the exponents to get the result. That is where Gauss's and Eisenstein's lemmas start. They compute an exponent for each (p/q) and (q/p) with -1 as the base. Say these are u and v respectively. Then (p/q)(q/p) = (-1)^{u + v}.

From the other direction, the product gives the desired result. That is the result Legendre observed when he conjectured the result depended on whether p and q were congruent to 1 or 3 modulo 4. Note that (p-1)(q-1)/4 is just the statement of the desired parity using p and q mod 4.

To connect the two directions are transformations from the sum of those exponents to that product that preserves parity.

[desiresjab]

From one direction, we really should be adding the exponents to get the result. That is where Gauss's and Eisenstein's lemmas start. They compute an exponent for each (p/q) and (q/p) with -1 as the base. Say these are u and v respectively. Then (p/q)(q/p) = (-1)^{u + v}.

From the other direction, the product gives the desired result. That is the result Legendre observed when he conjectured the result depended on whether p and q were congruent to 1 or 3 modulo 4. Note that (p-1)(q-1)/4 is just the statement of the desired parity using p and q mod 4.

To connect the two directions are transformations from the sum of those exponents to that product that preserves parity.

I think we could be on the right track with these thoughts. Hell, maybe you have already seen it through and through. All I know is I haven't.

[YesNo]

I don't know much about quadratic reciprocity. I haven't figured it out. Besides, once we get past quadratic reciprocity, there's cubic, and then quartic and then on and on for all I know. So we have only skimmed the subject.

I have been trying to put together the Artin's Conjecture puzzle. I'm starting with 2 and trying to find ways to get an infinite number of primes for which 2 is a primitive root. I think it is easy to get an infinite number of composite numbers for which 2 is a primitive root: just take an odd prime p for which 2 is a primitve root, then it should be a primitive root for pⁿ if (2/p) = -1 (mod p²). Actually, I am not sure that is right, but it doesn't solve Artin's Conjecture. That requires an infinite number of primes, not composites.

To keep my motivation, I have started asking questions on math.stackexchange centered around trying to show that there are an infinite number of Germain primes. These are primes p such that 2p + 1 is also a prime. They are easier to work with. If there are infinitely many of them that should solve Artin's Conjecture for a = 2.

[desiresjab]

I don't know much about quadratic reciprocity. I haven't figured it out. Besides, once we get past quadratic reciprocity, there's cubic, and then quartic and then on and on for all I know. So we have only skimmed the subject.

I have been trying to put together the Artin's Conjecture puzzle. I'm starting with 2 and trying to find ways to get an infinite number of primes for which 2 is a primitive root. I think it is easy to get an infinite number of composite numbers for which 2 is a primitive root: just take an odd prime p for which 2 is a primitve root, then it should be a primitive root for pⁿ if (2/p) = -1 (mod p²). Actually, I am not sure that is right, but it doesn't solve Artin's Conjecture. That requires an infinite number of primes, not composites.

To keep my motivation, I have started asking questions on math.stackexchange centered around trying to show that there are an infinite number of Germain primes. These are primes p such that 2p + 1 is also a prime. They are easier to work with. If there are infinitely many of them that should solve Artin's Conjecture for a = 2.

I have put my ten thousand hours of music in; I have put my ten thousand hours of writing prose and poetry in. With these enterprises I did much more than that. But I must be at about five thousand hours of math because the moves are not instinctive the way they would be with professional mathematicians. It takes me a long time, I make a lot of mistakes and move in with false assumptions quite often that later embarrass me.

For we amateurs maybe that is the way of it. Most unsloved problems scare me off because I know my inabilty to resist temptations. I only move in with a problem when it advances the cause of advancing my learning, or of it has a particular form that intrigues me. Sometimes they have been fun problems that I found here or there; I got into criptorithms for a while. Most of the time now I move in where I think I can learn as much as possible.


I have lived with Fermat's little theorem, Euler's phi function and quadratic reciprocity in my time. I am going to court Euler's divisor functions coming soon, because the few kisses I stole were not enough. That girl has a lot to say. Very centrally connected. After that I have to move in with complex numbers for a long time. Meta mathematics has been going on there since Gauss formalized the language.

I need two minds or more. The other side has now pulled me back to my fictional trilogy. That enterprise is where quadratic reciprocity is--awaiting the final clarity of events that make everything fit. The series is done except for some middle chapters left out in the inspirational heat of moving forward.

[YesNo]

On the one hand I am not trying to solve Artin's Conjecture. I'd be happy with the subjective process of understanding it before moving onto something else. On the other hand I feel like a teenager looking at the stack of books in the library and planning on how to read all of them. What probably counts is the subjectivity involved in understanding something.

Have you ever posted your question on a mathematics forum about the number of lattice points in Eisenstein's triangles? That sounds like an interesting puzzle. I haven't heard it mentioned except here, but then I don't know much about lattice points. People only started counting lattice points a few decades ago based on some cursory research I did into your problem.

Regarding those lattice points, if p = q then the two triangles should have the same number of lattice points since the slope of the diagonal would be 1 and it should go through all the lattice points on the diagonal. Letting q = p + 2 keeps that slope close to 1 and the diagonal misses all the lattice points. That's how I'm looking at the problem at the moment.