Cosmology

Quote:

---

Originally Posted by desiresjab

It is interesting. It is close to what I was doing last night on my own. Another interesting fact that could easily be overlooked is that when you square numbers mod (m) and every entry is duplicated once, it is an odd number and an even number which produce the same result in every case. What that means is factorization is not unique in this language. 5² and 6² both equal 3 (mod 11).

---

It looks like 3 is a quadratic residue because both 5 and 6 can be squared to give 3 mod 11. But then we know half of the elements will be quadratic residues and the other half non-residues. So there should be two elements when they are squared that give 3.

Another way of looking at this is to ask what solutions are there to the following polynomial: x² - 3 = 0 (mod 11) There should be 2 solutions and there are.

When we are looking at the residue classes mod 11, we aren't looking at integers any more. Instead we are working with equivalence classes of integers, or sets of integers. The integers would be in the integral domain Z, where there are primes because multiplicative inverses for all elements do not exist, but these residue classes are in the finite field Z/Z₁₁ where there aren't any primes anymore. In the finite field, the non-zero elements all have multiplicative inverses and so they would be units like 1 and -1 are in the integers Z. For example let n, not equal to 0, be an element in Z/Z₁₁, then since n^11-1 = 1 (mod 11), the multiplicative inverse of n is n⁹ (mod 11).

Quote:

---

Originally Posted by desiresjab

I was over at a math site recently and asked what they know over there. No responses yet, but I am sure they will not do as well as we are doing here. They love to talk a big game over there about advanced calculus and other impressive topics, but ol' QR is quite enough to stop all their chatter, espesially when I told them I was not interested in restatements of the law or facts surrounding it. Those surrounding facts I am interested in, but they don't need to know that, since we can get that right here and discover those facts for ourselves. I don't need them blabbimg restatements forever because they don't know what else to do. Maybe someone over there will come through yet. Don't hold your breath.

---

I got an account on https://math.stackexchange.com/ to get more information as well. It is good to have questions. The available answers aren't all the answers. Although QR is useful, what we really want is a quick way to evaluate (p/q) without having to consider (q/p). Quadratic reciprocity allows us to evaluate the one that is easiest to calculate, but perhaps there is a faster method. That sounds to me like what you are looking for.

Quote:

---

Originally Posted by desiresjab

When it comes to predicting whether an overlapping square will be even or odd, forget about it, at least so far with what we know.

---

That is an interesting question. One doesn't have to take the representatives for the equivalence classes from {0,1,...,p-1}. They could come from {-(p-1)/2,...,(p-1)/2}. The evenness and oddness of the result might change when using that set.

Quote:

---

Originally Posted by YesNo

It looks like 3 is a quadratic residue because both 5 and 6 can be squared to give 3 mod 11. But then we know half of the elements will be quadratic residues and the other half non-residues. So there should be two elements when they are squared that give 3.

Another way of looking at this is to ask what solutions are there to the following polynomial: x² - 3 = 0 (mod 11) There should be 2 solutions and there are.

When we are looking at the residue classes mod 11, we aren't looking at integers any more. Instead we are working with equivalence classes of integers, or sets of integers. The integers would be in the integral domain Z, where there are primes because multiplicative inverses for all elements do not exist, but these residue classes are in the finite field Z/Z₁₁ where there aren't any primes anymore. In the finite field, the non-zero elements all have multiplicative inverses and so they would be units like 1 and -1 are in the integers Z. For example let n, not equal to 0, be an element in Z/Z₁₁, then since n^11-1 = 1 (mod 11), the multiplicative inverse of n is n⁹ (mod 11).




I got an account on https://math.stackexchange.com/ to get more information as well. It is good to have questions. The available answers aren't all the answers. Although QR is useful, what we really want is a quick way to evaluate (p/q) without having to consider (q/p). Quadratic reciprocity allows us to evaluate the one that is easiest to calculate, but perhaps there is a faster method. That sounds to me like what you are looking for.



That is an interesting question. One doesn't have to take the representatives for the equivalence classes from {0,1,...,p-1}. They could come from {-(p-1)/2,...,(p-1)/2}. The evenness and oddness of the result might change when using that set.

---

Huh? I am not looking for a faster way to do anything. I have said so many times what I am looking for that I do not feel like saying it again right now.

Something I said before is coming true. Hardly anyone understands quadratic reciprocity. No one is answering on the math forum I visited. They love to showoff. If anyone knew, they would surely answer. I feel partially vindicated in that the problem truly is difficult. Learning enough about it to pass a course in elementary number theory is not that hard, but knowing what I am asking--now that is hard.

To address something you said: In fact, the parity would change when we go to negative representatives of the residue system. Good observation.

What question did you ask them?

Quote:

---

Originally Posted by YesNo

What question did you ask them?

---

I said I was only intersted in the mechanism itself that made 4n+3 primes behave as they do toward each other in QR, not restatements of the law or facts surrounding it, and that I insisted on seeing this in terms of the numbers themselves rather than a higher abstaction coming out of group symmetries and the like. I admitted this might be like standing outside a forest with a flashlight looking for something hidden behind a tree, but still I asked for an explanation in terms of the numbers.

I suspect that the basis of some higher abstraction proofs in abstract algebra is to show an anti-symmetry between the two subrings of squares, or something along those lines. Those proofs look down from above, I want to look from below in the guts of the machine.

In the meantime I have developed what may be a valid conjecture, that the two triangles WAY and YAX in Eisenstein's rectangle will contain an identical number of lattice points when p and q are twin primes, and only then. This may seem obviious but be very hard to prove. It is an ad hoc conjecture, just a problem, like so many that Erdos proposed, perhaps not important, but a fact nonetheless and an interesting challenge to recreate upon.

A weakness of the conjecture is that I suspect that the outside members of prime triplets which are large enough might also do this. Okay, I leave out the if and only if part of the conjecture.

After merely checking all of my experiments on graphing paper, I am positive the conjecture could be extended to all prime triplets from at least the second such triplet to infinity with a stipulation. That stipulation is that two of the prime triplets are 4n+1 numbers, because when two 4n+3 primes clash we know they will obviously have different numbers of lattice points in the two triangles because one will be even and the other odd. As long as our prime triplets contain two 4n+1 numbers, we will be okay, and I am pretty sure that conjecture would hold.

Quote:

---

Originally Posted by desiresjab

In the meantime I have developed what may be a valid conjecture, that the two triangles WAY and YAX in Eisenstein's rectangle will contain an identical number of lattice points when p and q are twin primes, and only then. This may seem obviious but be very hard to prove. It is an ad hoc conjecture, just a problem, like so many that Erdos proposed, perhaps not important, but a fact nonetheless and an interesting challenge to recreate upon.

A weakness of the conjecture is that I suspect that the outside members of prime triplets which are large enough might also do this. Okay, I leave out the if and only if part of the conjecture.

---

That sounds like an interesting problem. What do you mean by prime triplets? Three consecutive primes?

Quote:

---

Originally Posted by YesNo

That sounds like an interesting problem. What do you mean by prime triplets? Three consecutive primes?

---

There are two different brands for p, q and v, {n, n+2, n+6} and {n, n+4, n+6}.

The conjecture is that as long as at least two in either set are of the form 4n+1, they will obviously express not only the same parity with each other in all three combinatory pairs, but also have the same number of lattice points in WAY and YAX, as seen from Eisenstein's rectangle represented in Wiki-pejia.

It seems intuitively clear, but I don't know how to prove it. I am not saying that is the maximum boundary condition, either. That is, there may be wider gaps than six which allow for an identical number of lattice points in both triangles. It depends on only two things--the absolute cardinality of p,q and v, and whether only one of them is a 4n+3 number.

The larger the absolute magitude of the triplet, the closer any two of them compared to each other will be to the ratio of a square 1/1. Since this holds for even small triplets where the ratio is not as close to 1, it must be true for ones of larger absolute magnitude that meet the only other condition.

What I suspect could be proven with analytical methods is that for any gap, as wide as one wants to make it, WAY and YAX can still produce the same number of lattice points, as long as the primes involved are large enough and both are not type 4n+3. Similar things have been proven about primes and other intervals. This one feels intuitively right. It might well have been already proven, or at least conjectured.

After sleeping on your problem related to twin primes, I realized I couldn't solve it.

It would be easy to show that the parity in the two triangles of lattice points are the same, but not that there are exactly as many lattice points in both triangles. In the triplets, having two 4n + 3 primes would make the number of lattice points in the rectangle odd and so the parity in the triangles would be different.

One way of solving it might be to go through Eisenstein's construction using a prime p = 4m + 1 and then seeing if the lattice points remain the same using 4m + 3. I realize I don't understand Eisenstein's proof well enough to do this easily.

Also one might be able to generalize this to any two numbers whose difference is 2. Some of the points might be on the diagonal line separating the two triangles, but then they would either not be counted or counted in both triangles.

If you haven't published this problem, it might be worth doing so say on places like math.stackexchange.com. An interesting question is more valuable than a quick solution.

I was thinking more about the number of lattice points in the two triangles. Intuitively, it would seem that there should be the same number of lattice points in both triangles because the diagonal line divides the rectangle into two equal area triangles. With two 4m + 3 primes the total number of lattice points in the rectangle is odd, so one of the triangles should have an odd number of lattice points and the other an even number. There has to be a difference of at least 1 for those primes.

Do you have an example (two primes p and q) where the difference in the number of lattice points is greater than 1?

Maybe the more basic question is to find pairs of primes that divide the lattice points in the two triangles so that the difference in the number of lattice points in each triangle gets larger.

Quote:

---

Originally Posted by YesNo

I was thinking more about the number of lattice points in the two triangles. Intuitively, it would seem that there should be the same number of lattice points in both triangles because the diagonal line divides the rectangle into two equal area triangles. With two 4m + 3 primes the total number of lattice points in the rectangle is odd, so one of the triangles should have an odd number of lattice points and the other an even number. There has to be a difference of at least 1 for those primes.

Do you have an example (two primes p and q) where the difference in the number of lattice points is greater than 1?

Maybe the more basic question is to find pairs of primes that divide the lattice points in the two triangles so that the difference in the number of lattice points in each triangle gets larger.

---

Examples are easy to come by for 5 and 11, the triangles have 6 and 4 lattice points. The difference always has to be at least 2 for examples of the same parity.

For primes 5 and 15, they are 9 and 7.

It has not failed for twin primes or correctly composed prime triplets, and I have not found it to be true anywhere else.

* * * * *

What I have done with regards to QR is moved on to abstract algebra. Fortunately for myself, I can relate a great deal of what they are saying to what I already know of groups, rings and fields from number theory, otherwise I would be lost. In lecture 27 of 38 they finally got real close to QR.

Quote:

---

Originally Posted by desiresjab

Examples are easy to come by for 5 and 11, the triangles have 6 and 4 lattice points. The difference always has to be at least 2 for examples of the same parity.

---

That's good to know.

Quote:

---

Originally Posted by desiresjab

For primes 5 and 15, they are 9 and 7.

---

Although 15 is not prime, I would think this should work for numbers in general. Some of the lattice points could be on the diagonal line if one doesn't use primes.

Quote:

---

Originally Posted by desiresjab

It has not failed for twin primes or correctly composed prime triplets, and I have not found it to be true anywhere else.

---

I wonder if it is possible to pair the columns of lattice points. For example, the column where x = 1 would pair with the column where x = (p-1)/2. I suspect the number of lattice points from just these two columns would be the same in both triangles. Then proceed by induction, or some other means, to look at the other column pairs.

Quote:

---

Originally Posted by desiresjab

What I have done with regards to QR is moved on to abstract algebra. Fortunately for myself, I can relate a great deal of what they are saying to what I already know of groups, rings and fields from number theory, otherwise I would be lost. In lecture 27 of 38 they finally got real close to QR.

---

Which text are you using? Youtube may also have interesting reviews of algebra.

Quote:

---

Originally Posted by YesNo

That's good to know.



Although 15 is not prime, I would think this should work for numbers in general. Some of the lattice points could be on the diagonal line if one doesn't use primes.



I wonder if it is possible to pair the columns of lattice points. For example, the column where x = 1 would pair with the column where x = (p-1)/2. I suspect the number of lattice points from just these two columns would be the same in both triangles. Then proceed by induction, or some other means, to look at the other column pairs.



Which text are you using? Youtube may also have interesting reviews of algebra.

---

I meant 5 and 17. I believe there was one pair of primes that gave 9 and 6, but I cannot remember what the[pair was.

Presently, I do not have an abstract algebra text. I am doing everything off the internet.
In case you want a link to the thirty-eight lectures, I will provide it below. The instructor's name is Gross from Harvard, and this guy is exceptional.

Abstract algebra is a whole new language. They operate at a very high level of abstraction. If you miss somethibng, you have to go back. Everything is dependent on what is already supposed to have been learned. Very abstract thinks like manipulating and untangling compositions of functions. You have to know a kernel from an image, you constantly have to check to make sure what you are working on is actually associative or commutative, et al.

He may get to QR the next lecture I have up. What I can tell you is the solution wull be buried under even more layers of abstraction than I imagined.

I try to go too fast--six or seven lectures per day or more. That guarantees I will have to go back and do it agin. But by going ahead to the end, I know exactly what I should be concentrating on the second time around. The method is not as faulty as it seems.

https://www.youtube.com/watch?v=TsLW...=10#t=6.688125

I think I found it: https://www.youtube.com/watch?v=EPQg...8AC5CABC1321A3

Quote:

---

Originally Posted by YesNo

I think I found it: https://www.youtube.com/watch?v=EPQg...8AC5CABC1321A3

---

That is the right guy. Benedict Gross. I think he is chairman of the department of mathematics at Harvard. He never does get to QR, though there is a fair amount of material about squares, since that was a big subject of Gauss. The best minds today are still working on the stuff developed out of Gauss, who set the course for everything in the field and has never had a conjecture overturned anywhere in math or science but many verified.

Instead of numbers, these people study equivalence classes of numbers and of polynomials, both real and complex, through structures like groups, rings and fields. Every structure is carefully defined, and one has to know them apart and be aware of the criteria for being in one or the other during any process.

* * * * *

If a ground level view of the mechanism for 4n+3 primes in QR is possible, maybe someday I will find it. The tools and results of abstract algebra are often macroscopic but probably capable of focusing down to the particular mechanism responsible, too.

No sooner do I give up than I see what seems to be enough. It is like I said before, if p is a 4n+3 prime, p-1 will be divisible by 2 only once. The mechanism is best described for illustration in my own terminology as the difference between "barely even" and "highly even" even numbers. On one of those classes is where you end up once you subtract 1 from p.

I had to look at it long and painfully to finally confirm that what was staring me in the face was the actual mechanism itself. That is why I am dumb.

All that remains is to verify and confirm that one understands how we got to the point of (p-1)/2 in the first place. That is incredibly easy, already done. I am finished.