Cosmology

[desiresjab]

There is a lot to understand, but I try to think of these as pieces of a jig-saw puzzle. Here are the pieces so far in my quest to solve Artin's Conjecture, at least the part that says for any number greater than 1 there are infinitely many primes for which it is a primitive root.

Puzzle Piece 1: Two integers that are relatively prime have inverses with respect to each other. In particular (a,n) = 1 if and only if there exists x such that ax=1 mod n. This means we only have to look at relatively prime integers and φ(n) would represent how many there are. If p is a prime, then φ(p) = p - 1. For simplicity stick with primes and the numbers relatively prime to them.

Puzzle Piece 2: A primitive root a multiplied by itself has to generate all the residues mod n. In particular it can't stop generating a number different from 1 until it generated all of them. Further for any d > 1 dividing p - 1, a^(p-1)/d cannot equal 1 mod n. Otherwise it has stopped generating the residues and it is not a primitive root. So for d = 2, if a^(p-1)/2 = 1 mod n and therefore was a quadratic residue it would not be a primitive root.

Puzzle Piece 3: If Artin's conjecture is true, then for each a > 1 there exist infinitely many primes for which a is a quadratic nonresidue. The converse is false. However, maybe this is easier to solve if it hasn't already been solved.

Puzzle Piece 4: To simplify matters, let a = 3 which is one of the 4m+3 numbers. If p is another 4m+3 prime then QR can relate them so that calculations work faster, but I can't rely on calculations since I am working with infinitely many primes p. So far QR seems good for calculation, but nothing else.

Puzzle Piece 5 (the one I'm on now): Suppose p is a 4m+3 prime and p - 1 = 2r where r is another prime. Let a = 3. If (3|p) = -1, then I have handled the case when the divisor is 2: 3^(p-1)/2 = -1. Does this imply anything about the other divisor (p-1)/r? So, are there infinitely many primes where p-1 = 2r and r is prime and what additional conditions do I need to tell if 3 is a quadratic nonresidue?

At the moment I see QR's value in helping one calculate whether a number is a quadratic residue or not faster. I must be missing something important.

Philosophically, my own inclination is toward mathematics as we know it being necessary just as it is. God could not controvert or skirt this necessity, meaning God has limitations. A limted God was an idea of John Stuart Mill.

Very, very true, I could have set the bar anywhere, I could have chosen easier propositions. But I just happened to settle on QR because I knew it was hard, I did not understand it at the time and figured I should earn the right to make such a statement as God is constrained by mathematics.

I am so close now. I again sense Eisenstein's proof as the way forward. If one cannot see it in the numbers themselves, see it in the exponents represented by those dots and X's, then backwards extrapolate to the numbers.

[desiresjab]

I think a key point to realize about Eisenstein's proof, is that in his triangles AYX and WAY, the number of lattice points in them do not reperesent actual degree of the exponent on p or q, but have the right parity, which is all that matters--odd or even. For instance, the number of even lattice points in his big traingle, with 17 even lattice points, seems to represent the true exponent on -1 for p, whereas the triangles AYX and WAY merely give the right parity for p or q, which is suffiucent, indeed, but not quite the same thing.

In the event of two 4n+3 primes, the two small triangles will have opposite parity, which forces -1 as the final outcome of the operation.

I am conjecturing that the triangles AYX and WAY will never contain the same quantity of total lattice points, but their parity will be in accord when both are not 4n+3 types.

This will at first seem strange, as the two large triangles ABC and ADC always have the same total number of lattice points, of course. This does not mean one contains the same number of even or odd points as the other, however, for indeed they do not. Only their total number is equal.

Let's forget I made that conjecture, since it is false. You could say it is usually true but not always. I think they can be equal when the two primes are near the same size and their QR value is 1, and when, of course, they are both 4n+1 types only..

[desiresjab]

This goes to show how dangerous the conjecture game in mathematics is. I amended my above post four or five times until I finally saw the truth. The quantity of lattice points in the triangles AYX and WAY can be equal when the two primes are close enough in size to each other, and at least one is a 4n+1 type. I don't think that fact even has much significance. Red Herring.

[desiresjab]

Even so, the two triangles will have different quantities of even and odd lattice points, though the total number of lattices in each is the same. I am done with that. Can those two triangles ever have the same number of even points and the same number of odd points? I don't know, y'all, and I ain't gonna think about it. However, I think they cannot. Watch out! Watch out!

[desiresjab]

Hold it, dummy (speaking to myself). The natural exponent for -1, ie., the one which duplicates (p-1)/2 (q-1)/2, is found by summing the total number of lattice points in the triangles AXY and WAY. Ah, now that is good. We have gotten somewhere.

In the case of 11 and 13 for p and q, the exponent would be 30, which is even and therefore produces 1 when used as an exponent.

-1^{(11-1)/2 (13-1)/2}=-1³⁰=1

[YesNo]

Here are some links that I plan to look at more closely on Eisenstein's proof of QR to see if I can understand this. Do you have some links?

https://en.wikipedia.org/wiki/Proofs...ic_reciprocity

http://math.ucsb.edu/~jcs/QuadraticReciprocity.pdf

[desiresjab]

Here are some links that I plan to look at more closely on Eisenstein's proof of QR to see if I can understand this. Do you have some links?

https://en.wikipedia.org/wiki/Proofs...ic_reciprocity

http://math.ucsb.edu/~jcs/QuadraticReciprocity.pdf

I have used so many cites I could not begin to dig them up.

Make a p X q rectangle on graphing paper. Draw a diagonal carefully. 19 by 23 was the largest rectangle my paper allowed me to draw. There are enough primes below 23 to get the picture.

The exact number of lattice points corresponding to (p-1)/2 (q-1)/2 will be found in the triangles AYX and WAY within the rectangle (p/2)(q/2). Watch what happens in those two triangles as you construct rectangles for different primes and prime types. The total number of points in either triangle is the same, they do not have the same quantity of odds or evens.

The borders of (p/2)(q/2) are between lines on the graphing paper, ensuring that we have no lattice points on the perimeter. The perimeter of the invisibleble rectangle with lattice points on the perimeter of it would have dimensions (p-1)/2 (q-1)/2. This is the geometric connection between (p-1)/2 (q-1)/2 and (p/2)(q/2). I am not sure how clear that is. I am trying to bring you up to my current understanding on the problem.

I see it now. I see how those small triangles work and why they do. I am now about 98% satisfied with my undertstanding of Eisenstein's proof. It is no longer a mystery why the lattice points and the exponents match up.

God could only create a universe where QR is true, if our imaginations are asked to judge. No QR in our universe is every bit as absurd a notion to our brains as 2 is not the successor of 1. Over and out.

[YesNo]

I see the diagram. Also it makes sense that there is no lattice point on the diagonal line, y = (p/q)x, since p and q are primes. That is, an integer value for x would not make y an integer. I also see how there are (p-1)/2 (q-1)/2 lattice points in the two triangles, AYX and WAY. What I don't see is the connection between those lattice points and something that will discriminate between 1 or -1. All I can see is the overall count is correct. This seems to me like I am missing something.

I remember reading that Galileo pointed his telescope to Jupiter and asked one of his friends to look. Even though his friend was willing to agree with him, he didn't understand that what he was looking at were Jupiter's moons rather than more stars and so the evidence didn't convince him. I am sort of like that with this proof at the moment. Proofs are like spaghetti code until one understands them. Unraveling the spaghetti takes time. After understanding, one can try making a new proof that might be easier to understand. I hear there are hundreds of proofs for QR.

The Gauss Lemma makes more geometric sense to me at the moment than this one does. Start with Fermat's Theorem, a^p-1 = 1 mod p. The quadratic residues would have a^(p-1)/2 = 1 mod p as well. And so one already has one way to calculate whether a is a quadratic residue or not by repeated multiplication of a. The lemma replaces a with -1, which simplifies the calculation as far as the multiplication goes, but also complicates it since the exponent is no longer (p-1)/2, but the number of negative elements when reduced mod p to numbers between -(p-1)/2 and (p-1)/2. Finding that exponent is now the hard part.

Geometrically that makes sense to me and, by the way, it also helps solve the puzzle piece I was working on. It seems there are infinitely many primes p for which 3 is a quadratic nonresidue and hence a potential primitive root. I would now need to generalize that.

[desiresjab]

Sometimes it is difficult for me to determine if you are talking about QR or PR, or searching for a connection between the two. I have to go out on a limb and say provisionally I do not think the two are hugely connected. They are connected some way, however, because just about all number theoretic functions are connected, no matter how distantly. Their connection may even be important.

This could very well be a fault in my own vision. From looking at the problem so long my own way I may have developed myopia. My brain is open for business, though.

I know for a fact that Eisenstein went into his proof with much knowledge. He already knew the significance of (p-1)/2 and (q-1)/2, which is why he made the lattice points in AXY and WAY match up to them one-to-one.

At this point we have much knowledge, too. For instance, we do not have to consult anything to know that a 4n+1 prime will always have -1 as a quadratic residue and 4n+3 primes will never.

You seem to be aking: where is this information located in Eisenstein's rectangles? I have to say at this point I do not know if it even is. This is information we already know, and I am unaware of that information being graphically represented in Eisenstein's diagram at this point. I will search for its presence, however, for I am not fool enough to think there is nothing obvious I might be missing.

[desiresjab]

Gauss knew Eisenstein. He may have beeb Gauss's student. What I suspect is that Eisenstein found the most elementary proof possible. This what mathematicians always strive for. If one man's proof requires calculus and a second man's proof requires only algebra, the second proof is considered more elegant.

For Gauss to have been all around this proof only to have Eisenstein find and present it--did this rasp the old man? Gauss demoted Euler because he was so close to QR and did not get it. Yet he stood within inches (figuratively) of this be-all end-all of QR proofs.

Hundreds more proofs were to come, but we know none are as elegant as Eisenstein's. The fact that Wikipejia chose it is testament of this. Every other proof I have looked at is a devil, and requires higher concepts.

What was Gauss thinking when he made his famous comment about Eisenstein? Gauss made seven or eight proofs of QR in his lifetime. I would be willing to bet each illuminated a different aspect of it, or Gauss would not have bothered. The fact that he was still working on it throughout his lifetime probably means even he, the mightiest of mighty, felt he did not have full grasp of it. Why else would a man with so many other important things to get to still be fussing with QR decades after he solved it?

This means we sure as heck do not have to feel bad or guilty for only having partial understanding of this theory. Gauss had the telegraph to invent, conformal mapping to forrmalize, magnetism to overhaul, differential geometry to launch, yet he kept coming back to QR his entire life to produce more proofs. Ask youself, would he have done this if he had every bit of understanding he felt he needed on the topic?

He felt it was his crowning acheivement. This fellow who as a teenager cracked a seventeen hundred year old problem that had puzzled the ancients, who formalized modular arithmetic, who presented the first proof of the fundamental theorem of algebra, who built the algebraic structure for imaginary numbers--he considered QR the greatest (and perhaps the deepest) of his acheivements.

If we understood QR quickly and easily, something would be wrong. Powerhouses like Gauss and Euler and Legendre do not struggle mightily only for us to come along and breezily understand at will. Make no mistake about it, this stuff is hard. QR is a gateway to the really hard in number theory. It is always presented at the end of elementary number theory couses. After that you are no longer on the wading end of the pool--you swim or sink in those deep waters.

[Dreamwoven]

I wonder if this website would be helpful?
http://math.stackexchange.com/questi...olumn-pivoting

[YesNo]

I looked up "QR decomposition" and it seems to be concerned with factoring matrices. https://en.wikipedia.org/wiki/QR_decomposition It may be related, but I don't see how at the moment. Quadratic reciprocity, which we abbreviated here as QR, is about whether an integer x is a square modulo a prime p, that is, does there exist an integer r such that r² = x mod p? If so (x|p), the Legendre notation for whether x is a quadratic residue mod p, would equal 1.

What I am looking at is the Artin's conjecture which says given a number, m>1, there are infinitely many primes for which m is a primitive root. That is, multiply m by itself over and over again and all the elements of the reduced residue system mod that prime are generated. If m is a primitive root then it is also a quadratic non-residue, otherwise it would not generate all the elements, but stop half way through.

Desiresjab is interested in quadratic reciprocity and in particular Eisenstein's proof of it. I find that interesting also, because the more I learn about that the more I understand why Artin's conjecture is hard to solve.

Here is an outline of a proof of quadratic reciprocity using Eisenstein's lattice points: http://math.ucr.edu/home/baez/136/quadratic.pdf

The article doesn't prove anything. It just states the propositions, which is frustrating, but it only claimed to offer a "big picture". The proposition that gets me stuck is called in that paper "Baby Eisenstein's Lemma". It says that the number of points in the lattice in the lower triangle in Eisenstein's drawing has the same parity (even or odd) as the number of elements in Gauss Lemma that fall in the negative part of the reduced residue system from -(p-1)/2 to (p-1)/2). If we know how many there are then m is a quadratic residue if that number is even and a quadratic non-residue if that number is odd. So, it is not that they match one-to-one but that they have the same parity.

That's the clue I am following at the moment. It is only a parity issue between those lattice points and the elements in Gauss Lemma. However, I don't know how to prove that which means I don't understand it.

[desiresjab]

I looked up "QR decomposition" and it seems to be concerned with factoring matrices. https://en.wikipedia.org/wiki/QR_decomposition It may be related, but I don't see how at the moment. Quadratic reciprocity, which we abbreviated here as QR, is about whether an integer x is a square modulo a prime p, that is, does there exist an integer r such that r² = x mod p? If so (x|p), the Legendre notation for whether x is a quadratic residue mod p, would equal 1.

What I am looking at is the Artin's conjecture which says given a number, m>1, there are infinitely many primes for which m is a primitive root. That is, multiply m by itself over and over again and all the elements of the reduced residue system mod that prime are generated. If m is a primitive root then it is also a quadratic non-residue, otherwise it would not generate all the elements, but stop half way through.

Desiresjab is interested in quadratic reciprocity and in particular Eisenstein's proof of it. I find that interesting also, because the more I learn about that the more I understand why Artin's conjecture is hard to solve.

Here is an outline of a proof of quadratic reciprocity using Eisenstein's lattice points: http://math.ucr.edu/home/baez/136/quadratic.pdf

The article doesn't prove anything. It just states the propositions, which is frustrating, but it only claimed to offer a "big picture". The proposition that gets me stuck is called in that paper "Baby Eisenstein's Lemma". It says that the number of points in the lattice in the lower triangle in Eisenstein's drawing has the same parity (even or odd) as the number of elements in Gauss Lemma that fall in the negative part of the reduced residue system from -(p-1)/2 to (p-1)/2). If we know how many there are then m is a quadratic residue if that number is even and a quadratic non-residue if that number is odd. So, it is not that they match one-to-one but that they have the same parity.

That's the clue I am following at the moment. It is only a parity issue between those lattice points and the elements in Gauss Lemma. However, I don't know how to prove that which means I don't understand it.

The number of lattice points in Eisenstein's triangles AYX and WAY give the exact value of the exponents, not just the correct parity.

[desiresjab]

That is, they give the correct sum of total exponents. (12X10)/4=15+15

[desiresjab]

As far as I can tell quadratic reciprocity and QR as used in those matrices are different functions. I think QR means something else with regard to those matrices. I do not think it means quadratic reciprocity.