Cosmology

[desiresjab]

The elementary number theory text I got the most out of was H.L. Davenport's Higher Arithmetic. I always find it useful to keep a number of texts around in case one fails me. I found the section on primitive roots to be very lucid.

I do not think QR goes all the way back to Pythagoras, or anywhere near. I am almost certain the theory was unknown to the Greeks. I do not know of any mention of it in the ancient world, even by Diophantus. Unlike the compass and straight edge problem Gauss solved from the ancients, QR seems to have been discovered by Euler in the west and proven by Gauss.

[desiresjab]

Wayward Fact #1:

-1 is never a quadratic residue of primes of the form 4n+3, and always a quadratic residue of primes of the form 4n+1.

Wayward fact #2:

The behavior of 2 with respect to QR is different from the odd primes, as we might expect, but still fits into the theory consistently.

[YesNo]

Yes, I suspect QR is a relatively recent idea.

Regarding the first wayward fact, -1 is p - 1 mod p. One can tell if -1 is a quadratic residue for odd prime p by evaluating (-1|p) = (-1)^1/2(p-1). The exponent is even if p is of the form 4m + 1 and odd if p is of the form 4m + 3. It is the primitive root for only 2 and 3, so it is not considered in Artin's conjecture along with the perfect squares.

Here's a problem in Dickson's text (page 21): Show that the product of all primitive roots of a prime p > 3 is congruent to 1 mod p.

I can see that this makes sense, but I don't know how to prove it. For example, consider p = 5. The primitive roots are 2 and 3 and 2*3 = 1 mod p. One can write 3 = 2³ in order to combine it with 2 and then we have 2¹2³ = 2⁴ which should be 1 by Fermat's theorem. So it works for one case, but how would one show that in general? That's the one I'm stuck on at the moment.

[desiresjab]

Yes, I suspect QR is a relatively recent idea.

Regarding the first wayward fact, -1 is p - 1 mod p. One can tell if -1 is a quadratic residue for odd prime p by evaluating (-1|p) = (-1)^1/2(p-1). The exponent is even if p is of the form 4m + 1 and odd if p is of the form 4m + 3. It is the primitive root for only 2 and 3, so it is not considered in Artin's conjecture along with the perfect squares.

Here's a problem in Dickson's text (page 21): Show that the product of all primitive roots of a prime p > 3 is congruent to 1 mod p.

I can see that this makes sense, but I don't know how to prove it. For example, consider p = 5. The primitive roots are 2 and 3 and 2*3 = 1 mod p. One can write 3 = 2³ in order to combine it with 2 and then we have 2¹2³ = 2⁴ which should be 1 by Fermat's theorem. So it works for one case, but how would one show that in general? That's the one I'm stuck on at the moment.

That is an interesting problem. If I may make a suggestion. My intutition is that you need to pair the primitive roots up, since there is usually an even number of them, φ^(p-1) of them, actually. The trick is going to be something similar to what I did to prove Fermat's little Theorem. I believe you can pair the roots with their inverses for multiplication and get 1, since if a is a root, a^-1 is also.

Notice that in your example, 2 and 3 are inverses of each other (mod 5), or else their product would not equal 1. Hope that helps a little.

[desiresjab]

In the event that φ^(p-1) is not even, it should still work out. You simply multiply a times b, and that will be the inverse of the remaining c (mod p).

[desiresjab]

Sorry, duplicate post.

[desiresjab]

Now if you multiplied the entire residue system of a prime together, you should also get 1, right? For each element in the set has an inverse, which is also in the set, and there are an even number of elements in the complete reisdue system.

[desiresjab]

Come to think of it, the directly above task should actually be easier when φ^(p-1) is odd, since 1 is in every set, and must be paired with itself, leaving an even number of numbers to pair.

[desiresjab]

Sorry for spreading this out, but you have made me think. The evenness or oddness of the set probably does not matter. In those cases where it seems at first likely to interfere, I would almost bet that it will magically work out because one of the numbers in the set (besides 1) will be its own inverse. I do not have hardcore evidence or a proof, but my experience is telling me that. Anyway, I think I have said enough, if not too much.

[YesNo]

Thanks, desiresjab. It makes sense to pair the primitive roots with their inverses. It didn't occur to me that a primitive root's inverse is also a primitive root, but I think that should be the case.

I found another introductory textbook on the subject by Charles Vanden Eynden which I am also reading. When I get stuck with one, I move to the other.

[desiresjab]

Thanks, desiresjab. It makes sense to pair the primitive roots with their inverses. It didn't occur to me that a primitive root's inverse is also a primitive root, but I think that should be the case.

I found another introductory textbook on the subject by Charles Vanden Eynden which I am also reading. When I get stuck with one, I move to the other.

Pairing members of sets up is a common trick in the field. Do not be afraid of pen and paper. If you can get one move closer to the nut, then you only have to see two moves deep instead of three, etc.

I love the way modular arithmetic forces even the largest numbers to play its game. It puts numbers on the rack and extracts certain truths from them. It says to the gigantic prime: you are only 2 (mod 3), pal, now get up there.

[YesNo]

I just finished the following article giving a proof of Euler's theorem using pairing based on a number and its inverse. Being relatively prime to n is the same thing as having an inverse mod n. That is (a,n) = 1 <=> there exists x such that ax = 1 mod n. I'll have to keep that in mind. http://sites.millersville.edu/bikena...uler/euler.pdf

[desiresjab]

I just finished the following article giving a proof of Euler's theorem using pairing based on a number and its inverse. Being relatively prime to n is the same thing as having an inverse mod n. That is (a,n) = 1 <=> there exists x such that ax = 1 mod n. I'll have to keep that in mind. http://sites.millersville.edu/bikena...uler/euler.pdf

Yes indeed. There are so many of these little facts and theorems and criteria and lemmas and adjuncts that remembering them all when one of them happens to be relevant is a matter of experience as you see more and more of the mulitiplicty of connections between ideas at the level of numbers.

Only when I can see straight through QR to the intiuitive reason it must be so, could I possibly say God could not create a universe where QR is not as it is in our universe.

The one thing absolutely necessary to our universe or any universe is the same laws of mathematics everywhere. If I can see that about QR, I can say it, but I do not see it with that particular clarity yet, and furthermore do not know if I am capable of that. It is still a goal, though.

I figure God could not create a universe where 2 is not the successor of 1. When I can see the reasons for QR as clearly as 2 succeeding 1, I will know the limits of God, at least from my human perspective.

[desiresjab]

My suspicion is, for me the key is to "see through" why two 4n+3 primes behave the way they do, which I call irreciprocity.

If I can see why they cannot behave as 4n+1 primes or as a mixed pair, I sense I can see it all by the same method.

It might be that a formal proof exists which would satisfy me fully, but I have no access to it or would not have the tools to understand it. Many proofs enter the terrain of Group theory and Abstract algebra, and depend on quite a few other proofs. I need to upgrade there, but there is not time for everything.

I do not know if it is possible to see it the way I want to see it. I am not even sure that anyone does. Reading the words of math professors on the subject over at the n-category cafe, I can see that even among that level the grasp is dubious, depending on a particular proof usually. One senses the lack of a deep intuitive connection and understanding of why the numbers must behave as they do. Usually because they belong to some group, subgroup or coset, which is shown to be symmetrical or asymmetrical, as the purpose serves, etc., but which is rather far removed from ground level.

I may be trying to see something at ground level that is not visible from ground level.

There was another book I now remember. It was by a gentleman named Weils who was something of a modern giant. This book was a treatment of numbers from a group theoretical standpoint, and I did not take it seriously enough at the time. That might be all I need, not the whole course. The thing is, I already have a decent understanding of what those fields do and say. It is simply the strange notation I have not adapted to on my own. I am such a fuss budget and whiner before I settle down and adapt to what one obviously has to do.

[YesNo]

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.