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I'm not too sharp right now. I have come down with the flu or something. Reciprocity is easy to state in English. The first step to understanding it is to learn enough math (mostly modular) to understand the English.Quote:
Originally Posted by YesNo
Now of course it has a much bigger reason--the razzle dazzle of ciphers.
I'll be back after I throw up and sleep.
Get well!
The more I look at this the more I feel like I am still in the shallow end of the pool.
Edit: After reading a proof, different from Eisenstein's, I began to see how the primes p and q are connected: They are each related to their product, pq.
Also the proof depended upon something nearly obvious: if three integers are added together and their sum is an even number then either all three of the integers are even or only one of them is.
http://www.lehigh.edu/~shw2/q-recip/gauss5.pdf
Back with better health.Quote:
Originally Posted by YesNo
How Gauss determines that what me might call the overflow values are enough to determine the quadratic relationship is algebraic magic, of course, but not necessarily transparent. It gives another fact, but the torch in the fact is hard to light. There is extreme familiarity with modular operations and the Chinese remainder theorem and how they all apply. Seeing how every rule you need applies at the right time and place is always going to be the case. In one's personal investigations, if one misses one of these, a great deal of time can be lost but not necessarily wasted in chasing down proof along the way of something that boils down after all to a basic law of modular arithemtic the explorer has not yet assimilated into his mathematical vocabulary fully enough so that its applications and assumptions come naturally as they do in normal algebra.
No matter how simple they continue to try to make QR, it always turns out pretty complex, except in terms of the laws themselves, which are clear and easy to apply. In asking why the two species of primes themselves behave the way they do with themselves and with the other species, it is profitable to remember that they only do so in modular arithmetic, where QR is a theorem. The comparison to normal arithmetic in the Martinson link I gave earlier is still the most illuminative and suggestive article I have seen yet. For that reason, I believe a good review of Fermat's and LaGrange's sums of squares is in order. I did this in cursory fashion a few months back, without settling in for the full ride with different hosts.
Another nagging propositon I looked at only a few weeks ago is Bertrand's paradox. It deals with geometry and the power of the continuum infinity of points on the surface of a sphere. Someone gives a good explanation on YouTube of how it is mathematically possible to dissect a sphere of diamter X and reassemble the parts into two full and complete spheres of diameter X without adding any new material. No man or machine could actually make these slices and chops, but in theory it is feasible. Or is it an unresolved paradox of infinite set theory? It deserves a second dip. So many things to chase down.
To continue in the same vein, the quote But this is just the set of integers, lifted directly from the end of the Gauss proof is, in English, what always happens at the conclusion of proofs in this mode of math, now isn't it? The results of operations on a residue system are shown to be equivalent to another set of integers previously defined. When all you need is parity to prove your point the sets do not even need the same cardinality to yield their information. This is heading toward set theory and group theory. I am still stuck on the idea of an easier vantage to peer at the heart of the law and see naked numbers bathing.
Deciding how one can show some particular set equals another set or subset must be the normal way to proceed then, what you strive to frame your question in terms of in modular forests. That is my nutshell observation.
I looked around on YouTube and found this description of Bertrand's Paradox: https://www.youtube.com/watch?v=uI2FnUmBeeoQuote:
Originally Posted by desiresjab
It seems that the paradox is resolved once one defines what it means to "choose a chord at random". One of the choices started with a fixed point, another with a fixed diameter the chords had to cross and the third asked whether the midpoint of the chord was inside or outside an interior circle. Not all of the possible chords were permitted by the selection constraints in the first two examples. I suspect the third example did include all possible chords.
Bertrand's problem looks harder than I realized: https://en.wikipedia.org/wiki/Bertra..._(probability)
I don't think the problem is resolved, as I claimed above, by saying there are more chords in one of the three examples. One can assume there is one chord in each. Then what is the probability that its side is longer than the side of an inscribed equilateral triangle. I think you are right in looking at this as a problem of picking a point from an infinite number of points.
my recollection of time is that it began with the universe, (space-time). so the first thing we have to do is wrap our heads around the question: if space-time began with the universe, then how can we ask the question, "what was before?" before implies time, but it didn't exist.
A fascinating idea that i have wondered about is the Schrodinger's cat conundrum. if there has to be an observer before any event can take place, then there had to be an observer before the first two particles o matter inter-acted, did there not? This has led some to theorize that there was intelligence prior to matter, rather than the other way around.
The Schrodinger cat problem keeps bothering me as well. Basically, every time I think I understand what it is supposed to show, I doubt that I have it right. Although I haven't read much lately, I have Amit Goswami's "The Self-Aware Universe" on my desk. He promotes "idealist science" as opposed to "materialist science".
My current view is slightly different from "there was intelligence prior to matter". At the moment, I don't think unconscious matter exists. There is nothing but intelligence. What we see as matter is conscious at a lower level that appears unconscious at the macro level where we view it.
Of course, I might be completely wrong.
Let me know if this is the wrong place...
"The superfluid Universe":https://aeon.co/essays/is-dark-matte...tm_source=diggQuote:
We are used to thinking that quantum physics dominates only the microscopic realm. But the more physicists have learned about quantum theory, the more it has become clear that this isn’t so. Bose-Einstein condensates are one of the best-studied substances that allow quantum effects to spread widely through a medium. In theory, quantum behaviour can span arbitrarily large distances, provided it isn’t disturbed too much.
... and a toon: http://www.gocomics.com/bloom-county
Ta ! (short for tarradiddle),
tailor STATELY
I'm on this list and though I don't understand all the mathematical stuff the main players here write, I enjoy the discussions between desiresjab and YesNo. Read the thread and you will see what I mean.
I was offline for half a week with computer issues. The only thing to do was to go to pen and paper.
What do we know about squares in general? What is one fact of odd squares?
I don't know except the obvious in the integers, Z, that one gets an odd number when an odd number is squared and and even number when an even number is squared.Quote:
Originally Posted by desiresjab
I liked the cartoon.Quote:
Originally Posted by tailor STATELY
I didn't know that the theories around dark matter had two major variations, those promoting modified gravity and those promoting a cold particle. The superfluid idea is also new to me. Maybe it can bridge the ideas. The modified gravity reminds me of a talk by Rupert Sheldrake where he questioned whether physical constants, in particular G, were actually constant, but changed. If G changed that would be one way to get modified gravity.
I thought a very simple fact might be surprising. All odd squares are 4n+1 numbers. There are no 4n+3 squares; such an animal cannot exist.Quote:
Originally Posted by YesNo
An important fact for anyone trying to learn modular arithmetic has to do with symmetry. In normal arithmetic the negative and positive integers have symmetry across the point zero on the number line. That is, the absolute value of -5, for instance, is equal to 5. In modular arithmetic with primes, this is no longer true--
-5≡6 Mod 11.
Perfect multiples of the modulus have familiar symmetry across zero, but no other residue class does.
Under any prime modulus p, start squaring the positive integers n in succession. The series will always begin with the standard squares you are familiar with...1,4,9,16..., until n becomes greater than p1/2 (the square root of p), at which point n2 will wrap around the modulus to some value.
Quadratic reciprocity is about how two moduli wrap around each other under quadratic pressure.
We may further notice that when both p and q are primes of the species 4n+3, their squares always wrap around the other modulus so that p2≡2 (mod q) and q2≡1 (mod p), or vice versa. They will always express this relationship when squared within the modulus of the other. This is a fact of the universe, as inviolate as 2 is the successor of 1.
One always checks first to see if the larger of p and q simply reduces to a familiar square under the other as modulus. If so, the work is done. Otherwise one starts squaring n's to see if any wraps around to the value of p, under q as modulus, or vice versa.
Not surprisingly, then, two 4n+3 primes p and q wrap around each other just as their squares do. Either 4p+3≡1 (mod 4q+3) and 4q+3≡2 (mod 4p+3), or vice versa.
We must remember that if a≡b (mod m), then ap≡bp (mod m).
It appears my recent posts on 4n+3 only apply if one of the 4n+3 primes is 3 itself. More later. Three is not typical. Or is it?
That is a more interesting fact than the one I presented. Here's my proof of it since it is not immediately obvious:Quote:
Originally Posted by desiresjab
Assume there exists an odd square m2 congruent to 3 mod 4 to get a contradiction. There are two cases to consider: m is either congruent to 1 mod 4 or m is congruent to 3 mod 4.
Consider the first case, m ≡ 1 (mod 4). Then there exists r such that m = 4r + 1 and m2 = (4r + 1)(4r + 1) = 16r2 + 8r + 1 which is congruent to 1. So m is not congruent to 1.
Consider the second case m ≡ 3 (mod 4). Then there exists s such that m = 4s + 3 and m2 = (4s + 3)(4s + 3) = 16s2 + 24s + 9. Since 9 is congruent to 1 mod 4, m is not congruent to 3.
In all cases m is not congruent to 3 mod 4 and since this contradicts the assumption, the assumption is false.
I would say it is a fact of the axiom system and the set of elements one is using rather than a fact of the universe. One could change the axiom system or the set of elements and get something different. For example, Euclidean geometry need not have much to do with space in the universe around us, but the results would be inviolate facts within the axioms of Euclidean geometry. Only if one can't consistently change the axioms would it be possible to look at the results as relevant to the universe.Quote:
Originally Posted by desiresjab
I agree that the computationally hard part comes from the wrapping process.
You can do these things anywhere. Euclidian geometry would be an outside geometry in some universes. Its laws would remain true, just as the laws of non-Euclidian geometries are true for us.Quote:
Originally Posted by YesNo
The wrapping process of moduli can turn a 4n+1 square into a 4n+3 number since, for instance three is a square under some moduli.
I only need to pinpoint the mechanics that force (4n+3)2 to perform its consistent behavior under prime moduli, for the whole thing to shake out. It is a matter of mechanics. A mechanical detail is eluding me so far. That detail will clear up every question. I not only sense this is true, I know damned well it is. There is no doubt, either, that that detail is clearly available in group theory, which is why so many proofs rely on it.
I still believe it is something I can get from peering at the numbers. My investigations are going deeper underground where I need paper and pencil.
Remember the Martinson list for the sums of squares? He mentioned that any prime number generated by a sum of two squares was a 4n+1 number. He mentioned that the table would generate every prime of 4n+1 makeup. He did not mention that all odd numbers in the table were also 4n+1 numbers. Close inspection reveals that sums of two squares can only be 4n, 4n+2, or 4n+1 numbers.
On another note of interest. Breaking a large 4n+1 prime into its unique sum of two squares, is every bit as difficult as factoring. I have not delved deeply enough, but I wonder if any of the present encryption systems are taking advantage of this. A new function as the basis means no patent battles.
Anyway, I feel I am very close to the final solution with QR. I know where to dig and I think I know how to do it.
Here is a curious fact about 4n+3 primes. Look at seven and its squares with regard to other 4n+3 primes.
Because of what we already know, we can state unequivocally that no 4n+3 prime greater than 49 can ever wrap back to be a square (mod 7). Why? Because 72 is a normally occuring number (mod 59) and it will be the square between the two, since there can always and only be one square between 4n+3 primes.
72 is the naturally occuring quadtratic residue of every 4n+3 prime larger than 47, not the other way around, ever.
This idea has its way of working with 4n+1 primes and mixed couples, too. If the larger prime does not reduce back to a sqaure under the smaller prime, then the smaller one will not stretch to a sqaure either, by the rules.
Therefore, under any moduli if p2<q, p2 is always a natural residue, meaning there is no wrap around by the sqaure.
Among 4n+1 primes and mixed couples, this fact forces them to both be squares, and to never not be mutual squares, since they must act the same way as each other. There is only a question whenever q is < p2, or vice versa. Otherwise, the results are automatic. Of course, we still must explain why they behave the way they do when p2<q. It is always good to see the task more clearly.
And one further miscellaneous fact. All square n= the sum of the nth and (n-1)th triangular numbers.
This fact makes sense. Geometrically the two triangular numbers are on each side of a diagonal through a square matrix of points missing any lattice point. One triangular set of points will have n lattice points on a side and the other triangular set of points will have n-1 lattice points.Quote:
Originally Posted by desiresjab
I state so many things incorrectly on my way to getting them right, that I should not be writing here on the subject of QR until I finish with it.
I think I am going to find that understanding why it works will prove a lot more difficult than understanding the mere mechanics of it. Its outside mechanics are not so bad, but what makes its guts work is a lot tougher to see.
One reason to continue writing is to help clarify it for yourself. I don't mind reading it. It gives me something to think about. If it wasn't for you I wouldn't be thinking about any of this.
I looked at Eisenstein's proof on Wikipedia: https://en.wikipedia.org/wiki/Proofs...ic_reciprocity
Pieces of it are starting to click. I don't understand Eisenstein's lemma yet, but I think I see the geometric point. The goal is to show that the number of lattice points within AXYW has the same even or oddness (in the same residue class mod 2) as the points in ABC with the x coordinate an even number.
The reason the even x coordinates are needed is because he will consider the lattice points in XBCY and note that the lattice points in XBCZ are even because q-1 is an even number, each column has an even number of lattice points. Since XBCY and YCZ partition an even number of lattice points into two sets, the two sets have the same even or oddness. Then one can flip XCZ onto AXY. Now we already have the lattice points under even x coordinates. This flipping gives us the lattice points under odd x coordinates. So we have them all. Do this for AYW and we have all the lattice points, all (p-1)(q-1)/4 of them, in AXYW which is what we wanted.
The only piece missing is Eisenstein's lemma which is in the article, but I haven't finished understanding it yet.
Edit: I think I understand the lemma. It is interesting how it also uses the fact that when an even number is represented as the sum of two other integers, those two other integers are both either even or both odd. It cannot be the case that one is even and the other odd because their sum is even.
Yes, identical parity is a matter of (mod 2) with those columns. This is one of the first things Eisenstein makes clear.
P vs Q, when both are 4n+3 species. If the larger p wraps down to a square (mod q), then numbers under the smaller q may not square and then wrap to a square number (mod p).
11 (mod 7) wraps down to 4, a square (mod 7). Remember, there are only three different squares (mod 7), 1, 2 and 4. Any number squared (mod 7) wraps down to one of those three numbers.
Conversly, since 7 is >the square root of 11, then 4, 5, 6 7, 8, 9 or 10 squared would have to wrap back to 7, but something prevents them. I am now looking for exactly what prevents them from having reciprocity instead of irreciprocity. My goal is to see why two 4n+3 primes behave toward each other as they do. What prevents them from behaving as two 4n+1 primes or a mixed couple do? I feel I am on the right vein, looking for the exact spot to sink my pickaxe.
Your mention a few posts back that Gauss comes up with smaller numbers, but with correct parity, of course, was a good shout out.
It is not square vs square, in my current vision, but square vs some multiple of the other prime, which determines the precise behavior. This information is probably not in Eisenstein's rectangle, but I will not say for sure. There may be some reflection of it, but I think not as well, since there is a limit to the information the rectangle can contain.
If one looked at Eisenstein's lattice points in the rectangle AXYW, the only time that rectangle will have an odd number of lattice points, as it does in the p = 7 and q = 11 example, is when both p and q are congruent to 3 mod 4. Otherwise AXYW will have an even number of lattice points. Also the number of lattice points in AXYW is equal to the sum of the number of lattice points in the triangle AXY and the number of lattice points in the triangle AYW.
If we have an odd integer that is the sum of two other integers, then one and only one of those other integers can be odd, that is, one and only one of (p|q) and (q|p) can equal -1 and be a quadratic non-residue. The other has to be a quadratic residue.
This all tue, and these are the same things I keep repeating to myself. But I know there is a mechanism in back of it all which prevents two 4n+3 primes from having the same "character." I am still in search of the precise mechanism and now have only decent confidence that I will succeed.Quote:
Originally Posted by YesNo
There is a shortcut I am not sure you know. Suppose we have a Legendre symbol (p/q), and q is much larger. Instead of squaring numbers (mod q) to see if one is a square, we are to always permitted to invert the symbol to (q/p). Then all we have to do is reduce. We only need one prime's quadratic character to know the other's, and this works equally with both species. One hundred percent legal.
Another legal move takes advantage of the symbol's mulipicative properties. if p=a(b), then (a/q)(b/q) will always give the right answer.
There seems to be more going on here as you mention.
I am aware of the complete multiplicative nature of the Legendre symbol and that one can reduce the larger prime mod the smaller one. That reduced number will not likely be a prime so one would have to factor it. Also (a2|p)=1, so these even powers can be discarded.
I am looking at Vanden Eynden's "Number Theory". He proves QR using Gauss's methods. I'll see if I can find something more enlightening by using that proof.
By reducing, I only mean this: wrap the bigger number around the smaller modulus until the remainder is revealed, the way one can reduce 31 to 1 (mod 3). In our manipulations, it is always legal to invert the Legendre symbol, then reduce, since it is generally easier to "reduce" than to start squaring numbers to see if a square appears under the other modulus. I hope that was clear.Quote:
Originally Posted by YesNo
What I am looking for is probably not available, for I have never seen it mentioned that someone was explaining the mechanism of QR, they were only proving it. Gauss proved the law itself, and many others have since, but no one has explained it that I know of. If this is pulled off on a literature forum, it will be a coup for the ages.
We might as well try explaining it to each other. Whether it is quantum physics or number theory, it probably doesn't make complete sense even to the people who know what they're talking about.
I think I understood what you meant by reducing the top prime in the Legendre symbol, (p/q). One might as well start with p > q and then find what p is congruent to modulo q. That won't likely be a prime any more, but one can try factoring it to simplify the calculation even more.
After looking at Vanden Eynden's text, I found this relationship which might be interesting and is part of his (Gauss's) proof of QR.
Let p and q be odd primes with p > q and p is congruent to q mod 4. Then there exists some integer a such that p = q + 4a. Now that last equation implies the existence of three other congruence relationships.
1) If p = q + 4a, then p ≡ q (mod 4a). This is just the original congruence including a.
2) If p = q + 4a, then p = 4a + q and so p ≡ 4a (mod q). Now, 4a is linked to p via q.
3) If p = q + 4a, then -q = 4a + p(-1) and so -q ≡ 4a (mod p). Now 4a is linked to -q mod p.
In this way 4a is the link between p and q. One can get the QR result for the cases when p ≡ q (mod 4) by considering the following:
(p/q) = (4a/q) = (4a/p) = (-q/p) = (-1/p)(q/p)
The first part comes from p ≡ 4a (mod q). The second part was proved in the book and is non trivial, but can be assumed for the moment. The third part comes from 4a ≡ -q (mod p) and the last part comes from the multiplicative property of the Legendre symbol.
I'm not sure if this helps any, but it seemed interesting to me.
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).Quote:
Originally Posted by YesNo
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.
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.
Oh, and by reducing I mean simply to carry out the division mentally and find the remainder to see if it is a square. 31 reduces to 1 (mod 3), for instance. Specifically we could say, 10p+1=q, or (q-1)/10=p. This is in the neighborhood of what you were saying above.
A relevant concept I came up last year is "highly even" and "barely even" numbers. 4n+1 numbers minus one are all highly even and 4n+3 numbers minus one are all barely even. A barely even number is divisible by 2 only once, a highly even more than once. This falls right out of something called the ruler function, which I also discovered in my investigations.
These two types of numbers are obviously germane to QR. The higher degree of evenness of Eisenstein's rectangle when both triangles have an even number of lattice points must certainly be an important fact. When QR =-1, the overall rectangle has a downright paucity of factors of two, managing onlyfour of them.
The real problem with the above idea is that two (4n+1)-1 numbers which by definition have a higher degree of evenness can reject each other, so to speak, and mutually not be in each other's quadratic residue set.
I really enjoy our discussion here. I am quite lucky to find even one able volunteer willing to go along on this perilous mission with me. The biggest problem I have is that I am wearing myself out thinking about it.
The layout of this forum is actually superior to the math forum I visited when it comes to typing math. This forum at least allows exponents. The other forum still uses dumb up arrows for powers.
Damned duplicate posts!
the 7x11 rectangle with 60 interior lattice points is only divisible by 2 twice. Another conjecture gone awry, perhaps. So far, no matter what I try to connect the behavior to, it only looks promising for a while.
But wait. Are all such rectangles whose dimensions are (4j+3)(4k+3) divisible by two only twice? Is that the nature of them? Yes, of course. What else would I be talking about? (Beat my own forehead). The train is still on track. One wheel anyway.
This paucity of 2's may lead directly to the mother lode, the reciprocity mechanism.
Well, its better than the problem you had a while back, of disappearing posts!Quote:
Originally Posted by desiresjab