And one further miscellaneous fact. All square n= the sum of the nth and (n-1)th triangular numbers.
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.Originally Posted by desiresjab
My blog: https://frankhubeny.blog/
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.
Last edited by YesNo; 02-09-2016 at 12:42 PM.
My blog: https://frankhubeny.blog/
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.
My blog: https://frankhubeny.blog/
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.
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.
My blog: https://frankhubeny.blog/
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.
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.
My blog: https://frankhubeny.blog/
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).
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.
Last edited by desiresjab; 02-12-2016 at 10:10 PM.
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!
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