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Thursday, December 10, 2009

Algebraic integers vs Complex numbers

One of the weirdest things is where the ring of algebraic integers can be shown to contradict with the field of complex numbers.

Given

7(175x2 - 15x + 2) = (5a1(x) + 7)(5a2(x)+ 7)

where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

divide off the 7.

The problem is finding how the 7 multiplied, and it's easier to figure out in the complex plane.

So from the complex plane, first step is to normalize the functions, that is, have functions that equal 0, when x=0.

Looking at x=0, gives a2 + a = 0, so a1(0) = 0, or -1, and a2(0) = -1, or 0.

So I can let a1(0) = 0, and then introduce a new function b2(x), where:

b2(x) = a2(x) + 1, so a2(x) = b2(x) - 1, and making that substitution:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

where now a1(0) = b2(0) = 0, and with normalized functions, it's now clear that the first factor, which is (5a1(x) + 7), is the product of a factor multiplied by 7.

It might seem trivial, but notice that if the OTHER factor were such a product then you'd have:

7(175x2 - 15x + 2) = (5a1(x) + 1)(5b2(x)+ 14)

and if the 7 split up, so that each had sqrt(7), then:

7(175x2 - 15x + 2) = (5a1(x) + sqrt(7))(5b2(x)+ 2sqrt(7))

as I keep the appearance of the functions the same with those hypothetical's as the functional notation can hide things, but the 2 cannot.

So from the complex plane it is clear we need one more function--the other b--as we actually have something like:

7(175x2 - 15x + 2) = (5(7)b1(x) + 7)(5b2(x)+ 2)

And dividing off the 7 is easy then:

175x2 - 15x + 2 = (5b1(x) + 1)(5b2(x)+ 2)

Which it should be! Constant factors are supposed to be trivially removable.

But notice, you needed a1(x) = 7b1(x) for it to BE trivial.

Going back to the challenge then, and considering the ring of algebraic integers, it seems logical that in agreement with the field of complex numbers the ring of algebraic integers would give the same result, but as the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

that would indicate that just one root would have 7 as a factor. But there is the contradiction as provably if the roots are non-rational with an integer x, then NEITHER of the roots can have 7 as a factor in the ring of algebraic integers!

So maybe there is a mistake in the proof over the complex plane?

Not if 7*1 = 7, which it does even on the complex plane. The important thing here is not the functions--it's the two constants.

With normalized functions:

7(175x2 - 15x + 2) = (5a1(x) + 7)(5b2(x)+ 2)

can only be obtained one way. If you don't believe me, try some other way.

So how could a factor result emerge from the field? Because it's the distributive property, and is equivalent to something like:

7(x2 + 3x + 2) = (7x + 7)(x+2)

But now we can get to the truly weird.

The key here at this point is the source of the a's, as they are roots of:

a2 - (7x-1)a + (49x2 - 14x) = 0

But with x=1, that has roots 3 + sqrt(-26) and 3 - sqrt(-26), and you now know that one of them is a product of 7 and some other number, but which one?

There's no way to determine.

And even weirder, it's not possible to say that one in particular does as the square root is ambiguous. To understand THAT result consider:

5 + sqrt(4)

and that is 7 and 3, but 3 and 7 work too as sqrt(4) = -2 or 2.

Maybe, you say, no? Well (-2)(-2) = 4, so it is proven that -2 IS a solution to sqrt(4).

Historically some time ago some math people decided they didn't like that the square root gives two results, so they defined one away and people get taught that you take the positive, but that is a convention that usually doesn't seem to matter, but here it does.

It is one of those annoying math errors where a person may believe that sqrt(4) is just 2.

But sqrt(4) is 2 or -2 no matter what people believe.

With 3 + sqrt(-26) and 3 - sqrt(-26) in both cases you have two numbers, but the order simply shifts, where you just can't get rid of the square root, but you know that for one case 7 should be a factor.

How? Because of the proof above in the field of complex numbers that's how!

But the ring of algebraic integers will not allow 7 to be a factor of either root.

So you have just learned of a fascinating contradiction at the heart of modern number theory. A place where the ring of algebraic integers dares to contradict the field of complex numbers.


James Harris

Saturday, October 24, 2009

Connecting prime distribution to a continuous function

Going back into my archives to copy from prior posts I have a result from 2002 where a simple progression of formulas shows how the prime distribution is connected to a continuous function:

1. Prime counting function sieve form:

With natural numbers where p_j is the j_th prime:

P(x,n) = x - 1 - sum for j=1 to n of {P([x/p_j],j-1) - (j-1)}

where if n is greater than the count of primes up to and including sqrt(x) then n is reset to that count.

To see an example of it used click here.

The [] is the floor function and is actually redundant as you're to use natural numbers.

The sqrt() is the integer square root, for instance sqrt(10) = 3, as that's floor(sqrt(10)), and it's automatically positive because natural numbers are only the positive integers greater than 0.

2. Prime counting function difference equation form:

With natural numbers

P(x,y) = x - 1 - sum for j=2 to y of {(P([x/j],j-1) - P(j-1, sqrt(j-1)))(P(j, sqrt(j)) - P(j-1, sqrt(j-1)))}

where if y>sqrt(x), then P(x,y) = P(x,sqrt(x)).

And there is no longer need to tell the equation of any numbers that are prime. If you program it, you'll notice it plucks out primes on its own as it counts, as only when j is prime does:

P(j, sqrt(j)) - P(j-1, sqrt(j-1)) = 0


3. Continuous function:

In the complex plane

P'y(x,y) = -(P(x/y,y) - P(y, sqrt(y))) P'(y, sqrt(y))

and a continuous function is found easily enough from the difference equation from above. To see the full derivation click here.

The partial differential equation form is rather succinct like the sieve form, and I just gave the PDE versus talking about summation as here there is no actual prime count involved.

And that is why the prime distribution is connected to a continuous function: a sieve form as P(x,n) equation can go to a difference equation form, which leads to a partial differential equation.


James Harris

Sunday, May 17, 2009

Pell's Equation basics

The equation x2 - Dy2 = 1 is commonly called Pell's Equation by modern mathematicians and it has reportedly been studied for thousands of years.

It is generally considered with all integer solutions where none are zero, and there are classical methods for solving it where I've seen the most commonly mentioned one relies on continued fractions.

I have figured some things out about it that I find of interest:

1. For any positive integer D, if D+/-2 is a perfect square, then D+/-1 is the first solution to Pell's Equation for x, while if D+/-1 is a perfect square then x=2D-/+1 is a solution, and also if D is divisible by 4 and D+/-4 is a perfect square then x = (D+/-2)/2 is a solution.

e.g. With D=7, 7+2 is a square as it is 9, so x=8:

82 - 7*32 = 1.

And also, with D=14, adding 2 gives 16, so x=15:

152 - 14*42 = 1

In general, if D+/-2 is a perfect square, then D = n2 -/+ 2 for some natural number n, and

(n2 -/+ 1)2 - (n2 -/+ 2)n2 = 1

and notice n4 -/+ 2n2 + 1 - n4 +/- 2n2 = 1.

And, if D+/-1 is a perfect square, then D = n2 -/+ 1 for some natural number n, and

(2n2 -/+ 1)2 - (n2 -/+ 1)(2n)2 = 1

and notice 4n4 -/+ 4n2 + 1 - 4n4 +/- 4n2 = 1.

If D is divisible by 4 and D+/-4 is a perfect square then you have

(2n2 -/+ 1)2 - (4n2 -/+ 4)(n)2 = 1.

So there are trivially found answers when D+/-2, or D+/1, or D+/-4 with D divisible by 4 is a perfect square.


2. An alternate equation to Pell's Equation is j2 - Dk2 = -1, which is also commonly called the negative Pell's Equation. I have found that if you have a solution to the negative Pell's Equation you then immediately have a solution to Pell's Equation from: x = 2j2 + 1.

For instance,

297182 - 61*38052 = -1, so j = 29718, gives

x = 1766319049, from x = 2j2 + 1, and

17663190492 - 61*2261539802 = 1.

That is to show one of the larger solutions. A simpler example is,

22 - 5*12 = -1, gives x=2*22 + 1 = 9.

And 92 - 5*42 = 1, as required.

The requirement that x = 2j2+1, of course, means x must be odd. Further, it can be shown that when the negative Pell's Equation has a solution and x is the minimum solution for Pell's Equation, x = -1 mod D is required.


3. In addition to the negative Pell's Equation alternate there are at least 2 more basic alternates that will solve Pell's Equation with solutions to them:

j2 - Dk2 = 2, x = j2 - 1

and

j2 - Dk2 = -2, x = j2 + 1


4. While number theorists consider Pell's Equation only with integers, it can be considered with rationals, parameterized and related then to all the conic sections except the parabola:

Given x2 - Dy2 = 1, in rationals:

y = -2t/(D - t2)

and

x = (D + t2)/(D - t2)

and you get hyperbolas with D greater than 0, and ellipses with D less than 0, and the circle when D=-1, giving the well-known circle parameterization:

With x2 + y2 = 1, in rationals:

y = 2t/(1 + t2)

and

x = (1 - t2)/(1 + t2)


James Harris

Friday, April 17, 2009

Mystery with Pell's Equation parametric solution

What I've managed to find from a result that followed from my Quadratic Diophantine Theorem is a parametric solution to Pell's Equation in rationals:

With x2 - Dy2 = 1

I have proven:

y = 2[f2v - 1]/[D - (f2v - 1)2]

and

x = [D + (f2v - 1)2]/[D - (f2v - 1)2]

where f1f2 = D-1, and the f's are non-zero integer factors, while v is nonzero but is otherwise a free variable.

Which is, of course, looks similar to the parametric solution for circles in rationals discovered in antiquity:

x = (1 - t2)/(1 + t2)

y = 2t/(1 + t2)

See: http://mathworld.wolfram.com/Circle.html eqns. 16 & 17

It looks similar as it is the parametric solution for circles when D=-1, for ellipses with D<0, and of course, hyperbolas are given by D>0.

In considering what I see as the mystery of the quietness in the mathematical literature around this solution, it seems possible that the focus on Pell's Equation as a Diophantine equation, considering integers only, can be part of it.

Research on the web indicates that the parametric solution I've found was known much earlier, as Fermat was aware of it. So now I wonder why it gets so little mention and why wasn't the relation to rational parametric solutions for circles, ellipses and hyperbolas noted? Or was it? I'm continuing research.

There are other interesting things I've found.

For example, for any positive integer D, if D+2 is a perfect square, then D+1 is the first solution to Pell's Equation for x.

e.g. 82 - 7*32 = 1, 152 - 14*42 = 1

In general, if D+2 is a perfect square, then D = n2 - 2 for some natural number n, and

(n2 - 1)2 - (n2 - 2)n2 = 1

and notice n4 - 2n2 + 1 - n4 + 2n2 = 1.

But that is probably something so well-known that it's considered trivial.

Another thing is the results with the negative Pell's Equation and two other alternates shows that in general the alternates can be solved in order to solve Pell's Equation, as they must exist for prime D, and a method can be generalized for composite D as well.

e.g. For the famous D=61 case, the solution from the negative Pell's Equation is not as spectacular:

j2 - Dk2 = -1

297182 - 61*38052 = -1

gives x = 1766319049, from x = 2j2 + 1.

17663190492 - 61*2261539802 = 1

The difference in computation is staggering though as my example above suggests, as the first x for Pell's Equation is roughly the square of the first j for the alternates in all cases, and the same techniques that work for Pell's Equation work for the alternates, for instance, continued fractions.

The issue is I'm guessing not minor, as consider a paper I found with some digging on the web from Los Alamos National Laboratory researchers:

...Zeros of weight 1 3j- and 6j-coefficients are known to be related to the solutions of classic Diophantine equations. Here it is shown how solutions of the quadratic Diophantine equation known as Pell's equation are related to weight 2 zeros of 3j- and 6j-coefficients.

Link to full citation


James Harris

Sunday, April 12, 2009

Advancing tautological spaces methodology

It has been almost nine years since I came up with the idea I now call using tautological spaces as that happened back in December 1999, and since that time I've done some work on terminology (link is a reference for the rest of this page).

Now after my success with a general method to reduce binary quadratic Diophantine equations, I see a need to advance the methodology of tautological spaces where that is simple enough.

To handle quadratic Diophantines in general I see a need for a more general tautological space:

a1h + a2h +v1a3h +...+vn-2anh = 0(mod a1h + a2h +v1a3h +...+vn-2anh )

to handle conditionals with n variables, where I call h, the hyperdimension.

As an example consider n=4, where all the conditional variables are raised to the 1st power so h=1, as then the tautological space is described by:

T_S(1,1,{1,1},{1,1})

So it is a 6-dimensional tautological space a hyperdimension of 1, so there are two free variables which by my type of usage would be v1 and v2, so that looks like:

x + y + v1z + v2w = 0(mod x + y + v1z + v2w)

The v's give additional degrees of freedom to allow reducing out to a general solution like with my Quadratic Diophantine Theorem.

Actually I realized the need to extend tautological space terminology by wondering about considering solutions where there are more conditional variables than say, x, y and z.


James Harris

Edited 6/24/12 to correct leaving off hyperdimension which IS necessary.

All about identities

The role of identities in mathematics is so familiar that most people take it for granted while my findings show that identities can be seen in an entirely new light.

Identities are simple enough as they are always true and are the tautologies of mathematics.

For instance 1=1 and x=x are both identities.

Mathematics uses identities all the time as consider the very easy solution of the following equation:

x-2 = 0

as you can use the identity 2=2 to find that

x-2+2 =2, so x=2

and you have a solution that IS so trivial that most probably don't think of identities when they solve such an equation.

A slightly more complicated but still easy use of identities is with completion of the square, as for instance, given

x2 + 2xy = z2

adding the identity y2 = y2 gives you

x2 + 2xy + y2 = y2 + z2

which of course is just (x+y)2 = y2 + z2, allowing you to take the square root of both sides and solve to find

x = -y +/- sqrt(y2 + z2).

Despite their importance in algebra where you cannot really do algebra without identities, development of the algebra of identities has not been extensive up until now which can be seen by comparing from my own research where I rely on more complicated identities still like

x2 + y2 + vz2 = x2 + y2 + vz2

where to a large extent what I do is like what was shown before, except that the identities that I add end up being a lot more complicated, and are being used for more complex results.

That brings up the need for classification, where I've decided that the word "identity" doesn't do enough to cover everything so I introduced the phrase "tautological space", as with my use of identities the identity is not an afterthought at the end just to get to the answer, but a start of the argument and a dominant part of the proof.

Continuing now to classify all identities as tautological spaces, I've used the number of variables to give the dimension, so that with 4 variables you have a 4-dimensional tautological space.

The simplest identities like 2=2 that have no variables, have no dimension and I call them unary tautological spaces because they are all multiples of 1=1.

If any variables of the tautological space are raised to a power other than 1 then that is a hyperdimension and is shown by a set of natural numbers. Variables in a tautological space cannot be raised to a negative power, nor to 0 as that is redundant since instead a constant can be used.

For example with x2 + y2 + vz2 = x2 + y2 + vz2 you have a 4-dimensional tautological space with a hyperdimensional set of {2,2,1,2}.

Since the hyperdimensional set gives the number of dimensions, more compactly then you can describe a tautological space as follows.

T_S{2,2,1,2}

And that allows a complete classification of all possible identities.

Coming back to look this over I see that it does not, so I'm updating with a change in terminology.

The free variable which I traditionally call v is given by the 1, but I think I can use nesting to show where it is:

T_S(2,2,{1,2})

So that means a 4-dimensional tautological space where the conditional variables are all squared.


James Harris

Sunday, April 05, 2009

Rational parameterization conic sections result

It's worth pulling out into a separate post a wonderful unifying result for 3 of the 4 conic sections that follows from the parameterization of Pell's Equation:

Given x2 - Dy2 = 1, in rationals:

y = 2t/(D - t2)

and

x = (D + t2)/(D - t2)

showing it more traditionally versus the way that results from my own derivation.

You get hyperbolas with D>0, the circle with D=-1, and ellipses with D<0.

You can see the D=-1 case from a well-known mainstream source at the following link:
See: http://mathworld.wolfram.com/Circle.html eqns. 16 & 17

I actually feel very honored to have my own derivation leading to such a beautiful result, which allows categorization of ellipses and hyperbolas by a single number, which then is probably connected to the eccentricity.

So what mathematicians call "Pell's Equation" is in other areas a one-stop equation for generating 3 of the 4 conic sections in rationals just by fiddling with one variable--D.


James Harris

Tuesday, March 24, 2009

Dual factorizations with Pell's Equation

In rationals starting with Pell's equation:

x2 - Dy2 = 1

I have proven that

(D-1)j2 + (j+/-1)2 = (x+y)2

where j = ((x+Dy) -/+1)/D.

Here the +/- indicates that one variation will work so it is an OR and not an AND. Either plus OR minus will give a valid j.

It suffices to substitute out j, and simplify, which will give Pell's Equation again, showing the relations are valid.

Dual factorizations:

(x-1)(x+1) = Dy2

gives an opportunity to factor D, which I will declare to be an odd composite.

While

(D-1)j2 = (x+y - (j+/-1))(x+y + (j+/-1))

gives the opportunity factor D-1.

Focusing on the second factorization, note to generalize I need additional variables.

Introducing u and v, where j = uv, I further need factors f1 and f2 where f1f2 = D-1.

Then I have for generality

(x+y - (uv+/-1)) = uf1

and

(x+y + (uv+/-1)) = uv2f2

Note that f1 and f2 can be declared to be integers, and further that x and y uniquely determine u and v, as consider:

(x+y + (uv+/-1)) = (uv)vf2

where I remind that j = ((x+Dy) -/+1)/D = uv, so given any x and y in rationals, I can simply separate off all non-unit non-zero integer factors in common with D-1, for f1 in the first, and for f2 in the second, leaving u and v as rationals, and trivially solved and proven to exist as rationals.

Therefore, it is proven that for every x and y that fulfill Pell's Equation there is a u and v pair.

Given that there must exist x and y, such that (x-1)(x+1) = Dy2, non-trivially factors D, it is proven there exists a u and v pair unique to any such cases.

That completes the underlying mathematical proof showing the dual factorization and proving the existence of rational v, for a non-trivial factorization of a composite D.

It is now possible to move further into the utility of these expressions and reveal a direct factoring method.

There are then three equations available:

(x+y - (uv+/-1) = uf1

(x+y + (uv+/-1) = uv2f2

((x+Dy) -/+1)/D = uv

Considering x and y to be unknowns it follows then that x and y can be determined as functions of v. Doing so gives:

y = 2[f2v - 1]/[D - (f2v - 1)2]

and

x = [D + (f2v - 1)2]/[D - (f2v - 1)2]

where v must be nonzero.

But then for a non-trivial factorization, f2v - 1 must have that non-trivial factor as its factor.

If you consider x as a ratio of integers r and t, where x = r/t, for a non-trivial factorization of D, an odd composite, where D = g1g2 where the g's are integers, then a solution for x must exist with:

r = (g1 + g2)/2, and t = (g1 - g2)/2.

However, adding the numerator and denominator of x above gives 2D, meaning that they are not coprime, and one of the factors must be a non-trivial factor of D.

If that factor is g2, then a boundary condition is:

g2(g1 + g2) < 2D

so, g22 < sqrt(D), must be true, for that case, meaning D cannot be a square.

Now then, if f2v - 1 = g2, then

v < (sqrt(D) - 1)/f2

is a boundary condition.

That allows you to get search conditions for v, but it's not clear at this time if there is any significant benefit to using this approach.

I will note that if f2 - 1 <= g2, then there must be an integer v satisfying the condition which will non-trivially factor with f2. The bottom default case is f2 = 2, or 1, which of course, has the highest number of possible searches, so if there is a connection available it would be preferable with the largest f2 possible.

I had hopes that this approach solved the factoring problem, but had failed to simplify my equations sufficiently to see that the shared factor in common with D between the numerator and denominator of x, precluded the simple solution I thought was available.


James Harris

Saturday, January 24, 2009

Finding Pythagorean triples with Pell's equation

An earlier post I made makes mention of relating Pell's equation to Pythagorean triples, and it turns out that you can just solve for Pythagorean triples using solutions from certain Pell's equations:

Given x2 - Dy2 = 1, and x = 1 mod D or x = -1 mod D, from the prior post I reference above I have that

(D-1)j2 + (j+1)2 = (x+y)2

or

(D-1)j2 + (j-1)2 = (x+y)2

where

j = ((x+Dy)-1)/D or j = ((x+Dy)+1)/D

which gives a Pythagorean triple whenever D-1 is a square, for instance with D=2, I have a solution with x=17, y=12, as 172 - 2(12)2 = 1, so

j = ((17+2(12))-1)/2 = 20 is a solution giving:

202 + 212 = 292

Notice that different types of Pythagorean triples are included as D increases in size, as for instance, with D=5, you have triples of the form:

(2j)2 + (j+/-1)2 = (x+y)2

(where I use "+/-" as an OR as above as it's less tedious)

so given a solution to x2 - 5y2 = 1, you have solutions.

e.g. x=9, y = 4, works as 92 - 5(4)2 = 1, so j = (9 + 5(4) + 1)/5 = 6 is a solution. And then

122 + 52 = 132.

Notice too that when D-1 is not a square you get integer solutions to ellipses, and you also get the result that with a natural number n, there are always an infinity of integer, non-zero solutions to the equation: nu2 + v2 = w2

The question now arises, can you go in the other direction and solve for a Pell's equation using a Pythagorean triple?

The answer is, yes. Consider, 242 + 72 = 252.

If 7 is j+/-1, I can try j = 6 or 8, and find D=17 or D=10.

With j=6, D=17, I have j = ((x+Dy) -/+1)/D, so

17(6) = x+17y -/+ 1, so x+17y = 17(6)+/-1, and x+y = 25, so

16y = 17(6)+/-1 - 25, which doesn't give an integer y.

But trying j=8, D=10:

x+10y = 10(8)+/-1, so 9y = 10(8)+/-1 - 25, gives y = 6 as a solution, so x = 19, and

192 - 10(6)2 = 1.

So you can go in either direction.

Which then proves an infinite number of solutions to the more general form of Pell's Equation.


James Harris

Sunday, January 11, 2009

To discuss go to my math group

I am increasingly interested in the resources that My Math Group provides, as reality is that it's extremely difficult to get a comment on this blog itself, and besides, the format at the group is more naturally suited for discussion anyway.

So if you've wanted to talk about my research you have the opportunity available.


James Harris