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-   -   Is this solvable in general. (https://www.mersenneforum.org/showthread.php?t=22578)

jwaltos 2017-09-14 04:56

Is this solvable in general.
 
Soliciting advice on how to solve this equation:

-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c


Assuming a, b, c and d are known, what is the best method to resolve the remaining variables: x, y, such that the resulting number is a specific integer.

xilman 2017-09-14 06:27

[QUOTE=jwaltos;467756]Soliciting advice on how to solve this equation:

-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c


Assuming a, b, c and d are known, what is the best method to resolve the remaining variables: x, y, such that the resulting number is a specific integer.[/QUOTE]Where does this come from? It's far too complicated for it to be at all likely that you made it up on the spur of the moment so it likely arises in the course of some other algebraic manipulations. Knowing its origin could well make a solution easier to find.

BTW, it's not an equation (I don't see a '=' anywhere within it) but a formula. Someone has to do RDS's job for him these days ...

retina 2017-09-14 07:09

[QUOTE=jwaltos;467756]-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c


Assuming a, b, c and d are known, what is the best method to resolve the remaining variables: x, y, such that the resulting number is a specific integer.[/QUOTE]You can replace x and y with any value you like. Simplifying and coalescing all the constants gives this [b]formula[/b]:

f(x,y) = a*x^3 + (b+c*y)*x^2 + (d+e*y)*x + f*y^2 + g*y + h

S485122 2017-09-14 07:29

You both forgot the last part of the question : "such that the resulting number is a specific integer."

Rephrasing a bit [quote]Assuming a, b, c and d are known, what is the best method to resolve the remaining variables: x, y in order that the formula yields a specific integer A when
A=-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c[/quote]

Jacob

retina 2017-09-14 08:21

It looks not to dissimilar from an elliptic curve.

jwaltos 2017-09-14 12:18

Thank you for the responses.
I developed this equation as part of an integer factorization toolkit. Complex and rational values apply to `a,b,c,d,x,y` which can also resolve this equation to an integer value.

I can solve for the `a,b,c,d` values in a deterministic manner but resolving `x,y` for large values of `F` in polynomial time still evades me. I am somewhat stymied and vexed regarding how to solve this equation without using sieving or random processes for large values. Transformations into polar coordinates, complex analysis and differential geometry are valid solution paths. A suitable selection for the values `a,b,c,d,x` will algebraically factor the equivalence into (m*y+c1)*(n*y+c2). ie. a=0:b=1:c=35:d=40:x=1 -->(17*y+15)*(2130*y+97)

F=-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c
where F can be any integer.

Basically, my question is, what are the least number of variables within this equation that must be known and what are their numeric limits before it cannot be solved in polynomial time. Conversely, what must be known before this equation can be solved in polynomial time and what tools are required. ie LLL, infinite precision, etc...

Dr Sardonicus 2017-09-14 14:18

[QUOTE=jwaltos;467756]Soliciting advice on how to solve this equation:

-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c


Assuming a, b, c and d are known, what is the best method to resolve the remaining variables: x, y, such that the resulting number is a specific integer.[/QUOTE]
I second [b]xilman[/b]'s question, where did this function come from?

I'll call it F(a,b,c,d,x,y)

Knowing what [i]kind[/i] of critter this is, would be of considerable help in addressing the question of solving

F(a,b,c,d,x,y) = N

CRGreathouse 2017-09-14 16:17

[QUOTE=jwaltos;467769]Basically, my question is, what are the least number of variables within this equation that must be known and what are their numeric limits before it cannot be solved in polynomial time.[/QUOTE]

You're asking about a bivariate Diophantine cubic equation. But even bivariate Diophantine quadratics generally take more than polynomial time, so I see no reason to be that optimistic.

jwaltos 2017-09-14 18:02

[QUOTE=xilman;467758]Where does this come from? It's far too complicated for it to be at all likely that you made it up on the spur of the moment so it likely arises in the course of some other algebraic manipulations. Knowing its origin could well make a solution easier to find.

BTW, it's not an equation (I don't see a '=' anywhere within it) but a formula. Someone has to do RDS's job for him these days ...[/QUOTE]

This equation was derived in my attempts to develop a simple method of factoring integers that can be done in polynomial time without the use of quantum based systems OR to determine that the IFP cannot be resolved in poly time. Some of the literature I have cited in prior posts as well as member posts have contributed to the origin of that expression. It's a simple representative result.
By relaxing the condition in Matiyasevich's theorem for `integer only` solutions, Le Chatelier's principle could be invoked (by analogy) where poly time solutions can be made explicit.

And yes, like a blind squirrel searching for nuts, optimism does help but having a `nose` for certain things prevents that squirrel from starving.
I can't elaborate more without redundancy so I'll just say thanks to those who submitted their input and keep beavering away at this.

CRGreathouse 2017-09-14 20:30

[QUOTE=jwaltos;467787]This equation was derived in my attempts to develop a simple method of factoring integers that can be done in polynomial time without the use of quantum based systems OR to determine that the IFP cannot be resolved in poly time. Some of the literature I have cited in prior posts as well as member posts have contributed to the origin of that expression. It's a simple representative result.
By relaxing the condition in Matiyasevich's theorem for `integer only` solutions, Le Chatelier's principle could be invoked (by analogy) where poly time solutions can be made explicit.[/QUOTE]

I think you've reduced factoring to a problem harder than factoring. :unsure:

R. Gerbicz 2017-09-14 20:42

[QUOTE=jwaltos;467787]This equation was derived in my attempts to develop a simple method of factoring integers that can be done in polynomial time without the use of quantum based systems OR to determine that the IFP cannot be resolved in poly time. Some of the literature I have cited in prior posts as well as member posts have contributed to the origin of that expression. It's a simple representative result.
By relaxing the condition in Matiyasevich's theorem for `integer only` solutions, Le Chatelier's principle could be invoked (by analogy) where poly time solutions can be made explicit.

And yes, like a blind squirrel searching for nuts, optimism does help but having a `nose` for certain things prevents that squirrel from starving.
I can't elaborate more without redundancy so I'll just say thanks to those who submitted their input and keep beavering away at this.[/QUOTE]

I'd say you are overcomplicating this. We can reformulate the factorization problem with a non-linear integer optimization problem: for a given n>1 integer write

[CODE]
min x+y
subject to
x*y=n
x>=0
y>=0
x,y is integer
[/CODE]
then n is prime iff the opt=n+1, otherwise n is composite and we'll know a non-trivial factor of n. With this if the above problem is solvable in polynomial time then integer factorization problem is also in poly. (just call it also recursively for the d,n/d factor where d=x).

Using this in some case any solution will give a non trivial factorization, say for
(5*x+2)*(5*y+2)=n. (it'll give a solution if n=4 mod 5 and n has a d=2 mod 5 divisor). Why would be your longer and higher/ degree polynom is easier than mine?

jwaltos 2017-09-14 22:26

[QUOTE=retina;467759]You can replace x and y with any value you like. Simplifying and coalescing all the constants gives this [b]formula[/b]:

f(x,y) = a*x^3 + (b+c*y)*x^2 + (d+e*y)*x + f*y^2 + g*y + h[/QUOTE]

This approach, transforming the structure of the polynomial expression is necessary to better understand the context of what I am asking.

[QUOTE=R. Gerbicz;467798] Why would be your longer and higher/ degree polynom is easier than mine?[/QUOTE]
Because it is an optimal and very specific entry point to a difficult set of questions for which the conceptual tools may/may not exist to resolve them.. one way or the other. Every variable and operation has a very specific interpretation much like equations within general physics where dimensional analysis is applicable.
For example, the uniformization theorem and the geometrization conjecture are contexts that I have explored. As a non-specialist such topics are heavy going but worth it.

[QUOTE=CRGreathouse;467797]I think you've reduced factoring to a problem harder than factoring. :unsure:[/QUOTE]
I don't think so. I've come to appreciate that factoring is a very challenging problem requiring a special set of wits.
Sometimes making something more complicated initially will simplify things drastically..or create a huge intractable swamp. As Laplace once stated, "Read Euler.."

As will be noted through my posts I do have my weaknesses, however, what I have asked should be clear.

jwaltos 2017-09-14 23:43

I should have included this earlier as this is an aspect of testing and implementing certain results obtained relative to the question asked above;
Regarding computation, in addition to mainstream CAS's has anyone worked with ACL2 and/or Vermaseren's FORM?

science_man_88 2017-09-15 00:18

[QUOTE=jwaltos;467756]Soliciting advice on how to solve this equation:

-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c


Assuming a, b, c and d are known, what is the best method to resolve the remaining variables: x, y, such that the resulting number is a specific integer.[/QUOTE]

well the only parts that can change it from even to odd etc. ( assuming all variable are integer) are 19/2 * x^2 , 21/2 * x, 205 * a, 11 * c, 11 * d * y, 205 * b * y, the rest would all set it as even in the integers. so if for example a,b,c,d are all even then it will come down to the ones that contain only x in that list to determine parity. other considerations can be made of course.

CRGreathouse 2017-09-15 13:39

[QUOTE=science_man_88;467811]well the only parts that can change it from even to odd etc. ( assuming all variable are integer) are 19/2 * x^2 , 21/2 * x, 205 * a, 11 * c, 11 * d * y, 205 * b * y, the rest would all set it as even in the integers. so if for example a,b,c,d are all even then it will come down to the ones that contain only x in that list to determine parity. other considerations can be made of course.[/QUOTE]

If x is odd, you can multiply everything (including the hidden constant A that everything equals) by 2 to normalize the terms to integers. If x is even, then everything is already an integer.

CRGreathouse 2017-09-15 13:59

[QUOTE=jwaltos;467810]I should have included this earlier as this is an aspect of testing and implementing certain results obtained relative to the question asked above;
Regarding computation, in addition to mainstream CAS's has anyone worked with ACL2 and/or Vermaseren's FORM?[/QUOTE]

These really aren't the right tools for the sort of investigation you're doing.

You want Sage or MAGMA for state-of-the-art integral or S-integral computations on elliptic curves, or [url=http://johncremona.github.io/ecdata/]John Cremona's tables[/url] if you know the conductor and it's small enough (less than 400,000 at present). Look for integral_points/S_integral_points in Sage and IntegralPoints/SIntegralPoints in MAGMA.

For subexponential algorithms on Diophantine quadratics -- generalized Pell's equation and the like -- PARI/GP is state-of-the-art as far as I know. See the help for the functions quadunit, bnfisintnorm, and qfsolve.

Dr Sardonicus 2017-09-15 14:11

F(x,y,a,b,c,d) = -2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c

1) Does solving F(x,y,a,b,c,d) = N bear directly on factoring N?

2) If not, how does the value of N relate to the number to be factored?

3) Where does this function come from?

Without an answer to his last question -- which has already been asked twice, but not really answered -- it seems pointless to try to proceed. Phrases like "part of an integer factorization toolkit," "derived in my attempts to develop a simple method of factoring integers," "an optimal and very specific entry point to a difficult set of questions," or "Every variable and operation has a very specific interpretation much like equations within general physics where dimensional analysis is applicable" don't really address the question.

As examples of the sort of questions whose answers might actually be informative: Is F(x,y,a,b,c,d) obtained from a function in more variables by assigning values to some of them? If so, what was that function? Was F obtained from a homogeneous function?

jwaltos 2017-09-15 16:22

[QUOTE=CRGreathouse;467829]These really aren't the right tools for the sort of investigation you're doing.

You want Sage or MAGMA for state-of-the-art integral or S-integral computations on elliptic curves, or [URL="http://johncremona.github.io/ecdata/"]John Cremona's tables[/URL] if you know the conductor and it's small enough (less than 400,000 at present). Look for integral_points/S_integral_points in Sage and IntegralPoints/SIntegralPoints in MAGMA.

For subexponential algorithms on Diophantine quadratics -- generalized Pell's equation and the like -- PARI/GP is state-of-the-art as far as I know. See the help for the functions quadunit, bnfisintnorm, and qfsolve.[/QUOTE]



Thank you CR. I have used Sage but not MAGMA (yet). MAGMA has an implementation from Denis Simon which pertains to my question and I have looked at the originating paper. Years ago, John Cremona was gracious enough to answer a question of mine, again allied to the question here, providing guidance for a successful resolution to that query. I also use Pari and agree with your assessment regarding certain state-of-the-art functionality. I am usually thorough researching a topic but I always try to account for blind spots, hence my posting. Again, thank you for your insights and bringing them to my attention.




[QUOTE=Dr Sardonicus;467831]

1) Does solving F(x,y,a,b,c,d) = N bear directly on factoring N?

2) If not, how does the value of N relate to the number to be factored?

3) Where does this function come from?

[/QUOTE]


1. Yes.
2. See above.
3. I invented it.


Sardonicus, I sincerely appreciate the time taken to respond but I would be repeating myself if I were to address your questions fully. Sorry.


Thank you to those who responded to my question as it has helped me to better understand how others view what I am looking at.


Moderators, this thread can be closed anytime now. Thank you.

CRGreathouse 2017-09-15 17:21

[QUOTE=Dr Sardonicus;467831]3) Where does this function come from?

Without an answer to his last question -- which has already been asked twice, but not really answered -- it seems pointless to try to proceed. Phrases like "part of an integer factorization toolkit," "derived in my attempts to develop a simple method of factoring integers," "an optimal and very specific entry point to a difficult set of questions," or "Every variable and operation has a very specific interpretation much like equations within general physics where dimensional analysis is applicable" don't really address the question.

As examples of the sort of questions whose answers might actually be informative: Is F(x,y,a,b,c,d) obtained from a function in more variables by assigning values to some of them? If so, what was that function? Was F obtained from a homogeneous function?[/QUOTE]

[QUOTE=jwaltos;467847]3. I invented it.


Sardonicus, I sincerely appreciate the time taken to respond but I would be repeating myself if I were to address your questions fully.[/QUOTE]

I don't think you've made a [i]bona fide[/i] attempt to answer Dr Sardonicus' question, which is a pity -- it would probably help us answer your question. :down:

CRGreathouse 2017-09-15 17:34

[QUOTE=jwaltos;467843]I also use Pari and agree with your assessment regarding certain state-of-the-art functionality.[/QUOTE]

Note that qfsolve is relatively recent -- added in 2.8.1 -- though Denis SIMON's script from which it was adapted has been around for ages.

R. Gerbicz 2017-09-15 17:53

[QUOTE=CRGreathouse;467855]Note that qfsolve is relatively recent -- added in 2.8.1 -- though Denis SIMON's script from which it was adapted has been around for ages.[/QUOTE]

qfsolve solves a quadratic equation, so not playing here.

CRGreathouse 2017-09-15 18:20

[QUOTE=R. Gerbicz;467856]qfsolve solves a quadratic equation, so not playing here.[/QUOTE]

I mentioned that in my post:

[QUOTE=CRGreathouse;467829]These really aren't the right tools for the sort of investigation you're doing.

You want Sage or MAGMA for state-of-the-art integral or S-integral computations on elliptic curves, or [url=http://johncremona.github.io/ecdata/]John Cremona's tables[/url] if you know the conductor and it's small enough (less than 400,000 at present). Look for integral_points/S_integral_points in Sage and IntegralPoints/SIntegralPoints in MAGMA.

For subexponential algorithms on Diophantine quadratics -- generalized Pell's equation and the like -- PARI/GP is state-of-the-art as far as I know. See the help for the functions quadunit, bnfisintnorm, and qfsolve.[/QUOTE]

But since we're missing all but the final step in jwaltos' process, it seems likely (given the origin of the problem) that there will be quadratics at some point, so I thought it would be useful to mention tools for working with them. And indeed, he says he works with Pari:

[QUOTE=jwaltos;467843]I also use Pari and agree with your assessment regarding certain state-of-the-art functionality.[/QUOTE]

Dr Sardonicus 2017-09-15 18:58

[QUOTE=CRGreathouse;467851]I don't think you've made a [i]bona fide[/i] attempt to answer Dr Sardonicus' question, which is a pity -- it would probably help us answer your question. :down:[/QUOTE]There isn't any "probably" about it. Knowing the genesis of the formula would be indispensable in determining how to equate it to a given integer.

To be fair, it was [b]xilman[/b]'s question first. I have merely been pressing the OP for an answer. Instead, the response has been increasingly smarmy evasions. I therefore conclude this thread is for trolling, and will herewith stop feeding the troll.

CRGreathouse 2017-09-15 21:25

[QUOTE=Dr Sardonicus;467863]There isn't any "probably" about it. Knowing the genesis of the formula would be indispensable in determining how to equate it to a given integer.

To be fair, it was [b]xilman[/b]'s question first. I have merely been pressing the OP for an answer. Instead, the response has been increasingly smarmy evasions. I therefore conclude this thread is for trolling, and will herewith stop feeding the troll.[/QUOTE]

Sorry for the misattribution. Personally I'm less optimistic about the "probably": I can easily imagine knowing a convoluted background which gives little or no insight into how to use it. But I appreciate your open-mindedness/optimism.

jwaltos 2017-09-16 01:20

[QUOTE=Dr Sardonicus;467863]There isn't any "probably" about it. Knowing the genesis of the formula would be indispensable in determining how to equate it to a given integer.

To be fair, it was [b]xilman[/b]'s question first. I have merely been pressing the OP for an answer. Instead, the response has been increasingly smarmy evasions. I therefore conclude this thread is for trolling, and will herewith stop feeding the troll.[/QUOTE]

All right ********. Pretend it's a test and derive the Laplace transform from first principles as anyone with some mathematical background can do. No, I'm not trolling. If you feel like calling my responses smarmy evasions then you are either too ****** and/or too **** to think. *******.

I logged in again to delete the above..but no, this fits.

Because you need spoon feeding, here is a source that may help you but probably won't: [url]https://www.quora.com/Historically-how-and-why-was-the-Laplace-Transform-invented[/url]

ewmayer 2017-09-16 03:21

Please cool it, you two - the S/N ratio of the thread has dropped to near zero, further posts which do not appreciably raise it will be dealt with harshly.

jwaltos 2017-09-16 04:38

Sorry.

jwaltos 2017-09-18 22:31

Due to the sour note in some prior posts within this thread I'm including this avant propos.
If anyone has an issue with how or why I ask questions the way I do and has a need to trash talk, please PM me.
As always, constructive criticism is welcomed.
This is a two-part post.

1.
I had hoped that someone would notice that a scalar multiplier and constant simplified the initial polynomial.

factor((-2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c)*27000+12803);
->(30*x+900*b*y+11+900*a)*(6300*a+6073+6300*b*y+9210*x+27000*d*y-1800*x^2+27000*c)
Second, that treating the cubic as a constant will allow generalization of the polynomial to a general conic equation.

Without reservation, this polynomial will resolve any number N mod 27000 = 12803 as a product of sums. Modular sieves and methods to solve Thue equations
can be used to solve for the variables but these are not deterministic. Please note that this is a starting point.


2.
The following approximation is a curious result that I worked out in the early 90's. I believe the equivalence is dimensionally correct and numerically suggestive.
The relation can be expanded according to the Euler/de Moivre formula. If possible, is there a simple physical experiment that can be performed to meassure this?
(Maxwell noted that resistance has the same units of velocity, "The Scientific Papers of James Clerk Maxwell, Volume 2, p.126")
The following link, [url]http://vixra.org/pdf/1410.0105v1.pdf[/url], "A brief essay on numerology of the mass ratio of proton to electron." is instructive concerning
this genre of inquiry.

F=96 485.3329 s A / mol
h=6.62607004 × 10-34 m2 kg / s
G=6.67408 × 10-11 m3 kg-1 s-2
c=299,792,458 m/s
A=273.15 Celcius (ad hoc)
A~=274.212.. Celcius (ad hoc)

e=2.71828..
Pi=3.1415..

e^(-Pi)=(h*c^2)/(2*G)*F*(A~/273.15)

I'm presuming that this is a system at complete rest. Again, please note that this is a starting point.
"The Remarkable Sine Functions" by A. I. Markushevich is a reference text that I have used for 1. and 2.
"The Beat of a Different Drum: The Life and Science of Richard Feynman" by Jagdish Mehra is a reference for 2.

CRGreathouse 2017-09-19 14:31

[QUOTE=jwaltos;468056]Without reservation, this polynomial will resolve any number N mod 27000 = 12803 as a product of sums.[/QUOTE]

Interesting! What does it give for 14852125529093754492342250817558427131200867635803? (Smaller numbers available on request.)

[QUOTE=jwaltos;468056]Modular sieves and methods to solve Thue equations
can be used to solve for the variables but these are not deterministic.[/QUOTE]

Is the standard Tzanakis--de Weger method non-deterministic? :confused:

[QUOTE=jwaltos;468056]
The following link, [url]http://vixra.org/pdf/1410.0105v1.pdf[/url], "A brief essay on numerology of the mass ratio of proton to electron." is instructive concerning
this genre of inquiry.[/QUOTE]

I see there's a mistake in the table, I guess it hasn't been proofread.

That sort of numerology looks like a job for [url=http://mrob.com/pub/ries/]ries[/url] (in addition to Plouffe's inverter, which is already mentioned).

CODATA has an [url=https://physics.nist.gov/cgi-bin/cuu/Value?mpsme]updated value[/url] for mp/me but nothing is really affected, it's just a tweak to the trailing digits.

jwaltos 2017-09-19 15:56

[QUOTE=CRGreathouse;468099]Interesting! What does it give for 14852125529093754492342250817558427131200867635803? (Smaller numbers available on request.)

Is the standard Tzanakis--de Weger method non-deterministic? :confused:

I see there's a mistake in the table, I guess it hasn't been proofread.

That sort of numerology looks like a job for [url=http://mrob.com/pub/ries/]ries[/url] (in addition to Plouffe's inverter, which is already mentioned).

CODATA has an [url=https://physics.nist.gov/cgi-bin/cuu/Value?mpsme]updated value[/url] for mp/me but nothing is really affected, it's just a tweak to the trailing digits.[/QUOTE]


Tzanakis--de Weger method: I was unaware. Thanks.

Regarding the paper, it's vixra - great for non-peer reviewed works and sometimes helpful. I included it because what I have is pure speculation but it is curious.

I'm sending a PM.

jwaltos 2017-09-20 03:39

[QUOTE=CRGreathouse;468099]
Is the standard Tzanakis--de Weger method non-deterministic? :confused:.[/QUOTE]

I have borrowed this book a few times:

The Algorithmic Resolution of Diophantine Equations: A Computational Cookbook
By Nigel P. Smart

which contained the above method. The above method did not apply or I did not know how to apply it.

I found the following paper which helped. I would not have found it had you not made your comment.

On elliptic Diophantine equations that defy Thue's method: the case of the Ochoa curve
Benjamin M. M. de Weger and Roel J. Stroeker


As usual, attempting to clearly explain one's ideas to others usually helps one's own thought process. In this case, in addition to searching for viable algorithmic processes I will also be looking for identities.

Dr Sardonicus 2017-09-20 14:43

[QUOTE=jwaltos;467886]No, I'm not trolling.[/QUOTE]
OK, I'll take your word for it. Your posted apology for your flamage is accepted. I withdraw my accusation, and apologize for it.

Of course, I had noticed that

F = -2*x^3+(19/2-60*a-60*b*y)*x^2+(21/2+314*a+30*d*y+30*c+314*b*y)*x+2+205*a+11*c+900*b*y*c+420*a*b*y+210*a^2+210*b^2*y^2+11*d*y+900*a*d*y+900*b*y^2*d+205*b*y+900*a*c

was a quadratic in y, but had broken off pursuit of the matter because its discriminant was not a square, so that F had no polynomial factorization into factors which were linear in y, and because, in answer to my first question, you had indicated you were looking for representations by F [i]of the number to be factored[/i].

Be that as it may, the question of finding linear transforms of a quadratic which have linear factors is an intriguing one. And, as it turns out, there is a curious feature of F when viewed as a quadratic in y, which allows one to obtain such a transform. Its discriminant D is a quartic in x, with lead coefficient a perfect square. Thus there is a Laurent series for sqrt(D) beginning with the square root of the lead term. If R is the polynomial truncation of this series ("polynomial square root" of D), D - R^2 is a polynomial in x of degree at most 4/2 - 1 = 1.

But in this case (and this is the curious thing), D - R^2 is a [i]constant[/i],

D - R^2 = 89621/225*b^2 + 25606/15*d*b = 12803/225*b*(7*b+30*d).

Alas, setting this constant equal to 0 (b = 0 or 7*b + 30*d = 0) makes the coefficient of y^2 equal to 0, so does not lead to an algebraic factorization of F.

Trying again with G = p*F + Q, we again find the same curious thing with the discriminant (call it D2) and its "polynomial square root" R2. In this case,

D2 - R2^2 = (89621/225*p^2 - 840*q*p)*b^2 + (25606/15*p^2 - 3600*q*p)*d*b, or

D2 - R2^2 = b*p*(7*b + 30*d)*(12803*p - 27000*q)/225

If we set D2 - R2^2 = 0, D2 = R2^2 is a perfect square.

As with F, setting p = 0 or 7*p + 30*d = 0 makes the coefficient of y^2 equal to 0, so does not lead to an algebraic factorization.

However, setting

(*) 12803*p - 27000*q = 0

does lead to an algebraic factorization of G = p*F + q. One can take p = 27000, q = 12803.

In this case, one could obtain a factorization of N via a representations by F of (N - q)/p (rather than a representation of N by F).

So this is an actual derivation of a linear transform p*F + q of F with an algebraic factorization into polynomials that are linear in y. It is unique in the sense that (*) is the only linear relation between p and q which leads to a factorization by making the discriminant of p*F + q (viewed as a quadratic in y) the square of a polynomial, without making the coefficient of y^2 equal to 0. It hinges on an unusual feature of the discriminant D2 of p*F + q, viewed as a function of y: D2 is a quartic in x whose lead term is a perfect square. Thus, there is a "polynomial square root" R2. And

D2 - R2^2

is a [i]constant[/i], rather than a linear function in x. Setting this constant equal to 0 makes the discriminant equal to the square of its "polynomial square root," which in one case leads to an algebraic factorization.

I'm not sure what the fact that D2 - R2^2 is a constant (rather than linear in x) [i]means[/i], but I am not smart enough to know that it doesn't mean anything.

jwaltos 2017-09-20 22:52

Thanks for attempting to solve this and showing your methods.

One thing which you may not have tried is to find sequential solutions for F, that is,

...-3,-2,-1,0,1,2,3... or ...-3i,-2i,-1i,0,1i,2i,3i...(complex), and others.

You will find trivial factorizations of prime numbers which you can attempt to sequence as a solution subset..etc....

I found that by working with a pencil and paper only, I was able to establish patterns, boundary and initial conditions so that I could encapsulate the little mathematical universe I was working within, Mandelbrot sets et al. Once done I could then use things like cellular automata and other computational tools to model the behaviour of what I was looking at. After enough poking around I could then start to formalize things. This would be like looking at a stream then attempting to describe what you're looking at with Navier-Stokes.

It's messy stuff to the untrained eye but chaotic dynamics exists as a science for a reason ;)

Just added this as an aside: [url]http://www.ams.org/notices/200902/rtx090200226p.pdf[/url]

jwaltos 2017-09-21 01:30

Sardonicus,

Here is something else which is pertinent to my question, the Blissard (Umbral) Calculus and Abrahamson's non-standard analysis. Years ago I came across H. Dorrie's book, "100 Great Problems..." and attempted to solve all the questions. Bernoulli's Power Sum problem was one success and I expanded and generalized this problem with the methods I developed. I later found out that Blissard had developed what I came up with many years earlier. I was very impressed with what the librarian was able to dredge from the archives in university bowels regarding his published works. I prefer his insights over what is presently known as the Umbral Calculus. This methodology is a good tool when nothing else seems to work. The NSA (non-standard analysis) perspective can be useful if you are trying to be clever rigorously.

CRGreathouse 2017-09-21 14:12

Yes, I think everyone has been charmed by nonstandard analysis at one time or another. :smile:


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