20170914, 04:56  #1 
Apr 2012
Brady
384_{10} Posts 
Is this solvable in general.
Soliciting advice on how to solve this equation:
2*x^3+(19/260*a60*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. 
20170914, 06:27  #2  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
24624_{8} Posts 
Quote:
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 ... 

20170914, 07:09  #3  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·11·139 Posts 
Quote:
f(x,y) = a*x^3 + (b+c*y)*x^2 + (d+e*y)*x + f*y^2 + g*y + h Last fiddled with by retina on 20170914 at 07:10 

20170914, 07:29  #4  
Sep 2006
Brussels, Belgium
3×7×79 Posts 
You both forgot the last part of the question : "such that the resulting number is a specific integer."
Rephrasing a bit Quote:


20170914, 08:21  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·11·139 Posts 
It looks not to dissimilar from an elliptic curve.

20170914, 12:18  #6 
Apr 2012
Brady
2^{7}·3 Posts 
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+(19/260*a60*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... Last fiddled with by jwaltos on 20170914 at 12:47 Reason: clarification 
20170914, 14:18  #7  
Feb 2017
Nowhere
4,457 Posts 
Quote:
I'll call it F(a,b,c,d,x,y) Knowing what kind of critter this is, would be of considerable help in addressing the question of solving F(a,b,c,d,x,y) = N 

20170914, 16:17  #8 
Aug 2006
5985_{10} Posts 
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.

20170914, 18:02  #9  
Apr 2012
Brady
2^{7}·3 Posts 
Quote:
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. Last fiddled with by jwaltos on 20170914 at 18:14 

20170914, 20:30  #10  
Aug 2006
3^{2}×5×7×19 Posts 
Quote:


20170914, 20:42  #11  
"Robert Gerbicz"
Oct 2005
Hungary
31·47 Posts 
Quote:
Code:
min x+y subject to x*y=n x>=0 y>=0 x,y is integer 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? 

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