20090916, 13:36  #1 
Jan 2005
Transdniestr
503 Posts 
Diophantine Question
I have a 4th degree polynomial F(k) and I'm looking for a algorithm/heuristic to find solutions of the form: f(k) = r^2 where k, r, and F(x)'s coefficients are all integers.
(I'm looking for something better than setting r to particular values and solving the resulting quartic) I actually have several such similar polynomials (call them F(i)(k)) and my goal is to find k's such that F(i)(k) = r(i)^2 for several of these polynomials (again where all the variables/coefficients are all integers). My goal would be to find an x which solved several of these relations. Background: I'm trying to create an a.p. of 6 or more terms. (http://www.primepuzzles.net/puzzles/puzz_413.htm) x=n/d y=(n+k)/(d+k) When, a = n*(n+k)*(k+2*d) b = d*(d+k)*(2*n+k) a*b is a number such that ax+b/x, a+b and ay+b/y form an arithmetic progression of three terms. For a given x=n/d, I'm trying to find rational solutions for z(v) where z(v)a + b/z(v) = v(xa + b/x) for v = 2, 3, and 4 v = 2, 2, and 3 v = 3, 2, and 2 OR v= 4, 3, and 2 The quartic polynomials F(v)(k) evaluating to the square of an integer allows for rational solutions to z(v). Unfortunately, these polynomials are rather gnarly. For instance, F(2)(k) works out to be: (4*n^6  4*d*n^5  3*d^2*n^4 + 6*d^3*n^3  4*d^4*n^2)*k^4 + (8*n^7 + 8*d*n^6  26*d^2*n^5 + 10*d^3*n^4 + 6*d^4*n^3  12*d^5*n^2)*k^3 + (4*n^8 + 28*d*n^7  23*d^2*n^6  32*d^3*n^5 + 42*d^4*n^4  24*d^5*n^3  8*d^6*n^2)*k^2 + (16*d*n^8+ 16*d^2*n^7  52*d^3*n^6 + 20*d^4*n^5 + 12*d^5*n^4  24*d^6*n^3)*k + (16*d^2*n^8  16*d^3*n^7  12*d^4*n^6 + 24*d^5*n^5  16*d^6*n^4) At present, I'm just trying different k up to a threshold for each n/d and have found numerous 5 term sequences. (Actually, I found that simplifying the problem to use y=(n+k)/(d+k) resulted in finding many more solutions than when y some totally random rational < x). =============================================== Any pointers would be appreciated. Thanks 
20090916, 15:12  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:
are known. r^2 = F(k) is an elliptic (or hyperElliptic curve). While methods are known for finding integer points, they are generally adhoc. One general method is to find the Heegner points, but of course there is no general method for doing that either. Finding integer points on elliptic curves is a very very very DEEP subject. And of course, there will only be finitely many. There may be none if the rank of the curve is 0. 

20090916, 15:53  #3 
(loop (#_fork))
Feb 2006
Cambridge, England
3·5·7·61 Posts 
Yes, finding points on elliptic curves is a deep subject; it's just about practical for smallish curves, but the algorithms are real pigs to implement and you need lots of them. If you can afford $400, buy a copy of the Magma computer algebra system; for small examples you can use
http://magma.maths.usyd.edu.au/calc/ and say for example E:=EllipticCurve([0,0,1,7,6]); Rank(E); IntegralPoints(E); I can't quite remember the sequence of manipulations required to convert a quartic polynomial and a single rational point into an EllipticCurve and a map back to the y^2=quartic model. 
20090916, 16:51  #4 
Jan 2005
Transdniestr
503 Posts 
Thank you both for your feedback.
I wonder if a better tack would be to check if one (or more) of these polynomials (in three variables: n,d and k) can be factored into two smaller polynomials say g(n,d,k) and h(n,d,k). By setting g(n,d,k) = h(n,d,k), perhaps I make some progress simplifying it(say by using substitution). Is there a tool (preferably free or online) that can check if a multivariable polynomial is irreducible. I have PARI installed but it can only check/factor singlevariable polynomials. Could Magma handle something like that? I tried Wolfram Alpha but it exceeded the upper bound on query length. 
20090916, 17:00  #5 
Nov 2003
2^{2}·5·373 Posts 

20090916, 17:24  #6  
(loop (#_fork))
Feb 2006
Cambridge, England
1905_{16} Posts 
Quote:
Code:
P<n,d,k>:=PolynomialRing(Rationals(),3); F:=(4*n^6  4*d*n^5  3*d^2*n^4 + 6*d^3*n^3  4*d^4*n^2)*k^4 + (8*n^7 + 8*d*n^6  26*d^2*n^5 + 10*d^3*n^4 + 6*d^4*n^3  12*d^5*n^2)*k^3 + (4*n^8 + 28*d*n^7  23*d^2*n^6  32*d^3*n^5 + 42*d^4*n^4  24*d^5*n^3  8*d^6*n^2)*k^2 + (16*d*n^8+ 16*d^2*n^7  52*d^3*n^6 + 20*d^4*n^5 + 12*d^5*n^4  24*d^6*n^3)*k + (16*d^2*n^8  16*d^3*n^7  12*d^4*n^6 + 24*d^5*n^5  16*d^6*n^4); Factorisation(F); Code:
[ <d + 1/2*k, 1>, <n, 2>, <n + k, 1>, <n^5*d + 1/2*n^5*k  n^4*d^2 + 1/2*n^4*d*k + 1/2*n^4*k^2  3/4*n^3*d^3  15/8*n^3*d^2*k  1/2*n^3*d*k^2 + 3/2*n^2*d^4 + 5/4*n^2*d^3*k  3/8*n^2*d^2*k^2  n*d^5  1/4*n*d^4*k + 3/4*n*d^3*k^2  1/2*d^5*k  1/2*d^4*k^2, 1> ] Last fiddled with by fivemack on 20090916 at 17:25 

20090916, 21:04  #7  
Feb 2005
2^{2}·3^{2}·7 Posts 
Quote:
If is such that for some of its divisors: , we have for then for form a sequence of squares whose second differences equal the constant . For example, gives a sequence of squares whose second differences equal . Finding sequences of squares with constant second differences is a rather hard task (see the attached paper) and additional requirement of having difference of the special form makes it even harder. Last fiddled with by maxal on 20090916 at 21:24 

20090916, 22:38  #8 
Jan 2005
Transdniestr
503 Posts 
Thank you maxal. Very interesting paper.

20090917, 15:32  #9  
Feb 2005
2^{2}×3^{2}×7 Posts 
Quote:
While the sequence also has the second differences equal , it is a trivial and uninteresting sequence of this kind. Last fiddled with by maxal on 20090917 at 15:35 

20090920, 16:31  #10 
Jan 2005
Transdniestr
503 Posts 
Actually ther\ latter relation is precisely for what I'm looking to find solutions (for i=1,2 ... k where k>=6 and all the variables involved are integers). And, there's no need to square both sides of the equation since both sides of the equation must be positive.
I understand if it's not considered interesting by you. But, it's not trivial to calculate, is it? Sorry, if I misunderstood what you meant. Thanks. Last fiddled with by grandpascorpion on 20090920 at 16:50 
20090922, 19:32  #11 
Feb 2005
2^{2}·3^{2}·7 Posts 
I was discussing connection to the problem of finding sequence of squares with a constant second difference. A trivial solution to this problem is given by squares of the terms of an arithmetic progression. A nontrivial (and hardtofind) solution is a sequence of squares whose bases do not form an arithmetic progression.
In this respect, the sequence of squares is trivial while the sequence is nontrivial. Last fiddled with by maxal on 20090922 at 19:34 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Basic Number Theory 13: Pythagorean triples and a few Diophantine equations  Nick  Number Theory Discussion Group  2  20161218 14:49 
Diophantine Equation  flouran  Math  7  20091212 18:48 
Simple Diophantine equation  Batalov  Puzzles  3  20090410 11:32 
Diophantine problem  philmoore  Puzzles  8  20080519 14:17 