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Old 2009-09-16, 13:36   #1
grandpascorpion
 
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Jan 2005
Transdniestr

503 Posts
Default 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).

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Any pointers would be appreciated. Thanks
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