American Citizen on Mon, 04 Mar 2024 07:00:57 +0100 |
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Re: trying to parameterize solutions for Pythagorean ratios and Diophantine m-tuples |
So it is not necessary to use large u,v values for R = uP + vQ to find all the rational points from the basis {P,Q}
Randall On 3/3/24 04:41, Bill Allombert wrote:
On Sat, Mar 02, 2024 at 05:22:15PM -0800, American Citizen wrote:(3) 30*q*p^3 + 144*q^2*p^2 - 30*q^3*p = r^2 for p,q,r in Z This is almost an homogeneous quadratic equation in p,q made a square, but the product of the powers = 3 for each binomial term on the left. Is there a parametric solution for (2) given m,n or (3) ?Divide 3) by q^4 and set x=p/q, y =r/q^2, you get y^2 = 30*x^3 + 144*x^2-30*x then set y = Y/30 x = X/30 So Y^2 = X^3 + 144*X^2 - 900*X which is a rank-2 elliptic curve with full 2-torsion, so there are a lot of solutions. ? E=ellinit([0,144,0,-900,0]) ? elltors(E) %8 = [4,[2,2],[[-150,0],[0,0]]] ? ellrank(E) %9 = [2,2,0,[[-90,720],[45,585]]] For example [-90,720] is [p,q] = [3,-1], and [45,585] is [p,q] = [3, 2] and ellmul(E,[-90,720],5) gives [-1724297003000010/567013106089369,-858234297395033146517040/13501737061684979934653] leads to [57476566766667, -567013106089369] Cheers, Bill.