Re: trying to parameterize solutions for Pythagorean ratios and Diophantine m-tuples

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

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.