Re: Is it possible to have several solutions in this way to this equation using Pari/ɢᴘ ?

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

On Sun, Jan 19, 2025 at 08:05:09PM +0100, Laël Cellier wrote:
> first,
> beta=-(V\W);
> alpha=W*(V+W*beta);
> 
> is just 1 way to find a suitable solution alpha == w (v + w beta). I’m
> needing to find other ways in order to get different results.
> 
> Let’s give a numerical example :
> V=25 c=60 W=3 b=85 f=-1
> give :
> alpha=3
> beta=-8
> nfr=[-4, -1/9]~

So you want this system of equation to have solutions:

alpha^2*x^2+(2*alpha*beta-f*b)*x+(beta^2-c) = 0
alpha=W*(V+W*beta);

you can eliminate alpha and obtain

Q = subst(P,alpha,W*(V+W*beta))
Q=(W^4*beta^2+2*V*W^3*beta+V^2*W^2)*x^2+(2*W^2*beta^2+2*V*W*beta-b*f)*x+(beta^2-c)
To have rational solution the discriminant of Q needs to be a square
? poldisc(Q)
%3 = (-4*W^2*b*f+4*W^4*c)*beta^2+(-4*V*W*b*f+8*V*W^3*c)*beta+(b^2*f^2+4*V^2*W^2*c)

so we need to solve for (beta, D)

D^2= (-4*W^2*b*f+4*W^4*c)*beta^2+(-4*V*W*b*f+8*V*W^3*c)*beta+(b^2*f^2+4*V^2*W^2*c)

This is in general a conic which can be solved with qfsolve/qfparam
(or hyperellratpoints).

fun(V,W,b,c,f)=
{
  my(A= (-4*W^2*b*f+4*W^4*c), B = (-4*V*W*b*f+8*V*W^3*c)/2, C = b^2*f^2+4*V^2*W^2*c);
  my(M=[A, B, 0; B, C, 0; 0, 0, -1]);
  my(S=qfsolve(M));
  my(beta =S[1]/S[2]);
  my(alpha= W*(V+W*beta));
  my(X=nfroots(,alpha^2*x^2+(2*alpha*beta-f*b)*x+(beta^2-c)));
  [alpha,beta,X];
}

However in your example " V=25 c=60 W=3 b=85 f=-1 ",
the conic degenerates to
D^2 = 25*(30*beta+233)^2 
so you can pick whatever value you want for beta...
Maybe this is the actual mistery ?

Cheers,
Bill.