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

[Laël Cellier <lael.cellier@laposte.net>]

So yes, I want to solve a system of equations where alpha and beta can 
be set to any values as long as they’re valid for solving the first 
equation.

Are you meaning this/your code allows to get more than 2 solutions for 
the second equation ?

Cordialement,

Le 19/01/2025 à 21:39, Bill Allombert a écrit :
> 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.