Re: Mathematica "Reduce" function

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

On Wed, Mar 01, 2017 at 06:05:14PM +0100, Bill Allombert wrote:
> On Wed, Mar 01, 2017 at 05:45:07PM +0100, Pedro Fortuny Ayuso wrote:
> In your example, you can use polresultant:
> 
> ? P=polresultant(x^2 + 3*y^2 -4, 3*x^3 - 4*y^2 + x*y-1)
> %7 = 243*y^6-54*y^5-953*y^4+144*y^3+1300*y^2-96*y-575
> so if [x,y] is a solution, then 
> 243*y^6-54*y^5-953*y^4+144*y^3+1300*y^2-96*y-575 = 0
> then
> ? polrootspadic(P,3,10)
> %2 = []~

... so there are no solutions mod 3^k when k is stricty larger than
the p-adic valuation of the discriminant of P:
? valuation(poldisc(P),3)
%2 = 9

(this is conservative, in this case k>3 is sufficient).

Cheers,
Bill.