Re: How to find a solution to this equation so the result is a perfect square ?

[Denis Simon <denis.simon@unicaen.fr>]

Hi,
Let a = 53919893334301279589334030174040979997534463873026188556492192357233
and N = 20422354090808007360481140437163150058277495681745577657420405131241

Let Z^2 be the final square.
Basicly, what you want to solve is
Y^2+a*X=N*Z^2
which is equivalent to
Y^2-N*Z^2 = 0 mod a.
Hence, what you need is to find a square root of N modulo a.
This task is difficult and usually requires the factorisation of a.

If s denotes such a square root, then you can choose Z randomly and set
Y = lift(Mod(s*Z,a));
X = (Y^2-N*Z^2)/a;

Denis SIMON.


----- Mail original -----
> De: "Laël Cellier" <lael.cellier@laposte.net>
> À: "pari-users" <pari-users@pari.math.u-bordeaux.fr>
> Envoyé: Jeudi 13 Février 2025 11:33:24
> Objet: How to find a solution to this equation so the result is a perfect square ?

> Bonjour,
> simple question, I’ve the following expression : (y² +
> X×53919893334301279589334030174040979997534463873026188556492192357233)÷20422354090808007360481140437163150058277495681745577657420405131241
> for example,
> X=102111770454040036802405702185845298473875351094786902380116798296 and
> y=21 leads to
> 269599466671506397946670150870282914247971805150249613489807500649 which
> is the perfect square of 519229685853482762853049632922093
> 
> I want to find 1 or more set of integer x and y such as the end result
> is a perfect square. But how to do it if the divisor is different than
> 20422354090808007360481140437163150058277495681745577657420405131241 ?
> Cordialement,