| hermann on Mon, 27 Apr 2026 16:52:52 +0200 |
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| how to best determine a^2+b^2=-1 (mod p) for odd prime p? |
sosp1(p)={
[M,V]=halfgcd(lift(sqrt(Mod(-1,p))),p);
return([V[2],M[2, 1]]);
}
Today I heard in a seminar about existence of a^2+b^2=-1 (mod p) for any
odd prime p.
Asking gemini it said what I implemented, but I used sqrt(Mod(-1-x^2,p))
for
Mod(-1-x^2,p) that is a quadratic residue instead of "Tonelli Shanks"
(perhaps
that is used in GP sqrt?).
sosm1(p)={
while(1,x=random(p);w=Mod(-1-x^2,p);if(kronecker(lift(w),p)==1,
return([x,lift(sqrt(w))])));
}
Works, and since half of numbers generated by random(p) are quadratic
residues of p,
expected number of loop executions is 2. Question is, whether there is a GP builtin function to compute a,b? I did not find some in GP doc. Regards, Hermann.