an²+bn+c = 0 mod p

[bhelmes]

Let f(n)=an²+bn+c = 0 mod p
with n element of N and p prime.
n=?

Is there a better way than calculating f(n) for n=0 ...p-1

Thanks if you spend me some lines.

[Batalov]

Solve it like you were taught in elementary school, only remember that operations are mod p.
Instead of "there is no square root for negative numbers -> there is no solution" you will use
"D is not a quadratic character -> there is no solution".

[Nick]

Originally Posted by bhelmes

Is there a better way than calculating f(n) for n=0 ...p-1

Yes, in this subforum already!
https://www.mersenneforum.org/showthread.php?t=22058

[0scar]

Originally Posted by Nick

Yes, in this subforum already!
https://www.mersenneforum.org/showthread.php?t=22058

Great explanation from the given subforum.
About efficiently computing square roots modulo a large prime (hundreds of digits or more), there are two good algorithms by Tonelli-Shanks and by Cipolla.

[bhelmes]

Originally Posted by Nick

Yes, in this subforum already!
https://www.mersenneforum.org/showthread.php?t=22058

I did not understand how to find the square of the discriminant b²-4ac mod 4an.
b²-4ac is a quadratic residue, but not always a square.

[Nick]

Originally Posted by bhelmes

I did not understand how to find the square of the discriminant b²-4ac mod 4an.

See section 2.3.2 in the 2nd edition of the book by Crandall and Pomerance:
"Prime Numbers , a computational perspective"

[bhelmes]

Let an²+bn+c = 0 mod p

If I want to calculate the discriminant,
can I first calculate a*=a mod p, b*=b mod p and c*=c mod p
and then discr=(b*)²-4a*c* ?

[Nick]

Originally Posted by bhelmes

Let an²+bn+c = 0 mod p

If I want to calculate the discriminant,
can I first calculate a*=a mod p, b*=b mod p and c*=c mod p
and then discr=(b*)²-4a*c* ?

Yes (assuming you mean discr=(b*)²-4a*c* mod p).
That is what we mean by Propostion 22 here

[bhelmes]

@Nick, thanks a lot for your work and help.

an useful implementation in c and gmp of tonelli-shanks algorithm is:
http://www.codecodex.com/wiki/Shanks-Tonelli_algorithm


Let : 2ax+b≡y(mod4an)
I did not understand, how you solve it. gcd (2a,4an)=2a>1 so the inverse does not exist.

Dividing by 4a seems to be possible x/2=(y-b)*(4a)⁻¹ mod (n) with a=/=0
multiplying with 2 is x =(y-b)*(2a)⁻¹ mod n

Is that ok ?

[Nick]

We have \(2ax+b-y\equiv 0\pmod{4an}\) and \(\gcd(2a,4an)=2a\neq 1\).
If 2a does not divide b-y then there is no solution.
If instead b-y=2at for some integer t then \(x\equiv -t\pmod{2n}\).