Given a quadratic polynomial ax^2+bx+c, with a, b and c being integers, is there a characterization of all primes p for which the equation ax^2+bx+c≡0(modp) has solutions?

Bayobusalue

Bayobusalue

Answered question

2022-11-03

Primes modulo which a given quadratic equation has roots
Given a quadratic polynomial a x 2 + b x + c, with a, b and c being integers, is there a characterization of all primes p for which the equation
a x 2 + b x + c 0 ( mod p )
has solutions?

Answer & Explanation

Savanna Smith

Savanna Smith

Beginner2022-11-04Added 17 answers

Step 1
I never noticed this one before.
x 3 x 1 0 ( mod p )
has one root for odd primes p with ( 23 | p ) = 1..
x 3 x 1 0 ( mod p )
has three distinct roots for odd p with ( 23 | p ) = 1 and p = u 2 + 23 v 2 in integers.
x 3 x 1 0 ( mod p )
has no roots for odd p with ( 23 | p ) = 1 and p = 3 u 2 + 2 u v + 8 v 2 in integers (not necessarily positive integers).
Here we go, no roots (mod2), but a doubled root and a single (mod23), as
x 3 x 1 ( x 3 ) ( x 10 ) 2 ( mod 23 ) .
Step 2
Strange but true. Easy to confirm by computer for primes up to 1000, say.
The example you can see completely proved in books, Ireland and Rosen for example, is x 3 2 ,, often with the phrase "the cubic character of 2" and the topic "cubic reciprocity." 2 is a cube for primes p = 2 , 3 and any prime p 2 ( mod 3 ) . Also, 2 is a cube for primes p 1 ( mod 3 ) and p = x 2 + 27 y 2 in integers. However, 2 is not a cube for primes p 1 ( mod 3 ) and p = 4 x 2 + 2 x y + 7 y 2 in integers. (Gauss)
3 is a cube for primes p = 2 , 3 and any prime p 2 ( mod 3 ) . Also, 3 is a cube for primes p 1 ( mod 3 ) and p = x 2 + x y + 61 y 2 in integers. However, 3 is not a cube for primes p 1 ( mod 3 ) and p = 7 x 2 + 3 x y + 9 y 2 in integers. (Jacobi)

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