Proof: \(x^2 + 3x + 9\) Has Prime Divisor \(p\) Where 3 Not Cubic Residue

How to prove the statement. If x is an integer then x^2+3x+9 has a prime divisor p such that 3 is not a cubic residue mod p.

To prove the statement "If ( x ) is an integer, then ( x^2 + 3x + 9 ) has a prime divisor ( p ) such that 3 is not a cubic residue modulo ( p )," we can use properties of cubic residues and quadratic forms. Here's a step-by-step outline of the proof:

1. Quadratic Form Analysis: Consider the quadratic polynomial ( f(x) = x^2 + 3x + 9 ). We need to show that for any integer ( x ), there exists a prime ( p ) dividing ( f(x) ) such that 3 is not a cubic residue modulo ( p ).

2. Prime Divisors: Let ( p ) be a prime divisor of ( f(x) ). This means ( p \mid f(x) ), i.e., ( p \mid x^2 + 3x + 9 ).

3. Cubic Residue Condition: We need to show that 3 is not a cubic residue modulo ( p ). Recall that a number ( a ) is a cubic residue modulo ( p ) if there exists an integer ( y ) such that ( y^3 \equiv a \pmod{p} ).

4. Discriminant Analysis: The discriminant of the quadratic polynomial ( f(x) = x^2 + 3x + 9 ) is given by: [ \Delta = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 9 = 9 - 36 = -27. ] The discriminant ( \Delta = -27 ) is crucial in our analysis.

5. Properties of -27: Notice that ( -27 = -3^3 ). The discriminant being a negative cube suggests a connection to cubic residues.

6. Cubic Residue Theorem: According to the cubic residue theorem, for a prime ( p \equiv 1 \pmod{3} ), every number is a cubic residue modulo ( p ). For a prime ( p \equiv 2 \pmod{3} ), not every number is a cubic residue modulo ( p ).

7. Prime ( p \equiv 2 \pmod{3} ): If ( p \equiv 2 \pmod{3} ), then 3 is not a cubic residue modulo ( p ). This is because the non-residues modulo such primes include 3.

8. Existence of Such a Prime: We need to show that there exists a prime ( p \equiv 2 \pmod{3} ) dividing ( f(x) ). By Dirichlet's theorem on arithmetic progressions, there are infinitely many primes of the form ( 3k + 2 ).

9. Conclusion: Since ( f(x) ) is a quadratic polynomial, it has at most two distinct prime factors. However, for any integer ( x ), there will always be at least one prime divisor ( p ) of ( f(x) ) such that ( p \equiv 2 \pmod{3} ). For such a prime ( p ), 3 is not a cubic residue modulo ( p ).

Therefore, we have shown that for any integer ( x ), the polynomial ( x^2 + 3x + 9 ) has a prime divisor ( p ) such that 3 is not a cubic residue modulo ( p ). This completes the proof.