Prime Numbers that Can Never Divide (n^2 n 2)
While prime numbers are essential in number theory, some interesting questions arise when we consider the divisibility properties of polynomials. For example, let's explore the expression n2 n 2. We will investigate which prime numbers cannot divide this expression for any integer n.
Theoretical Background
This is an elementary exercise that leverages the properties of prime numbers and quadratic reciprocity. Let's break it down:
Understanding the Expression n2 n 2
For any integer n, the expression n2 n 2 takes on various values. Our goal is to find which prime numbers p can never divide n2 n 2.
Quadratic Residues and Disciminants
A polynomial (f(x)) has no roots modulo (p) if its discriminant is not a square modulo (p). For the polynomial (n^2 n 2), the discriminant is 12 - 4 cdot 2 -7. We need to find when the Legendre symbol (left(frac{-7}{p}right)) is (-1).
The Legendre Symbol and Quadratic Reciprocity
The Legendre symbol (left(frac{a}{p}right)) is defined for an odd prime (p) as:
[left(frac{a}{p}right) begin{cases} 1 text{if } a text{ is a quadratic residue modulo } p -1 text{if } a text{ is a non-residue modulo } p 0 text{if } a equiv 0 pmod{p} end{cases}]We can use the Legendre symbol to determine when (-7) is a non-residue modulo (p). Specifically, we have:
[left(frac{-7}{p}right) -1^{(p-1)/2} cdot left(frac{p}{7}right)]By the law of quadratic reciprocity, we get:
[left(frac{p}{7}right) (-1)^{(p-1)(7-1)/4} cdot left(frac{7}{p}right) left(frac{7}{p}right)]Thus, we have:
[left(frac{-7}{p}right) -1^{(p-1)/2} cdot left(frac{7}{p}right)]Quadratic Residues Modulo 7
The quadratic residues modulo 7 are 0, 1, 2, 4. Therefore, (left(frac{7}{p}right) 1) if (p equiv 1, 2, 4 pmod{7}) and (left(frac{7}{p}right) -1) if (p equiv 3, 5, 6 pmod{7}).
To find when (left(frac{-7}{p}right) -1), we consider the following cases:
If (left(frac{7}{p}right) 1), then (-1^{(p-1)/2} cdot 1 -1) when ((p-1)/2) is odd, i.e., (p equiv 3, 5, 7k 3 pmod{14}). If (left(frac{7}{p}right) -1), then (-1^{(p-1)/2} cdot (-1) -1) when ((p-1)/2) is even, i.e., (p equiv 1, 11, 7k 1 pmod{14}).This gives us the prime numbers that do not divide (n^2 n 2).
Computational Verification
Using a computer program to check all primes up to 100, we found the following prime numbers that do not divide (n^2 n 2): 3, 5, 13, 17, 19, 31, 41, 47, 59, 61, and 73, 83, 89, 97.
Concluding Remarks
While there is no known simple formula to determine if a prime number will divide a given polynomial for all integers, computational methods and number theory tools like the Legendre symbol and quadratic reciprocity can provide valuable insights.