Solving Quadratic Congruences: A Closer Look at n^2 ≡ 2000 (mod 2009)

In this article, we will delve into the process of solving the quadratic congruence n^2 ≡ 2000 (mod 2009). We will explore various methods and techniques in order to determine if there exists an integer n that satisfies the given congruence. We will also discuss the role of quadratic residues, Legendre symbols, and Jacobi symbols in this context.

Simplifying the Problem

To begin, let's rewrite the given equation as a more manageable form:

n^2 - 2000 ≡ 0 (mod 2009)

or equivalently,

n^2 ≡ 2000 (mod 2009)

Calculating 2000 mod 2009

Since 2000 is less than 2009, we have:

2000 ≡ 2000 (mod 2009)

Checking for Quadratic Residues

Next, we need to check whether 2000 is a quadratic residue modulo 2009. First, we factorize 2009:

2009 7^2 × 41

We will check if 2000 is a quadratic residue modulo 7 and 41.

Checking modulo 7

First, calculate 2000 mod 7:

u02C52000 div 7 ≈ 285.71 u21D2 2000 7 × 285 - 5 u21D2 2000 ≡ 5 (mod 7)

Now, check if 5 is a quadratic residue modulo 7 by checking the squares of 0, 1, 2, 3, 4, 5, 6:

- 0^2 ≡ 0 (mod 7)

- 1^2 ≡ 1 (mod 7)

- 2^2 ≡ 4 (mod 7)

- 3^2 ≡ 2 (mod 7)

- 4^2 ≡ 2 (mod 7)

- 5^2 ≡ 4 (mod 7)

- 6^2 ≡ 1 (mod 7)

Since 5 does not appear in this list, 5 is not a quadratic residue modulo 7. Therefore, there is no solution for n^2 ≡ 2000 (mod 2009).

Checking modulo 41

To check if 2000 is a quadratic residue modulo 41, we first calculate 2000 mod 41:

u02C52000 div 41 48 u21D2 2000 ≡ 32 (mod 41)

Now, check if 32 is a quadratic residue modulo 41 by checking the squares of 0, 1, 2, ..., 20 (since 20^2 is already the same as 2^2 modulo 41):

- 0^2 ≡ 0 (mod 41)

- 1^2 ≡ 1 (mod 41)

- 2^2 ≡ 4 (mod 41)

- 3^2 ≡ 9 (mod 41)

- 4^2 ≡ 16 (mod 41)

- 5^2 ≡ 25 (mod 41)

- 6^2 ≡ 36 (mod 41)

- 7^2 ≡ 1 (mod 41)

- 8^2 ≡ 4 (mod 41)

- 9^2 ≡ 18 (mod 41)

- 10^2 ≡ 29 (mod 41)

- 11^2 ≡ 34 (mod 41)

- 12^2 ≡ 30 (mod 41)

- 13^2 ≡ 19 (mod 41)

- 14^2 ≡ 9 (mod 41)

- 15^2 ≡ 36 (mod 41)

- 16^2 ≡ 3 (mod 41)

- 17^2 ≡ 40 (mod 41)

- 18^2 ≡ 2 (mod 41)

- 19^2 ≡ 44 ≡ 3 (mod 41)

- 20^2 ≡ 8 (mod 41)

Since 32 does not appear in this list, 32 is not a quadratic residue modulo 41.

Conclusion Using Jacobi Symbol

Noting that 2009 7^2 × 41, if the given congruence has a solution, then n^2 ≡ 2000 (mod 7) must also have a solution. Reducing 2000 modulo 7 results in:

n^2 ≡ 5 (mod 7)

However, the squares modulo 7 are 0, 1, 2, 4. Therefore, we have reached a contradiction and the original congruence has no solution.

Checking Using Legendre and Jacobi Symbols

First, note that 2009 7^2 × 41 is not prime, which makes the problem slightly more complex. We can use the Jacobi symbol to determine if 2000 is a quadratic residue of 2009. The Jacobi symbol follows these rules:

Jacobi Symbol Definition

[left(frac{a}{n}right) left(frac{a}{p_1}right)^{alpha_1} cdots left(frac{a}{p_k}right)^{alpha_k}]

where (n p_1^{alpha_1} cdots p_k^{alpha_k}) for primes (p_i) and positive integers (alpha_i). The Legendre symbol (left(frac{a}{p}right)) is defined as:

[left(frac{a}{p}right) left{ begin{array}{rl} 0 text{if } a ≡ 0 pmod{p} 1 text{if } a ot ≡ 0 pmod{p} text{ and for some integer } m, a ≡ m^2 pmod{p} -1 text{if } a ot ≡ 0 pmod{p} text{ and there is no such } m end{array} right.]

Applying Jacobi Symbol

Applying this to our problem:

[left(frac{2000}{2009}right) left(frac{2000}{7}right)^2 left(frac{2000}{41}right) left(frac{2000}{41}right)]

Since (2000 mod 7 5) and (2000 mod 41 32), we have:

[left(frac{2000}{2009}right) left(frac{5}{7}right)^2 left(frac{32}{41}right)]

Since (left(frac{5}{7}right) -1) and for 2000 to be a quadratic residue of 2009, it must also be a quadratic residue of all of its prime factors, we can conclude that 2000 is not a quadratic residue of 2009.