Dirichlet's theorem for modulus eight: Difference between revisions

Revision as of 20:56, 7 May 2009

Statement

For any odd natural number $a$ , there are infinitely many primes $p$ such that:

$p \equiv a (\mod 8)$ .

Facts used

Congruence condition for two to be a quadratic residue
Congruence condition for minus one to be a quadratic residue
Nonconstant polynomial with integer coefficients and nonzero constant term takes infinitely many pairwise relatively prime values

Proof

We need to consider four congruence classes modulo $8$ : $- 3, - 1, 1, 3$ . We do these case by case.

Congruence class of $- 1$ modulo $8$

By fact (1), $2$ is a quadratic residue modulo $p$ if and only if $p \equiv \pm 1 (\mod 8)$ . In particular, a prime $p$ can divide $n^{2} - 2$ for some natural number $n$ if and only if $p \equiv \pm 1 (\mod 8)$ .

Consider now the polynomial $f (x) = (2 x + 1)^{2} - 2 = 4 x^{2} + 4 x - 1$ . For any natural number $n$ , all prime divisors of $f (n)$ are congruent to $\pm 1 (\mod 8)$ . However, $f (n)$ itself is $- 1$ modulo $8$ , so $f (n)$ must have at least one prime divisor that is $- 1$ modulo $8$ . By fact (3), there are infinitely many pairwise relatively prime values of $f (n)$ , yielding infinitely many distinct primes that are $- 1$ modulo $8$ .

Congruence class of $3$ modulo $8$

By facts (1) and (2), we can deduce that $- 2$ is a quadratic residue modulo $p$ if and only if $p \equiv 1, 3 (\mod 8)$ . In particular, a prime $p$ can divide $n^{2} + 2$ for some natural number $n$ if and only if $p \equiv \pm 1 (\mod 8)$ .

Consider now the polynomial $f (x) = (2 x + 1)^{2} + 2 = 4 x^{2} + 4 x + 1$ . For any natural number $n$ , all prime divisors of $f (n)$ are congruent to $1, 3 (\mod 8)$ . However, $f (n)$ itself is $3$ modulo $8$ , so $f (n)$ must have at least one prime divisor that is $3$ modulo $8$ . By fact (3), there are infinitely many pairwise relatively prime values of $f (n)$ , yielding infinitely many distinct primes that are $3$ modulo $8$ .

@@ Line 17: / Line 17: @@
 ===Congruence class of <math>-1</math> modulo <math>8</math>===
-By fact (2), <math>2</math> is a quadratic residue modulo <math>p</math> if and only if <math>p \equiv \pm 1 \pmod 8</math>. In particular, a prime <math>p</math> can divide <math>n^2 - 2</math> for some natural number <math>n</math> if and only if <math>p \equiv \pm 1 \pmod 8</math>.
+By fact (1), <math>2</math> is a quadratic residue modulo <math>p</math> if and only if <math>p \equiv \pm 1 \pmod 8</math>. In particular, a prime <math>p</math> can divide <math>n^2 - 2</math> for some natural number <math>n</math> if and only if <math>p \equiv \pm 1 \pmod 8</math>.
 Consider now the polynomial <math>f(x) = (2x + 1)^2 - 2 = 4x^2 + 4x - 1</math>. For any natural number <math>n</math>, all prime divisors of <math>f(n)</math> are congruent to <math>\pm 1 \pmod 8</math>. However, <math>f(n)</math> itself is <math>-1</math> modulo <math>8</math>, so <math>f(n)</math> must have at least one prime divisor that is <math>-1</math> modulo <math>8</math>. By fact (3), there are infinitely many pairwise relatively prime values of <math>f(n)</math>, yielding infinitely many distinct primes that are <math>-1</math> modulo <math>8</math>.