[Number Theory] Proving special cases of Dirichlet's Theorem

Let's prove that there are infinitely many primes \(p\) such that \(p \equiv 3~ (mod~4)\).

Let \(p_1, p_2, ... p_k\) be a finite list of primes congruent to 3 mod 4, and construct a new integer

\[4 \times (p_1, p_2, ... p_k) -1 \]

that is 3 mod 4, and not divisible by any of the \(p_i\)'s, \(( 0<i \leq k)\).

Note that \(4 \times (p_1, p_2, ... p_k) -1 \) is an odd integer, so it can only have prime factors that are 1 or 3 mod 4. Let's suppose that all its prime factors were 1 and 4. Then

\[1\cdot 1 \cdot 1 \cdots 1=1 ~mod~ 4,\]

so \(4 \times (p_1, p_2, ... p_k) -1 \) would also be 1 mod 4, but this is a contradiction, because our integer is 3 mod 4 by construction. So there must be at least one prime factor that is 3 mod 4. But since \(4 \times (p_1, p_2, ... p_k) -1 \) is not divisible by any \(p_i\), which are primes congruent to 3 mod 4, we have thus found a new prime \( p_{k+1} = 3~mod~ 4\) that is not in our original list \(p_1, p_2, ... p_k\).

\(\therefore\) there must be an infinitude of primes that are congruent to \(3 ~mod ~4\).

Another special case is that there exists infinitely many primes \(p\) such that \(p \equiv 1~(mod~4)\). This case can be shown to be true in the same way, by constructing an integer \( (p_1p_2 \cdots p_t)^2 +1\), and following the same procedure above.

In fact, Dirichlet's theorem states that for \((k,m)=1\), any coprime integers \(m\) and \(k\), there exists infinitely many primes \(p\) such that \(p \equiv m ~(mod ~k)\). His own proof, consisting of Dirichlet L-functions, is often considered to have contributed to the origins of Analytic Number Theory.

Note by Tasha Kim
1 week, 2 days ago

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