# [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
3 months, 1 week ago

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