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

Let's prove that there are infinitely many primes pp such that p3 (mod 4)p \equiv 3~ (mod~4).

Let p1,p2,...pkp_1, p_2, ... p_k be a finite list of primes congruent to 3 mod 4, and construct a new integer

4×(p1,p2,...pk)14 \times (p_1, p_2, ... p_k) -1

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

Note that 4×(p1,p2,...pk)14 \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

1111=1 mod 4,1\cdot 1 \cdot 1 \cdots 1=1 ~mod~ 4,

so 4×(p1,p2,...pk)14 \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×(p1,p2,...pk)14 \times (p_1, p_2, ... p_k) -1 is not divisible by any pip_i, which are primes congruent to 3 mod 4, we have thus found a new prime pk+1=3 mod 4 p_{k+1} = 3~mod~ 4 that is not in our original list p1,p2,...pkp_1, p_2, ... p_k.

\therefore there must be an infinitude of primes that are congruent to 3 mod 43 ~mod ~4.

Another special case is that there exists infinitely many primes pp such that p1 (mod 4)p \equiv 1~(mod~4). This case can be shown to be true in the same way, by constructing an integer (p1p2pt)2+1 (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(k,m)=1, any coprime integers mm and kk, there exists infinitely many primes pp such that pm (mod k)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 Bright Glow
3 years, 2 months ago

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