How to prove there is infinite prime numbers of form $5n+3$ without Dirichlet theorem?

Question

Is there a nice elementary way to prove there is infinite prime numbers of form $5n+3$ (also for $5n+2$) with $n\in \mathbb{N}$?

I know how to do it for primes of form $pn+1$ for any prime $p\geq 3$ but not in this case.

Answer 1

Here is an elementary proof of a partial result described in comments - there are infinitely many primes of form $5k+2$ or $5k+3$.

For the sake of contradiction, we assume there are only finitely many primes $p_1,p_2,\dots,p_n$ of form $5k+2$ or $5k+3$ with $k \geq 0$. Then consider $$N=5\cdot p_2\cdots p_n+2.$$

($p_1=2$ is not included in the product). Then $N$ is not divisible by any of $2,3$ or $5$. Now if all factors of $N$ were of form $5k+1$ or $5k+4$, then $N$ would be of this form too, impossible. So there is a prime $q \mid N$ such that $q=5k+2$ or $q=5k+3$, but $q \neq 2,3$, and $q$ cannot be any of $p_i$, a contradiction.

This implies that primes of at least one of the two forms occur infinitely often.

Answer 2

$\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}$Okay, let's write out the Dirichlet proof for $N=5$ stripped down to as little analysis as I can. Define four periodic functions $\chi_r(n)$ by the following table: $$\begin{array}{|c|cccc|} \hline n \bmod 5= & 0&1&2&3&4 \\ \hline \chi_0(n)= &0&1&1&1&1 \\ \chi_1(n)= &0&1&i&-i&-1 \\ \chi_2(n)= &0&1&-1&-1&1 \\ \chi_3(n)= &0&1&-i&i&-1 \\ \hline \end{array}$$ Define $L(s, \chi_r) := \sum_{n=1}^{\infty} \tfrac{\chi_r(n)}{n^s}$ for $s>1$.

We now consider the behavior of the $L(s, \chi_r)$ as $s \to 1^+$. When $r=0$, we have $$L(s, \chi_0) = \sum_{m=0}^{\infty} \left( \tfrac{1}{(5m+1)^s} + \tfrac{1}{(5m+2)^s} + \tfrac{1}{(5m+3)^s} + \tfrac{1}{(5m+4)^s} \right) > \sum_{m=0}^{\infty} \frac{4}{5} \int_{5m+1}^{5m+6} \frac{dx}{x^s} = \frac{4/5}{s-1}.$$ Thus, $\lim_{s \to 1^+} L(\chi_0, s) = \infty$.

On the other hand, I claim that $\lim_{s \to 1^+} L(\chi_r, s)$ is a finite, nonzero, complex number for $r=1$, $2$, $3$. Indeed, we have $$\mathrm{Re}\ L(s, \chi_{1 \ \mbox{or} \ 3}) = \sum_{m=0}^{\infty} \left( \tfrac{1}{(5m+1)^s} - \tfrac{1}{(5m+4)^s} \right) \geq \sum_{m=0}^{\infty} \left( \tfrac{1}{5m+1} - \tfrac{1}{5m+4} \right)>0 \ \mbox{and}$$ $$L(s, \chi_2) = \sum_{m=0}^{\infty} \left( \tfrac{1}{(5m+1)^s} - \tfrac{1}{(5m+2)^s} - \tfrac{1}{(5m+3)^s} +\tfrac{1}{(5m+4)^s} \right)>\sum_{m=0}^{\infty} \left( \tfrac{1}{5m+1} - \tfrac{1}{5m+2} - \tfrac{1}{5m+3} +\tfrac{1}{5m+4} \right)>0.$$

Therfore, as $s \to 1^+$, the quantity $\log L(s, \chi_0)$ goes to $\infty$, but $\log L(s, \chi_r)$ stays bounded for $r=1$, $2$, $3$.

Now, note that each of the $\chi_r$ is multiplicative, so we can factor $$L(s, \chi_r) = \prod_p \left( 1+ \frac{\chi_r(p)}{p^s} + \frac{\chi_r(p)^2}{p^{2s}} + \cdots \right) = \prod_p (1-\chi_r(p) p^{-s})^{-1}.$$ Taking logs of both sides, we have $$\log L(s, \chi_r) = \sum_p \log (1-\chi_r(p) p^{-s})^{-1} = \sum_p \sum_{j=1}^{\infty} \frac{\chi_r(p)^j}{j p^{js}}.$$ The last sum is absolutely convergent for $s>1$, so we may reorder it however we like. The contribution of the terms with $j \geq 2$ is bounded as $s \to 1^+$. We deduce that, as $s \to 1^+$, the sum $\sum_p \tfrac{\chi_0(p)}{p^s}$ goes to $\infty$, whereas the sums $\sum_p \tfrac{\chi_r(p)}{p^s}$ for $r=1$, $2$, $3$ stay bounded.

Now, let $a$ and $b$ be integers with $a \not\equiv 0 \bmod 5$. We can check by hand that $$\frac{1}{4} \sum_{r=0}^3 \chi_r(a)^{-1} \chi_r(b) = \begin{cases} 1 & a \equiv b \bmod 5 \\ 0 & a \not \equiv b \bmod 5 \end{cases}.$$ Thus, $$\frac{1}{4} \sum_{r=0}^{3} \chi_r(a)^{-1} \sum_p \tfrac{\chi_r(p)}{p^s} = \sum_{p \equiv a \bmod 5} \frac{1}{p^s}.$$

Combining this with the results above, we see that $$ \sum_{p \equiv a \bmod 5} \frac{1}{p^s} \to \infty \ \mbox{as} \ s \to 1^+.$$

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To generalize this argument to other $N$, you need to

• Construct $\phi(N)$ many multiplicative functions $\chi_r : \ZZ \to \CC$ which are periodic mod $N$ and $0$ on the residue classes not relatively prime to $N$.

• Check the identity $\tfrac{1}{\phi(N)} \sum_r \chi_r(a)^{-1} \chi_r(b) = 1$.

• Show that $L(1, \chi_r) \neq 0$.

Once again, for any particular $N$, this is a finite computation.

Finally, a suggestion. If I were presenting this argument to the students, I would just use $N=4$ and the table $$\begin{array}{|c|cccc|} \hline n \bmod 4 = & 0& 1& 2 & 3 \\ \hline \chi_0(n) = & 0&1&0&1 \\ \chi_1(n) = & 0&1&0&-1 \\ \hline \end{array}.$$ Then all the computations are smaller, and the verification that $L(1, \chi_1) \neq 0$ is the familiar sum $$\frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots = \frac{\pi}{4}.$$ Then you could display the character tables for other values of $N$ at the end of the talk and challenge them to fill in the details.