# Show that there are infinitely many primes congruent to 1 or 4 or to 2 or 3 modulo 5.

We are given the following Dirichlet character: $\epsilon (n) = \begin{cases} 1 & \text{if } n\equiv 1,4 \pmod 5,\\ -1 & \text{if } n\equiv 2,3 \pmod 5. \end{cases}$

It is known that $\epsilon$ is multiplicative, i.e., $\epsilon(nm) = \epsilon(n)\epsilon(m)$ for any pair of relatively-prime integers $n$ and $m$.

I'm trying to define an appropriate $L$-series to show that there are infinitely many primes congruent to 1 or 4 modulo 5, and that there are infinitely many primes congruent to 2 or 3 modulo 5.

The $L$-series which I defined is $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4} + 1 + \cdots$, but I'm stuck at proving with it.

• – Aravind Oct 6 '15 at 11:10
• I'm not sure that it's the same question. Similar but different. – wannabe programmer Oct 6 '15 at 11:18

My answer comes from Section 8.4 of An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991 by I. Niven, H. S. Zuckerman, H. L. Montgomery.

First, we show that that there are infinitely many primes by using properties of the Riemann zeta function. We do that to provide a model and to use some of the findings for the answer to your question.

By combining the formula for the logarithm of the zeta function with an estimate for the zeta function, we find that $$\sum_{n=2}^\infty \frac{\Lambda(n)}{\log n} \frac1{n^s} = \log \frac1{s - 1} + O(s - 1)$$ uniformly for $1 < s \le 2$ where $\Lambda(n)$ is the von Mangoldt function, equal to $\log p$ if $n = p^k$ where $p$ is a prime and $k \ge 1$, or $0$ otherwise. In that sum, the contribution made by the higher powers of the primes is \begin{align} \sum_p \sum_{k=2}^\infty \frac1k \frac1{(p^k)^s} & \le \sum_p \sum_{k=2}^\infty \frac1{p^{ks}}\\ & = \sum_p \frac1{p^{2s} (1 - 1/p)} &&\text{(sum of a geometric series)}\\ & \le \sum_p \frac1{p (p - 1)}\\ & = \text{some constant} \tag1 \end{align} uniformly for $s \ge 1$. Hence $$\sum_p \frac1{p^s} = \log \frac1{s - 1} + O(1) \tag2$$ for $s > 1$. If there were only finitely many primes, then the sum on the left would tend to a finite limit as $s$ tends to $1$ from above. But that contradicts that the right side tends to infinity as $s$ tends to $1$ from above. Hence, there are infinitely many primes.

Along with $\epsilon$, we need the principal Dirichlet character $$\epsilon_0(n) = \begin{cases} 1 & \text{if n \equiv 1, 2, 3 or 4 \pmod 5, or}\\ 0 & \text{otherwise.} \end{cases}$$
Consequently, $$\frac{\epsilon_0(n) + \epsilon(n)}2 = \begin{cases} 1 & \text{if n \equiv 1 or 4 \pmod 5, or}\\ 0 & \text{otherwise, and} \end{cases} \tag3$$ $$\frac{\epsilon_0(n) - \epsilon(n)}2 = \begin{cases} 1 & \text{if n \equiv 2 or 3 \pmod 5, or}\\ 0 & \text{otherwise.} \end{cases} \tag4$$ Let $\chi$ denote either $\epsilon_0$ or $\epsilon$. Because $|\chi| \le 1$, the Dirchlet $L$-series $$L(s,\chi) = \sum_{n=1}^\infty \frac{\chi(n)}{n^s}$$ is absolutely convergent for $s > 1$. Because $\chi$ is totally multiplicative, we can use a special case of the Euler product to write $$L(s,\chi) = \prod_p \frac 1{1 - \chi(p)/p^s}$$ for $s > 1$. Taking logarithms and arguing as in the proof of the formula for the logarithm of the zeta function, we deduce that $$\log L(s, \chi) = \sum_{n=1}^\infty \frac{\Lambda(n)}{\log n} \frac{\chi(n)}{n^s}$$ for $s > 1$. From the estimate $(1)$, it follows that $$\sum_p \frac{\chi(p)}{p^s} = \log L(s, \chi) + O(1) \tag5$$ for $s > 1$. By the identities $(3)$ and $(4)$, we conclude that $$\sum_{\substack{p \\ p \equiv 1 \text{ or } 4}} \frac1{p^s} = \frac{\log L(s, \epsilon_0) + \log L(s, \epsilon)}2 + O(1)$$ $$\sum_{\substack{p \\ p \equiv 2 \text{ or } 3}} \frac1{p^s} = \frac{\log L(s, \epsilon_0) - \log L(s, \epsilon)}2 + O(1)$$ for $s > 1$.
It remains to determine the behavior of $\log L(s, \epsilon_0)$ and of $\log L(s, \epsilon)$ as $s$ tends to $1$ from above. If we take $\chi = \epsilon_0$ in $(5)$, we find that the sum on the left differs from that in $(2)$ only in that the prime $5$ is missing. Hence from $(2)$ we deduce that $$\log L(s, \epsilon_0) = \log \frac1{s - 1} + O(1)$$ for $s > 1$.
As for $L(s, \epsilon)$, we note first that the Dirichlet series $\sum \epsilon(n)/n^s$ is absolutely convergent only for $s > 1$. We now show that this series is conditionally convergent for $0 < s \le 1$. To that end, observe that the coefficient sum $\sum_{n \le x} \epsilon(n)$ takes only the values $0$ and $\pm 1$ and hence is uniformly bounded. If $s$ is fixed, $s > 0$, then the sequence $1/n^s$ tends to $0$ monotonically. Hence by Dirichlet's test, the series $\sum \epsilon(n)/n^s$ converges. Indeed, that series is uniformly convergent for $s \ge \delta > 0$. Because each term is a continuous function of $s$, it follows that the sum $L(s, \epsilon) = \sum \epsilon(n)/n^s$ is a continuous function of $s$ for $s > 0$. In particular, $L(s, \epsilon)$ tends to the finite limit $L(1, \epsilon)$ as $s$ tends to $1$. Moreover, if we group the terms in $L(1, \epsilon)$ two a time, we can use the alternating series test to see that $1/2 - 1/12 < L(1, \epsilon) < 1/2$. Hence, $L(1, \epsilon) > 0$, so that $\log L(s, \epsilon)$ tends to the finite limit $\log L(1, \epsilon)$ as $s$ tends to $1$ from above. As $\log L(s, \epsilon) = O(1)$ uniformly for $s \ge 1$, on combining our estimates we find that $$\sum_{\substack{p \\ p \equiv 1 \text{ or } 4}} \frac1{p^s} = \frac12 \log \frac1{s - 1} + O(1)$$ $$\sum_{\substack{p \\ p \equiv 2 \text{ or } 3}} \frac1{p^s} = \frac12 \log \frac1{s - 1} + O(1)$$ for $s > 1$. Because the right sides tend to infinity as $s$ tends to $1$ from above, it follows that the sums on the left contain infinitely many terms. Thus, there are infinitely many primes congruent to $1$ or $4$ and to $2$ or $3$ modulo $5$.