[proofplan] We argue by contradiction, supposing that the set of primes $p \equiv 3 \pmod 4$ is finite, and manufacturing a divergent subseries of $\sum 1/n$ on one hand from a convergent quantity on the other. The tool is the completely multiplicative non-principal character $\chi(n) = \left(\frac{-1}{n}\right)$ mod $4$: its alternating $L$-series $L = \sum \chi(n)/n$ converges. Pairing $L$ with the finite Euler factors $(1 + 1/p_j)$ at the supposed finitely many primes $p_j \equiv 3 \pmod 4$ strips those primes out of the sum while preserving convergence. The resulting sum ranges over integers whose prime factors avoid $\{p_1, \ldots, p_k\}$ and, if those were the only primes $\equiv 3 \pmod 4$, must restrict further to integers built only from $2$ and primes $\equiv 1 \pmod 4$; such a sum is an infinite sum of positive terms that dominates a divergent series, contradicting its value as a finite quantity. [/proofplan] [step:Define the character $\chi$ and verify that $L$ converges] Consider the function \begin{align*} \chi : \mathbb{Z}_{\geq 1} &\to \{-1, 0, 1\} \\ n &\mapsto \left(\frac{-1}{n}\right), \end{align*} defined to be the Jacobi symbol for odd $n \geq 1$ and extended by $\chi(n) = 0$ when $n$ is even (equivalently, using the convention that the Jacobi symbol with an even lower argument vanishes). By [Properties of the Jacobi Symbol](/theorems/1722), $\chi$ is completely multiplicative on $\mathbb{Z}_{\geq 1}$, and for odd $n$, \begin{align*} \chi(n) &= (-1)^{(n-1)/2} = \begin{cases} +1 & \text{if } n \equiv 1 \pmod 4, \\ -1 & \text{if } n \equiv 3 \pmod 4. \end{cases} \end{align*} Thus on the odd integers $\chi$ alternates $+1, -1, +1, -1, \ldots$ as $n = 1, 3, 5, 7, \ldots$. Define \begin{align*} L &= \sum_{n = 1}^{\infty} \frac{\chi(n)}{n} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots. \end{align*} The terms $1/(2m+1)$ are positive, decreasing, and tend to $0$; by the [Alternating Series Test](/theorems/???), $L$ converges. Moreover $L > 1 - 1/3 = 2/3 > 0$, so $L \neq 0$. [guided] The dual role of $\chi$ is essential: as a function of $n$, $\chi$ is completely multiplicative (so formal identities involving Euler factors apply), and as a sequence, $\chi(n)/n$ alternates in sign on the odd integers (so Leibniz-type convergence is available). Complete multiplicativity means $\chi(mn) = \chi(m) \chi(n)$ for all $m, n \geq 1$, not merely for coprime pairs; this follows from the fact that $\chi$ is defined by a congruence condition modulo $4$ that is preserved under multiplication. Why is $L$ convergent but not absolutely convergent? The absolute series $\sum |\chi(n)|/n = \sum_{n \text{ odd}} 1/n$ diverges (it is essentially half of the harmonic series). So we need signed cancellation, provided by the [Alternating Series Test](/theorems/???): the terms $1/(2m+1)$ for $m = 0, 1, 2, \ldots$ are positive, strictly decreasing, and vanish in the limit, so the alternating sum converges. [/guided] [/step] [step:Introduce the finite-set-of-primes hypothesis and form the associated product] Suppose for contradiction that the set \begin{align*} \{p \text{ prime} : p \equiv 3 \pmod 4\} &= \{p_1, p_2, \ldots, p_k\} \end{align*} is finite, where $k \geq 1$ (the case $k = 0$ would mean no primes $\equiv 3 \pmod 4$ exist, contradicting the single example $p = 3$). Each $p_j$ is odd, so $\chi(p_j) = -1$. Define \begin{align*} A &= \prod_{j=1}^{k} \left(1 + \frac{1}{p_j}\right) \cdot L. \end{align*} This is a finite product of real numbers times the finite real number $L$, so $A$ is a well-defined finite real number. [/step] [step:Show that $(1 + 1/p_j) L$ removes multiples of $p_j$ from the sum] Fix $p = p_j$ and consider, as an absolutely convergent manipulation on partial sums first and then passing to the limit, the product $(1 + 1/p) L$. We claim \begin{align*} \left(1 + \frac{1}{p}\right) L &= \sum_{\substack{n \geq 1 \\ (n, p) = 1}} \frac{\chi(n)}{n}. \end{align*} Split the sum defining $L$ according to divisibility by $p$: \begin{align*} L &= \sum_{(n, p) = 1} \frac{\chi(n)}{n} + \sum_{p \mid n} \frac{\chi(n)}{n}. \end{align*} In the second sum, write $n = p m$ with $m \geq 1$; then $\chi(n) = \chi(p) \chi(m) = -\chi(m)$ by complete multiplicativity and $\chi(p) = -1$. So \begin{align*} \sum_{p \mid n} \frac{\chi(n)}{n} &= \sum_{m \geq 1} \frac{-\chi(m)}{p m} = -\frac{1}{p} L. \end{align*} Therefore \begin{align*} L &= \sum_{(n, p) = 1} \frac{\chi(n)}{n} - \frac{L}{p}, \qquad \text{i.e.,} \qquad \left(1 + \frac{1}{p}\right) L = \sum_{(n, p) = 1} \frac{\chi(n)}{n}, \end{align*} as claimed. The rearrangement is legitimate because $L$ converges and we are merely regrouping a convergent series into two sub-series indexed by complementary sets; the identity $\sum_{p \mid n} \chi(n)/n = -L/p$ was obtained by the change of index $n = pm$, which is a bijection between $\{n \geq 1 : p \mid n\}$ and $\{m \geq 1\}$. [guided] This is the analytic analogue of the Euler-product manipulation $\zeta(s) (1 - p^{-s}) = \sum_{(n, p) = 1} n^{-s}$: pairing the full sum with the single Euler factor strips the prime $p$ out of the summation index. Here the Euler factor is $(1 + 1/p)$ rather than $(1 - 1/p)$ because $\chi(p) = -1$: the local factor for a multiplicative character at a prime $p$ is $(1 - \chi(p)/p)^{-1}$, so formally, \begin{align*} (1 - \chi(p)/p)^{-1} \sum_{(n, p) = 1} \frac{\chi(n)}{n} &= \sum_{n \geq 1} \frac{\chi(n)}{n} = L, \end{align*} and when $\chi(p) = -1$, $(1 - \chi(p)/p)^{-1} = (1 + 1/p)^{-1}$. Rearranging gives the formula. We do the full rearrangement by hand here rather than by appealing to an Euler-product theorem so that we control exactly what convergence hypothesis is being used. The identity $\sum_{p \mid n} \chi(n)/n = -L/p$ is a one-step manipulation of a convergent series under a bijection of indices, valid without any absolute-convergence assumption. [/guided] [/step] [step:Iterate over $p_1, \ldots, p_k$ to identify $A$ as a sum restricted by coprimality] Iterating the identity of Step 3, applied successively to $p_1$, then $p_2$, and so on, on the sum $\sum_{(n, p_1 \cdots p_{j-1}) = 1} \chi(n)/n$ (which has the same multiplicative character structure in $n$), we obtain \begin{align*} A &= \prod_{j=1}^{k} \left(1 + \frac{1}{p_j}\right) \cdot L = \sum_{\substack{n \geq 1 \\ (n, p_1 \cdots p_k) = 1}} \frac{\chi(n)}{n}. \end{align*} At each step, the same argument applies because each $p_j$ is a prime with $\chi(p_j) = -1$, and the sum at that stage has been restricted to indices coprime to $p_1, \ldots, p_{j-1}$, among which $p_j$ still appears multiplicatively. Hence $A$ equals a convergent real sum indexed by integers $n \geq 1$ coprime to $p_1 p_2 \cdots p_k$. [/step] [step:Reduce to a sum over integers whose odd prime factors are all $\equiv 1 \pmod 4$] By the finiteness hypothesis, every prime $p \equiv 3 \pmod 4$ lies in $\{p_1, \ldots, p_k\}$. Therefore an integer $n$ with $(n, p_1 \cdots p_k) = 1$ has no prime factor $\equiv 3 \pmod 4$; its prime factors are either $2$ or primes $\equiv 1 \pmod 4$. For such $n$, we compute $\chi(n)$ explicitly. If $n$ is even, then $\chi(n) = 0$, and such $n$ contribute $0$ to the sum. If $n$ is odd, then every prime factor of $n$ is $\equiv 1 \pmod 4$, so $n$ itself is $\equiv 1 \pmod 4$ (a product of integers $\equiv 1 \pmod 4$ is $\equiv 1 \pmod 4$); hence $\chi(n) = +1$. Restricting the sum to odd $n$, coprime to $p_1 \cdots p_k$, with all prime factors $\equiv 1 \pmod 4$: \begin{align*} A &= \sum_{\substack{n \geq 1 \\ (n, 2 p_1 \cdots p_k) = 1 \\ \text{all primes } q \mid n \text{ satisfy } q \equiv 1 \pmod 4}} \frac{1}{n}. \end{align*} Since the finiteness hypothesis states that every prime $\equiv 3 \pmod 4$ is in $\{p_1, \ldots, p_k\}$, the "coprime to $p_1 \cdots p_k$" constraint already rules out factors $\equiv 3 \pmod 4$. The remaining constraints are that $n$ is odd and all its prime factors are $\equiv 1 \pmod 4$, i.e., the sum is \begin{align*} A &= \sum_{n \in S} \frac{1}{n}, \qquad S = \{n \geq 1 : \text{every prime factor of } n \text{ is} \equiv 1 \pmod 4\}. \end{align*} [/step] [step:Derive a divergence contradiction] All terms in the sum defining $A$ are now positive, and the set $S$ is closed under multiplication. In particular, $S$ contains $1$ and every prime $q \equiv 1 \pmod 4$. By [Dirichlet's Theorem on primes in arithmetic progressions](/theorems/1751) (specialised to the progression $a = 1$, $N = 4$), there are infinitely many primes $q \equiv 1 \pmod 4$; moreover, the sum $\sum_{q \equiv 1 \pmod 4} 1/q$ diverges. This divergence alone forces $A \geq \sum_{q \in S, q \text{ prime}} 1/q = +\infty$, contradicting the finiteness of $A$ established in Step 2. Therefore the assumption that there are only finitely many primes $p \equiv 3 \pmod 4$ is false. [guided] We have arrived at an identity $A = \sum_{n \in S} 1/n$ in which the left-hand side is a finite real number (a finite product times the convergent series $L$) and the right-hand side is a sum of positive reciprocals indexed by a rich set of positive integers. Why must $\sum_{n \in S} 1/n$ diverge? Because $S$ contains every prime $q \equiv 1 \pmod 4$, and the reciprocal sum of primes in any non-trivial arithmetic progression $\{qN + a : (a, N) = 1\}$ diverges — this is the quantitative form of [Dirichlet's Theorem on primes in arithmetic progressions](/theorems/1751). One can argue without invoking Dirichlet's theorem directly by observing that the entire sub-series $\sum_{n \in S} 1/n$ contains $\sum_{q \equiv 1 \pmod 4, q \text{ prime}} 1/q$ as a sub-sum, and this prime-reciprocal subsum in the progression $\equiv 1 \pmod 4$ is known to diverge. The original sketch phrases this as "dominates $\sum 1/m$ over a set of positive density", reflecting the fact that the density of integers in $S$ is positive (though not equal to $1$); any such positive-density subset of $\mathbb{Z}_{\geq 1}$ has a divergent reciprocal sum, by an elementary comparison with $\sum 1/m$. The contradiction — finite $A$ equalling an infinite series of positive terms — refutes the initial finiteness hypothesis. Therefore the set $\{p \text{ prime} : p \equiv 3 \pmod 4\}$ is infinite. [/guided] [/step] [step:Conclude] The contradiction derived in Step 6 shows that the set of primes $p \equiv 3 \pmod 4$ cannot be finite. Hence there are infinitely many primes congruent to $3$ modulo $4$. [/step]