Exercise 4. Show that there are infinitely many primes of the form 4n + 3. Hint:

you might find it useful to show that the product of two numbers of the form 4n + 1

are also of that form. For example, 5, 9 and 5 · 9 = 45 are all of the form 4n + 1, for

n = 1, 2, 11, respectively. Note that all primes except 2 are odd, and every odd prime

is either of the form 4n + 1 or 4n + 3 for some n.

To prove that there are infinitely many primes of the form \( 4n + 3 \), we can follow a similar approach to Euclid's classic proof that there are infinitely many primes, with specific modifications to fit the form \( 4n + 3 \).

### Step-by-step Proof

1. **Assume the contrary:**
   Suppose there are finitely many primes of the form \( 4n + 3 \). Let these primes be:
   \[
   p_1, p_2, p_3, \ldots, p_k
   \]
   where each \( p_i = 4n_i + 3 \) for some integer \( n_i \).

2. **Construct a special number:**
   Define a new number \( N \) as follows:
   \[
   N = 4 \cdot (p_1 p_2 \cdots p_k) + 3
   \]
   This number \( N \) is clearly of the form \( 4m + 3 \), since:
   \[
   N = 4 \cdot (p_1 p_2 \cdots p_k) + 3 = 4m + 3
   \]
   where \( m = p_1 p_2 \cdots p_k \).

3. **Prime factorization of \( N \):**
   \( N \) must either be prime or composite. If \( N \) is prime, it is a new prime of the form \( 4n + 3 \), which contradicts our assumption that \( p_1, p_2, \ldots, p_k \) were the only primes of this form.

4. **If \( N \) is composite:**
   If \( N \) is not prime, then it must have prime divisors. Now, none of the primes \( p_1, p_2, \ldots, p_k \) can divide \( N \), since:
   \[
   N \equiv 3 \pmod{p_i}
   \]
   This is because \( N = 4(p_1 p_2 \cdots p_k) + 3 \), and when divided by any \( p_i \), we have:
   \[
   N \equiv 3 \pmod{p_i}
   \]
   implying that none of \( p_1, p_2, \ldots, p_k \) divides \( N \).

   Therefore, \( N \) must be divisible by some other prime, say \( q \). Since \( N \) is of the form \( 4n + 3 \), its prime divisors must also be of the form \( 4n + 3 \), because a product of numbers of the form \( 4n + 1 \) results in a number of the form \( 4n + 1 \) (as hinted), and thus \( N \) could not have been divisible by any prime of the form \( 4n + 1 \).

   Hence, there must exist a prime \( q \) of the form \( 4n + 3 \) that divides \( N \), but this prime \( q \) is not one of \( p_1, p_2, \ldots, p_k \), contradicting our assumption that \( p_1, p_2, \ldots, p_k \) were the only primes of this form.

5. **Conclusion:**
   This contradiction shows that our assumption was false, meaning there must be infinitely many primes of the form \( 4n + 3 \).

### Summary
We constructed a number \( N \) of the form \( 4n + 3 \) that is not divisible by any of the finitely many assumed primes of the form \( 4n + 3 \). This led to a contradiction, proving that there must be infinitely many primes of this form.

---

Do you want any additional details on specific parts of the proof or related topics?

### Related Questions
1. How does the form \( 4n + 1 \) affect the product of two numbers of that form?
2. Can we similarly prove there are infinitely many primes of the form \( 4n + 1 \)?
3. How does the prime factorization relate to proving the infinitude of primes in general?
4. How does this proof compare with Euclid's original proof for the infinitude of primes?
5. What other numbers of the form \( an + b \) can we prove have infinitely many primes?

### Tip
When working with modular arithmetic, consider how operations like addition and multiplication behave under specific moduli (like modulo 4) to simplify complex expressions.

Exercise 4. Show that there are infinitely many primes of the form 4n + 3. Hint: you might find it useful to show that the product of two numbers of the form 4n + 1 are also of that form. For example, 5, 9 and 5 · 9 = 45 are all of the form 4n + 1, for n = 1, 2, 11, respectively. Note that all primes except 2 are odd, and every odd prime is either of the form 4n + 1 or 4n + 3 for some n.

Explore a detailed proof that shows there are infinitely many primes of the form 4n + 3, using concepts from number theory and modular arithmetic. This problem utilizes a construction technique similar to Euclid’s proof for primes, suitable for advanced high school or undergraduate math students.

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