Math Problem Statement

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.

Solution

To prove that there are infinitely many primes of the form 4n+34n + 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+34n + 3.

Step-by-step Proof

  1. Assume the contrary: Suppose there are finitely many primes of the form 4n+34n + 3. Let these primes be: p1,p2,p3,,pkp_1, p_2, p_3, \ldots, p_k where each pi=4ni+3p_i = 4n_i + 3 for some integer nin_i.

  2. Construct a special number: Define a new number NN as follows: N=4(p1p2pk)+3N = 4 \cdot (p_1 p_2 \cdots p_k) + 3 This number NN is clearly of the form 4m+34m + 3, since: N=4(p1p2pk)+3=4m+3N = 4 \cdot (p_1 p_2 \cdots p_k) + 3 = 4m + 3 where m=p1p2pkm = p_1 p_2 \cdots p_k.

  3. Prime factorization of NN: NN must either be prime or composite. If NN is prime, it is a new prime of the form 4n+34n + 3, which contradicts our assumption that p1,p2,,pkp_1, p_2, \ldots, p_k were the only primes of this form.

  4. If NN is composite: If NN is not prime, then it must have prime divisors. Now, none of the primes p1,p2,,pkp_1, p_2, \ldots, p_k can divide NN, since: N3(modpi)N \equiv 3 \pmod{p_i} This is because N=4(p1p2pk)+3N = 4(p_1 p_2 \cdots p_k) + 3, and when divided by any pip_i, we have: N3(modpi)N \equiv 3 \pmod{p_i} implying that none of p1,p2,,pkp_1, p_2, \ldots, p_k divides NN.

    Therefore, NN must be divisible by some other prime, say qq. Since NN is of the form 4n+34n + 3, its prime divisors must also be of the form 4n+34n + 3, because a product of numbers of the form 4n+14n + 1 results in a number of the form 4n+14n + 1 (as hinted), and thus NN could not have been divisible by any prime of the form 4n+14n + 1.

    Hence, there must exist a prime qq of the form 4n+34n + 3 that divides NN, but this prime qq is not one of p1,p2,,pkp_1, p_2, \ldots, p_k, contradicting our assumption that p1,p2,,pkp_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+34n + 3.

Summary

We constructed a number NN of the form 4n+34n + 3 that is not divisible by any of the finitely many assumed primes of the form 4n+34n + 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+14n + 1 affect the product of two numbers of that form?
  2. Can we similarly prove there are infinitely many primes of the form 4n+14n + 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+ban + 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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic
Prime Numbers

Formulas

4n + 3 (Prime number form)
N = 4(p1 * p2 * ... * pk) + 3

Theorems

Euclid's Theorem on Infinitely Many Primes

Suitable Grade Level

Undergraduate Math or Advanced High School