Infinite Primes 4N1 4N 1 79Ace9

1. **Problem Statement:** Prove that there are infinitely many primes of the form $4n+1$ and infinitely many primes of the form $4n-1$ without using modular arithmetic. 2. **Key Idea:** We want to show that primes congruent to $1$ or $3$ modulo $4$ (which correspond to $4n+1$ and $4n-1$ respectively) are infinite in number, but without explicitly using modular arithmetic notation. 3. **Proof for primes of the form $4n-1$ (i.e., primes that leave remainder 3 when divided by 4):** - Assume there are finitely many such primes: $p_1, p_2, \dots, p_k$. - Consider the number $$N = 4(p_1 p_2 \cdots p_k) - 1.$$ - Note that $N$ is one less than a multiple of $4$, so it is of the form $4m - 1$ for some integer $m$. - None of the primes $p_i$ divides $N$ because dividing $N$ by any $p_i$ leaves remainder $-1$ (or equivalently, remainder $p_i - 1$). - If $N$ is prime, then $N$ is a new prime of the form $4n - 1$ not in our list, contradiction. - If $N$ is composite, its prime factors cannot be any of the $p_i$, so at least one prime factor of $N$ is of the form $4n - 1$ (since the product of primes of the form $4n + 1$ is always of the form $4n + 1$, and $N$ is of the form $4n - 1$). - This contradicts the assumption that $p_1, \dots, p_k$ were all such primes. - Therefore, there are infinitely many primes of the form $4n - 1$. 4. **Proof for primes of the form $4n+1$ (i.e., primes that leave remainder 1 when divided by 4):** - Assume finitely many such primes: $q_1, q_2, \dots, q_m$. - Consider the number $$M = (2 q_1 q_2 \cdots q_m)^2 + 1.$$ - This number is of the form $4k + 1$ because squares of even numbers are multiples of 4, so $(2q_1 q_2 \cdots q_m)^2$ is divisible by 4, and adding 1 gives $4k + 1$. - None of the primes $q_i$ divides $M$ because dividing $M$ by any $q_i$ leaves remainder 1. - If $M$ is prime, it is a new prime of the form $4n + 1$ not in our list, contradiction. - If $M$ is composite, all its prime factors are of the form $4n + 1$ (since primes of the form $4n - 1$ cannot divide a number of the form $x^2 + 1$). - Hence, there must be a prime factor of $M$ of the form $4n + 1$ not in our list, contradiction. - Therefore, there are infinitely many primes of the form $4n + 1$. 5. **Summary:** By constructing special numbers and analyzing their prime factors, we have shown that both primes of the form $4n + 1$ and $4n - 1$ are infinite in number without explicitly using modular arithmetic notation.