Difficulty in understanding the proof of infinitude of primes in a certain arithmetic progression

Question

Let $m$ as a fixed odd prime. How to show there are infinitely many primes of the form $2km+1$ (for some positive integer $k$).

Can someone please help? Any help would be appreciated. Thanks in advance.

Answer 1

The only one written out with sufficient detail for me to reverse-engineer the proof is the Landry one. Here is my take on it (in English):

First fix $m$ as an odd prime. We will show there are infinitely many primes of the form $2km+1$ (for some positive integer $k$). Assume there are only finitely many (we reach a contradiction). Let $\theta$ be the largest one. Let $x$ be the product of all primes of the form $2km+1$. Then

• Claim 1: $x^m+1$ is not divisible by any primes of the form $2km+1$.

• Claim 2: $\frac{x^m+1}{x+1}$ is a positive integer.

• Claim 3: All prime divisors of $\frac{x^m+1}{x+1}$ are of the form $2km+1$.

Assuming these three claims: From Claim 1 and 2 it follows that the integer $\frac{x^m+1}{x+1}$ is not divisible by any primes of the form $2km+1$. So $\frac{x^m+1}{x+1}$ must be divisible by some prime that is not of the form $2km+1$. But this contradicts Claim 3. $\Box$

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Proof of Claim 1: Let $z$ be a prime of the form $2km+1$. By construction, $x^m$ is divisible by $z$. Thus, $x^m+1$ cannot be divisible by $z$.

Proof of Claim 2: We have for any real number $y$: $$ (1+y+y^2+...+y^{m-1})(1-y) = 1-y^m$$ Choosing $y=-x$ (where $-x$ is an integer) gives $$ \underbrace{(1+(-x) +(-x)^2 + ... + (-x)^{m-1})}_{integer}(1+x)=1-(-x)^m = 1+x^m$$ where the final equality holds because $m$ is odd, so $(-1)^m=-1$.

Proof of Claim 3: I don't know. But, nothing in the above has used the assumption that $m$ is prime (we have only used that $m$ is an odd positive integer). So, Claim 3 must somehow use the assumption that $m$ is prime.