My question is about the proof of this theorem by Dirichlet:
Let $n\in \mathbb{N}$, $n\geq 2$. Then there are infinitely many primes in the sequence
$(kn+1)_{k\in \mathbb{N}}$.
I was reading a proof in the german textbook "Lehrbuch der Algebra" by Fischer, which stated the following:
Suppose we have found $r$ many primes $p_1,\dots, p_r$ already, then we set $x:=n\cdot p_1 \cdot \dots \cdot p_r$. Then $\Phi_n(x) \notin \{0,\pm 1\}$ and we can get a new prime by considering a prime divisor $p$ of $\Phi_n(x)$.
I understand the following steps showing that $p$ is actually a new prime which is in the sequence but I don't see why $\Phi_n(x) \notin \{0,\pm 1\}$ must hold.
For example Washington gives a similiar proof in his book "Introduction to Cyclotomic fields", but he uses a constant $c$ and claims that we can choose $c$ in such way that $\Phi(cx) \notin \{0,\pm 1\}$, which makes sense to me.
So to phrase my question explicitly:
How can I show that $\Phi_n(x) \notin \{0,\pm 1\}$?
Maybe I have overlooked something and the question is trivial, however I would appreciate any help.
Edit: If $r=0$, $x:=n$.
Best Answer
By definition, the zeroes of $\Phi_n$ are roots of $1$, so they have modulus $1$. In particular, $\vert \Phi_n(0)\vert =1$, and $\Phi_n(0)=\pm 1$, since it is an integer.
We then have $\Phi_n(x)=xP(x)\pm 1$ for some $P\in\mathbb{Z}[X]$. Assume now that $\Phi_n(x)=0,1,-1$, then $xP(x)=0,\pm 1,\pm 2$. We cannot have $P(x)=0$ since $\Phi_n$ is the monic polynomial of smallest degree which annihilates a primitive root of $1$. Thus $P(x)\neq 0$, and $\vert P(x)\vert \geq 1$. It follows that $\vert x\vert\leq \vert xP(x)\vert \leq 2$. This is not possible since $p_1\geq 3$.