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Let $a,n$ be positive integers with $(a,n) = 1$. Then by Dirichlet's theorem on primes in an arithmetic progression we know that there are infinitely many primes $p$ satisfying $p\equiv a\pmod n$.

Dirichlet's theorem uses analytic number theory. I was wondering if there exists an elementary proof (not only meaning not using complex functions, but also relatively easy) of the existence of any (not infinitely many) prime in such an arithmetic progression. I couldn't come up with a proof myself.

doetoe
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    Note that proving that even one prime exists in an arithmetic sequence is actually equivalent to proving that the sequence has infinitely many. – abiessu Jun 17 '20 at 01:10
  • @abiessu I was afraid of that. Is it easy to see that that is the case? – doetoe Jun 17 '20 at 01:12
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    Yes. Consider the sequence $nx+a$ where we prove that a prime exists for $x=x_0$. Next consider the sequence $nx+a+nx_0+n$, our proof should apply to this sequence as well, and so on to infinity. – abiessu Jun 17 '20 at 01:15
  • Selberg gave an "elementary" proof, but it's not clear whether it meets the "relatively easy" criterion. – NickD Jun 17 '20 at 01:20
  • There are of course elementary proofs for particular values of $a$ and $n$; e.g., here – J. W. Tanner Jun 17 '20 at 01:22
  • "elementary proofs" and "easy proofs" are in general not the same. Some "big guns" are probably inevitable for the proof of Dirichlet's theorem. – Peter Jun 17 '20 at 08:51

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I don't know if this answers your question, but there is a simple proof that for all $n\geqslant 2$, there exists an infinite number of primes $p$ such that $p\equiv 1[n]$. Let $\Phi_n$ the $n$-th cyclotomic polynomial. We first prove that if $p$ is a prime number such that there exists $a\in\mathbb{Z}$ such that $p\mid\Phi_n(a)$ and $p\nmid \Phi_d(a)$ for all $d\mid n$ with $d<n$, then $p\equiv 1[n]$. Since $$ X^n-1=\prod_{d\,|\, n}\Phi_d $$ $p\mid a^n-1$ and thus the order of $a$ in $(\mathbb{Z}/p\mathbb{Z})^*$ divides $n$. Let $d$ be a divisor of $n$ with $d<n$. We have $$ \overline{a}^d-1=\prod_{k\,|\, d}\overline{\Phi_k(a)} $$ in $\mathbb{Z}/p\mathbb{Z}$. But by hypothesis $\overline{\Phi_k(a)}\neq 0$ for all $k\mid d$ (because $k\mid n$ and $k<n$). Since $\mathbb{Z}/p\mathbb{Z}$ is a field, the above product is non-zero and thus $\overline{a}^d-1\neq 0$. This means that the order of $a$ in $(\mathbb{Z}/p\mathbb{Z})^*$ is $n$, and thus $n\mid p-1$ that is to say $p\equiv 1[n]$.

Now let $n\geqslant 2$ and let us suppose there is a finite number of primes $p_1,\ldots,p_s$ such that $p_i\equiv 1[n]$ for all $i$ and let $N=np_1\cdots p_s$. Let $Q=\prod_{d|N,d<N}\Phi_d$, then $Q\wedge\Phi_N=1$, by Bezout's theorem there exist $U,V\in\mathbb{Q}[X]$ such that $U\Phi_N+VQ=1$. Let $a\in\mathbb{Z}$ such that $aU,aV\in\mathbb{Z}[X]$. There is an infinite number of such $a$ so we can chose one such that $\Phi_N(a)\notin\{-1,0,1\}$. Let $p$ be a prime divisor of $\Phi_N(a)$, then $\overline{a}^N=1$ in $\mathbb{Z}/p\mathbb{Z}$, thus $p\nmid a$ and $p\nmid Q(a)$, otherwise $p$ would divide $aU\Phi_N(a)+aVQ(a)=a$ which is not. Because of the lemma, $p\equiv 1[N]$ and since $n|N$, $p\equiv 1[n]$ but $p$ is not among the $p_i$ because $p\geqslant 1+N>p_i$ for all $i$.

user26857
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Tuvasbien
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  • Thanks Tuvasbien. It is the kind of proof I was looking for, but since this kind of proof seems to only work for some specific arithmetic sequences (see e.g. the theorem referenced in https://math.stackexchange.com/q/3214509/9911), I was wondering if we could still have something simple in the general case if we only require the existence of a prime, rather than an infinitude. The remark of abiessu however shows that general existence and infinitude are equivalent. – doetoe Jun 18 '20 at 07:23
  • Although there are simple proofs that $\pi(x)\gg\frac{x}{\ln x}$, I don't think that such a proof exists in the general case of arithmetic progressions. Indeed the existence of an infinite number of primes is very easy, but the infinity of primes in arithmetic progressions needed the proof of Dirichlet. Furthermore there is no "good" relation between $\pi(x)$ and $\pi(x,q,a)$ that would let us deduce the arithmetic progression theorem from the infinitude of prime numbers, even $$ \pi(x)=\sum_{a\in(\mathbb{Z}/q\mathbb{Z})^{\times}}\pi(x,q,a) $$ is absolutely not useful. – Tuvasbien Jun 18 '20 at 15:35