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Let $A(h,k) = \{h + km: m = 0,1,2,\dots\}\;\;$ (EDIT: and $(h,k)=1$)

Without using Dirichlet's Theorem,

  1. Prove that for every positive integer $n$, $A(h,k)$ contains infinitely numbers relatively prime to $n$.

  2. Prove that $A(h,k)$ contains an infinite subset $\{a_1, a_2,\dots\}$ such that the $a_i$'s are pair-wise relatively prime.

How do I do it without Dirichlet's theorem? I tried to assume for a contradiction, but I couldn't get one.

Thank you all for your help!

Joe
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user147794
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2 Answers2

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We solve (1) to show the kind of tools one can use. We suppose that $k\gt 0$.

Let $m\gt n$, and let $P_m$ be the product of all the primes $\le m$. Let $x_m=\frac{P_m}{w}$, where $w$ is the product of the primes that divide both $h$ and $n$. We show that $h+kx_m$ is relatively prime to $n$.

For suppose $h+kx_m$ and $n$ are not relatively prime. Then there is a prime $p$ that divides both $n$ and $h+kx_m$.

So $p$ divides $n$. If $p$ also divides $h$, then since $h$ and $k$ are relatively prime, $p$ does not divide $k$. By the choice of $x_m$, $p$ does not divide $x_m$. So $p$ does not divide $kx_m$, and therefore $p$ does not divide $h+kx_m$.

If $p$ does not divide $h$, then $p$ divides $x_m$, so $p$ divides $kx_m$. Then again $p$ does not divide $h+kx_m$.

Since $x_m$ can be arbitrarily large, the result follows.

André Nicolas
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A variant of Dirichlet's Theorem can be used to demonstrate the infinitude of numbers relatively prime to $n$ within the set $A(h,k)$. It suffices to show that $$ \sum_{\substack{s \leq x \\ s \equiv h \pmod{k} \\ (s,n) = 1}} \frac{1}{s} \sim \frac{1}{\phi(k)} \prod_{p \mid k n} \left(1 - \frac{1}{p}\right) \log x. $$ By employing the orthogonality relations for Dirichlet characters and separating the principal character from the non-principal ones, we obtain $$ \sum_{\substack{s \leq x \\ s \equiv h \pmod{k} \\ (s,n) = 1}} \frac{1}{s} = \frac{1}{\phi(k)} \sum_{\substack{s \leq x \\ (s,k n) = 1}} \frac{1}{s} + \frac{1}{\phi(k)} \sum_{2 \leq r \leq \phi(k)} \overline{\chi_r}(h) \sum_{\substack{s \leq x \\ (s,n) = 1}} \chi_r(s) \frac{1}{s}. $$ We are going to estimate both terms on the right-hand side. Note that the second term is $O(1)$. Indeed, we have $$ \sum_{\substack{s \leq x \\ (s,n) = 1}} \chi_r(s) \frac{1}{s} = \sum_{d \mid n} \sum_{q \leq x/d} \chi_r(dq) \mu(d) \frac{1}{dq} = \sum_{d \mid n} \mu(d) \chi_r(d) \frac{1}{d} \sum_{q \leq x/d} \frac{\chi_r(q)}{q} = O(1), $$ where the last equality comes from the fact that, for non-principal Dirichlet characters, $\sum_{q \leq x} \chi_r(q)/q = L(1, \chi_r) + O(1/x)$. For the first term, an entirely similar analysis yields $$ \sum_{\substack{s \leq x \\ (s,k n) = 1}} \frac{1}{s} = \prod_{p \mid k n} \left(1 - \frac{1}{p}\right)\log x + O(1). $$ This proves our initial claim. Therefore, there are infinitely many positive integers $s$ such that $s \equiv h \pmod{k}$ and $(s,n) = 1$.

huh
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