2.1 The LTE lemma

The next result, which seems to appear for the first time in 1878 in the work of Lucas [Reference Lucas4, Section XIII], will play a key part in the proof of Theorem 1.1. For a proof, see, for instance, [Reference Manea5, Theorem 1].

Lemma 2.1 (LTE)

Let $a \ne b \in \mathbb {Z}$ , $q \geqslant 3$ prime such that $q \nmid ab$ and $q \mid a-b$ , and $m \in \mathbb {Z}_{\geqslant 1}$ . Then

$$ \begin{align*}v_q \left( a^{m} - b^{m} \right) = v_q(a-b) + v_q(m).\end{align*} $$

2.2 Tools for values of cyclotomic polynomials

The following lemma is quite well known and may follow from the fact that $\Phi _m(X)$ is a reciprocal polynomial for instance. We provide here a slightly different proof.

Lemma 2.2 Let $n \in \mathbb {Z}_{\geqslant 2}$ and $m \in \mathbb {Z}_{\geqslant 1}$ . Then

$$ \begin{align*}\prod_{k=1}^{\infty} \left( 1 - {\frac{1}{n^k}}\right) n^{\varphi(m)} \leqslant \Phi_m(n) \leqslant \prod_{k=1}^{\infty} \left( 1 - {\frac{1}{n^k}}\right)^{-1} n^{\varphi(m)}.\end{align*} $$

Proof By Möbius inversion,

$$ \begin{align*}\Phi_m(n) = \prod_{k \mid m} \left(n^{k} - 1 \right)^{\mu(m/k)} = \prod_{k \mid m} n^{k\mu(m/k)} \ \prod_{k \mid m}\left(1 - {\displaystyle\frac{1}{n^k}} \right)^{\mu(m/k)}\end{align*} $$

and

$$ \begin{align*}\log \left( \prod_{k \mid m} n^{k\mu(m/k)}\right) = \log n \sum_{k \mid m} k \mu \left( \frac{m}{k} \right) = \varphi(m) \log n,\end{align*} $$

where we used the Dirichlet convolution identity $\mathrm {id} \star \mu = \varphi $ , so that

$$ \begin{align*}\Phi_m(n) = n^{\varphi(m)} \prod_{k \mid m} \left( 1 - {\frac{1}{n^k}}\right)^{\mu(m/k)}\end{align*} $$

and the asserted bounds follow easily.

The next result is well known (see [Reference Washington8, Lemma 2.9] for instance).

Lemma 2.3 Let q be a prime. Then, for all $n \in \mathbb {Z}$ and $m \in \mathbb {Z}_{\geqslant 1}$ such that $q \nmid mn$ , we have

$$ \begin{align*}q \mid \Phi_m(n) \iff m = \mathrm{ord}_q(n).\end{align*} $$

2.3 Proof of Theorem 1.1

Write $n = q^{\nu } a b$ with $\nu \in \mathbb {Z}_{\geqslant 0}$ , $(a,b)=1$ such that $p \mid a \Longrightarrow p^{k} \equiv 1 \; (\mathrm {mod}\ q)$ and $p \mid b \Longrightarrow \left ( p \ne q \right .$ and $\left. p^{k} \not \equiv 1 \; (\mathrm {mod}\ q) \right )$ . Note that $\sigma _k \big ( q^{\nu } \big ) = 1 + q^k + \dotsb + q^{k \nu } \equiv 1 \; (\mathrm {mod}\ q)$ , and hence $v_q \left ( \sigma _k \big ( q^{\nu } \big ) \right ) =0$ . Thus,

(2.1) $$ \begin{align} v_q \left( \sigma_k(n)\right) = v_q \left( \sigma_k(a)\right) + v_q \left( \sigma_k(b)\right). \end{align} $$

If $q \geqslant 3$ , $q \ne p$ , and $q \mid p^k - 1$ , we apply Lemma 2.1 with $a=p^k$ , $b=1$ , and $m= \alpha +1$ , yielding

$$ \begin{align*}v_q \left( \sigma_k \left( p^{\alpha}\right) \right) = v_q \left( \frac{p^{k(\alpha+1)}-1}{p^k-1} \right)= v_q \left( p^{k(\alpha+1)} - 1 \right) - v_q \left( p^{k} - 1 \right)= v_q(\alpha + 1)\end{align*} $$

so that

(2.2) $$ \begin{align} v_q \left( \sigma_k(a)\right) = \sum_{p^{\alpha} \| a} v_q(\alpha+1) =\sum_{\substack{p^{\alpha} \| n \\ p^{k} \equiv 1 \; (\mathrm{mod}\ q)}} v_q(\alpha+1). \end{align} $$

To evaluate $v_q \left ( \sigma _k(b)\right )$ , we assume throughout that $p^k \not \equiv 1 \; (\mathrm {mod}\ q)$ and first note that, for all $p^{\alpha } \| b$ , we have

$$ \begin{align*}v_q \left( \sigma_k \big( p^{\alpha} \big) \right) = v_q \left( p^{k(\alpha+1)}-1 \right)\end{align*} $$

since $q \nmid p^{k}-1$ . Also notice that this latter condition implies $\varpi _1 \geqslant 2$ and $\varpi _k \geqslant 2$ .

1. ⊳ If $ \varpi _k \nmid \alpha + 1 $ , the Euclidean division of $\alpha + 1$ by $\varpi _k$ yields $\alpha + 1 = h \varpi _k + r$ with $1 \leqslant r < \varpi _k$ , so that

   $$ \begin{align*}p^{k(\alpha+1)}-1 = \left( p ^{k \varpi_k} \right)^{h} \times p^{rk} - 1 \equiv p^{rk} - 1 \not \equiv 0 \; (\mathrm{mod}\ q)\end{align*} $$
   otherwise we would have $\varpi _k \leqslant r$ since $r \geqslant 1$ , which would contradict the inequality $r < \varpi _k$ . We infer that, in this case, we have
   $$ \begin{align*}v_q \left( \sigma_k\big( p^{\alpha} \big)\right) = 0.\end{align*} $$

2. ⊳ If $ \varpi _k \mid \alpha + 1 $ , write $\alpha + 1 = h \varpi _k$ with $h \in \mathbb {Z}_{\geqslant 1}$ . Since $p^{k \varpi _k} \equiv 1 \; (\mathrm {mod}\ q)$ , Lemma 2.1 yields

   $$ \begin{align*} v_q \left( \sigma_k\big( p^{\alpha} \big)\right) &= v_q \left( p^{k h \varpi_k} - 1 \right) = v_q \left( p^{k \varpi_k} - 1 \right) + v_q(h) \\ &= v_q \left( p^{k \varpi_k} - 1 \right) + v_q(\alpha+1) - v_q(\varpi_k) \\ &= v_q \left( p^{k \varpi_k} - 1 \right) + v_q(\alpha+1), \end{align*} $$
   where we used $\varpi _k \mid q-1$ and hence $ v_q(\varpi _k) = 0 $ . Therefore,
   $$ \begin{align*} v_q \left( \sigma_k(b)\right) &= \sum_{\substack{p^{\alpha} \| b \\ \varpi_k \mid \alpha + 1}} \left( v_q(\alpha+1) + v_q \left( p^{k \varpi_k} - 1 \right) \right) \\ & = \sum_{\substack{p^{\alpha} \| n \\ p \ne q \\ p^{k} \not \equiv 1 \; (\mathrm{mod}\ q) \\ \varpi_k \mid \alpha + 1}} \left( v_q(\alpha+1) + v_q \left( p^{k \varpi_k} - 1 \right) \right). \end{align*} $$
   We now distinguish between two cases:

   1. (a) If $q \mid p^{\varpi _k} - 1$ , Lemma 2.1 implies that

      $$ \begin{align*}v_q \left( p^{k \varpi_k} - 1 \right) = v_q \left( p^{\varpi_k} - 1 \right) + v_q(k).\end{align*} $$
      Also note that, since $1 \equiv p^{k \varpi _k} \equiv p^{\varpi _k} \; (\mathrm {mod}\ q)$ , we derive $\varpi _1 \mid \varpi _k (k-1)$ , and since $\varpi _1 = \varpi _k d_k $ , where $d_k := \left ( \varpi _1,k\right )$ , we get $d_k \mid k-1$ . Since $d_k \mid k$ , we obtain $d_k = 1$ and then $\varpi _k = \varpi _1$ . Hence,
      $$ \begin{align*}v_q \left( p^{k \varpi_k} - 1 \right) = v_q \left( p^{\varpi_1} - 1 \right) + v_q(k).\end{align*} $$

   2. (b) Assume $q \nmid p^{\varpi _k} - 1$ . Using again $\varpi _1 = \varpi _k d_k $ , we derive $ k \varpi _k = \frac {k \varpi _1}{d_k} := m \varpi _1$ , where $m:= \frac {k}{d_k}$ . Since $q \mid p^{\varpi _1} - 1$ , Lemma 2.1 yields

      $$ \begin{align*} v_q \left( p^{k \varpi_k} - 1 \right) &= v_q \left(p^{m \varpi_1} - 1\right) \\ &= v_q \left( p^{\varpi_1} - 1 \right) + v_q(m) \\ &= v_q \left( p^{\varpi_1} - 1 \right) + v_q (k) - v_q (d_k). \end{align*} $$
      Furthermore, since $\varpi _1 \mid q-1$ , we have $q \nmid \varpi _1$ , so that $ v_q (d_k) = 0 $ .

   Hence,

   $$ \begin{align*}v_q \left( \sigma_k(b)\right) = \sum_{\substack{p^{\alpha} \| n \\ p \ne q \\ p^{k} \not \equiv 1 \; (\mathrm{mod}\ q) \\ \varpi_k \mid \alpha + 1}} \left( v_q(\alpha+1) + v_q(k) +v_q \left( p^{\varpi_1} - 1 \right) \right).\end{align*} $$
   Now
   $$ \begin{align*}v_q \left( p^{\varpi_1} - 1 \right) = v_q \left( \prod_{d \mid \varpi_1} \Phi_d (p) \right) = \sum_{d \mid \varpi_1} v_q \left( \Phi_d (p) \right)\end{align*} $$
   and since $q \ne p$ and $q \nmid d$ for all $d \mid \varpi _1$ because $\varpi _1 \mid q-1$ , Lemma 2.3 yields $v_q \left ( \Phi _d (p) \right ) \ne 0 \iff d= \mathrm {ord}_q(p)= \varpi _1$ , so that
   $$ \begin{align*}v_q \left( p^{\varpi_1} - 1 \right) = v_q \left( \Phi_{\varpi_1} (p) \right).\end{align*} $$

Finally, we get

(2.3) $$ \begin{align} v_q \left( \sigma_k(b)\right) = \sum_{\substack{p^{\alpha} \| n \\ p \ne q \\ p^k \not \equiv 1 \; (\mathrm{mod}\ q) \\ \varpi_k \mid \alpha + 1}} \left( v_q(\alpha+1) + v_q(k) + v_q \left( \Phi_{\varpi_1} (p) \right)\right) \end{align} $$

and we complete the proof by reporting (2.2) and (2.3) in (2.1).

2.4 Proof of Corollary 1.3

First note that, when $p^k \equiv 1 \, (\mathrm {mod}\ q)$ , then $\varpi _k = 1$ , so that Theorem 1.1 may be restated as

$$ \begin{align*}v_q \left( \sigma_k(n)\right) = \sum_{\substack{p^{\alpha} \| n \\ p \ne q \\ \varpi_k \mid \alpha + 1}} v_q(\alpha+1) + \sum_{\substack{p^{\alpha} \| n \\ p \ne q \\ p^k \not \equiv 1 \; (\mathrm{mod}\ q) \\ \varpi_k \mid \alpha + 1}} \Big( v_q(k) + v_q \left( \Phi_{\varpi_1} (p) \right) \Big)\end{align*} $$

and therefore,

(2.4) $$ \begin{align} v_q \left( \sigma_k(n)\right) & \leqslant \sum_{\substack{p^{\alpha} \| n \\ p \ne q \\ \varpi_k \mid \alpha + 1}} \left( v_q(\alpha+1) + v_q(k) \right) + \sum_{\substack{p^{\alpha} \| n \\ p \ne q \\ p^k \not \equiv 1 \; (\mathrm{mod}\ q) \\ \varpi_k \mid \alpha + 1}} v_q \left( \Phi_{\varpi_1} (p) \right) \\ & := S_1+S_2, \notag \end{align} $$

say.

1. ⊳ The sum $S_1$ does not exceed

   $$ \begin{align*}S_1 \leqslant \frac{1}{\log q} \left( \sum_{p^{\alpha} \| n} \log (\alpha + 1) + \log k \sum_{p \mid n} 1 \right) = \frac{\log \tau(n) + \omega(n) \log k}{\log q}\end{align*} $$
   and the bounds [Reference Nicolas and Robin6, Théorème 1] and [Reference Robin7, Théorème 11] imply that
   (2.5) $$ \begin{align} S_1 < \frac{\left( {1.385} \log k + {1.066} \right) \log n}{\log q \log \log n}. \end{align} $$

2. ⊳ It remains to estimate $S_2$ . Let us first note that the condition $p^k \not \equiv 1 \; (\mathrm {mod}\ q)$ is equivalent to $\varpi _1 \mid q-1$ and $\varpi _1 \nmid k$ . Also recall that $M_{q,k} = \max \varphi (d)$ , where the maximum is taken over the divisors $ d$ of $q-1 $ such that $d \nmid k$ , with the convention $M_{q,k} = 0$ if this set is empty. Hence, in view of the previous remark, we get the bound $\varphi (\varpi _1) \leqslant M_{q,k}$ for all primes $p \mid n$ satisfying $p^k \not \equiv 1 \; (\mathrm {mod}\ q)$ .

   Using Lemma 2.2, we derive for $n \geqslant 3$ ,

   (2.6) $$ \begin{align} S_2 & \leqslant \frac{1}{\log q} \sum_{\substack{p \mid n \\ p^k \not \equiv 1 \; (\mathrm{mod}\ q)}} \log \left( \Phi_{\varpi_1} (p) \right) \notag \\ & \leqslant \frac{1}{\log q} \sum_{\substack{p \mid n \\ p^k \not \equiv 1 \; (\mathrm{mod}\ q)}} \Biggl( \log \left( p^{\varphi(\varpi_1)}\right) - \sum_{k=1}^{\infty} \log \left( 1 - \frac{1}{p^k}\right) \Biggr) \notag \\ & \leqslant \frac{1}{\log q} \log \left( \prod_{\substack{p \mid n \\ p^k \not \equiv 1 \; (\mathrm{mod}\ q)}} p^{\varphi(\varpi_1)} \right) + \frac{1}{\log q} \sum_{p \mid n} \sum_{k=1}^{\infty} \left( \frac{1}{p^k} + \frac{1}{p^{2k}}\right) \notag \\ & \leqslant \frac{1}{\log q} \log \left( \prod_{\substack{p \mid n \\ p^k \not \equiv 1 \; (\mathrm{mod}\ q)}} p^{M_{q,k}} \right) + \frac{1}{\log q} \sum_{p \mid n} \left( \frac{1}{p-1} + \frac{1}{p^2-1} \right) \notag \\ &< \frac{M_{q,k}}{\log q} \log \left( \prod_{p \mid n} p \right) + \frac{1}{\log q} \sum_{p \mid n} \left( \frac{1}{p} + \frac{4}{p^2} \right) \notag \\ & \leqslant \frac{M_{q,k} \log \gamma(n)}{\log q} + \frac{1}{\log q} \left( \log \log \log n + 2 + 4 \log \zeta(2) \right) \notag \\ & < \frac{M_{q,k} \log \gamma(n)}{\log q} + \frac{4(\log \log \log n + 1)}{\log q}, \end{align} $$

where we used [Reference Bordellès2, Exercise 61, p. 337] stating that $\sum _{p \mid n} 1/p < \max \left ( 1, \log \log \log n + 1 \right ) $ when $n \geqslant 3$ . The first inequality in Corollary 1.3 follows from inserting (2.5) and (2.6) into (2.4). The second inequality comes from the well-known bounds $\varphi (\varpi _1) \leqslant \varpi _1 - 1 \leqslant q-2$ , since $\varpi _1 \geqslant 2$ , implying that $M_{q,k} \leqslant q-2$ .