Correction to: The arithmetic of Carmichael quotients

1 Erratum to: Period Math Hung (2015) 71:11–23 https://doi.org/10.1007/s10998-014-0079-3

2 Replacement of [3, Proposition 4.3]

We first illustrate a counter-example for [3, Proposition 4.3]. Take $m=273 = 3 \times 7 \times 13$. Using the notation in [3, Proposition 4.3], we have $d=3$ and $d^{\prime }=1$. That is, the homomorphism $\phi _m$ defined there is surjective. However, for any positive integer a coprime to m, $C_m(a)$ is divisible by 3, because $6\mid \lambda (m)$ and then $9\mid a^{\lambda (m)}-1$. This leads to a contradiction.

Proposition 4.3 and its proof in [3] should be replaced by Proposition 1.1 below. Fortunately, this does not affect other results and arguments in [3], although Proposition 4.3 in [3] was quoted several times there.

Assume that positive integer m has the prime factorization $m=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$. In [1, Proposition 4.4], the Euler quotient has been used to define a homomorphism from $(\mathbb {Z}/m^{2}\mathbb {Z})^{*}$ to $(\mathbb {Z}/m\mathbb {Z},+)$, whose image is $d\mathbb {Z}/m\mathbb {Z}$, where

$$\begin{aligned} d=\prod _{i=1}^{k}d_{i} \quad \text {and} \quad d_{i}=\left\{ \begin{array}{ll} \gcd (p_{i}^{r_{i}}, 2\varphi (m)/\varphi (p_i^{r_i})) &{} \text {if}\, p_{i}=2\ \text {and}\ r_{i}\ge 2,\\ \gcd (p_{i}^{r_{i}}, \varphi (m)/\varphi (p_i^{r_i})) &{} \text {otherwise}.\\ \end{array} \right. \end{aligned}$$

(1.1)

In fact, the above $d,d_i$ are equivalent to those $d,d_i$ defined in [3], respectively.

By [3, Proposition 2.2 (2)], the Carmichael quotient $C_{m}(x)$ induces a homomorphism

$$\begin{aligned} \phi _m:(\mathbb {Z}/m^{2}\mathbb {Z})^{*}\rightarrow (\mathbb {Z}/m\mathbb {Z},+), x\mapsto C_m(x), \end{aligned}$$

where $C_m(x) = (x^{\lambda (m)}-1)/m$ and $\lambda (m)$ is the Carmichael function.

Proposition 1.1

Let $m=p_{1}^{r_{1}}\cdots p_{k}^{r_{k}}$ be the prime factorization of $m \ge 2$. For $1\le i\le k$, put

$$\begin{aligned} d^{\prime }_{i}=\left\{ \begin{array}{ll} \gcd (p_{i}^{r_{i}}, 2\lambda (m)/\lambda (p_i^{r_i})) &{}\quad \text {if}\, p_{i}=2\ \text {and}\ r_{i}= 2,\\ \gcd (p_{i}^{r_{i}}, \lambda (m)/\lambda (p_i^{r_i})) &{}\quad \text {otherwise}. \end{array} \right. \nonumber \end{aligned}$$

Let $d^{\prime }=\prod _{i=1}^{k}d^{\prime }_{i}$. Then the image of the homomorphism $\phi _m$ is $d^{\prime }\mathbb {Z}/m\mathbb {Z}$.

Proof

We show the desired result case by case.

(I) First we prove the result for the case $k=1$, that is $m=p^r$, where p is a prime and r is a positive integer.

Suppose that $p=2$. If $r=2$, then $C_m(3)=2$, and for any positive integer n we have $C_m(2n+1)= n(n+1)$, which is even, so the image of $\phi _m$ is $2{\mathbb Z}/m{\mathbb Z}$. On the other hand, if $r = 1$ or $r \ge 3$, since $C_2(3)=1$ and $C_8(3)=1$, by using [3, Proposition 2.8] we see that $C_m(3)$ is an odd integer, so the image of $\phi _m$ is ${\mathbb Z}/m{\mathbb Z}$.

Now, assume that $p>2$. Note that $C_p(p+1) \equiv -1 \pmod p$, by [3, Proposition 2.8] we have $C_m(p+1) \equiv -1 \pmod p$, which implies that $p \not \mid C_m(p+1)$. Thus, there exists a positive integer n such that $n C_m(p+1) \equiv 1 \pmod m$. Then, by [3, Proposition 2.2 (1)] we deduce that $C_m((p+1)^n) \equiv 1 \pmod m$. So, the image of $\phi _m$ is ${\mathbb Z}/m{\mathbb Z}$.

(II) To complete the proof, we prove the result when $k \ge 2$.

For simplicity, denote $m_i = m/p_i^{r_i}$ and $n_i = \lambda (m) / \lambda (p_i^{r_i})$ for each $1\le i \le k$, and then let $m_i^{\prime }$ be an integer such that $m_i^{2}m_i^{\prime } \equiv 1 \pmod {p_i^{r_i}}$. By [3, Proposition 2.7], we have

$$\begin{aligned} C_m(a) \equiv \sum _{i=1}^{k} m_i m_i^{\prime } n_i C_{p_i^{r_i}}(a) \pmod {m}. \end{aligned}$$

(1.2)

So, for each $1\le i \le k$, $C_m(a) \equiv m_i m_i^{\prime } n_i C_{p_i^{r_i}}(a) \pmod {p_i^{r_i}}$. If $p_i=2$ and $r_i=2$, note that for any odd integer $a>1$, $C_4(a)$ is even, then we see that $d_i^{\prime } \mid n_i C_{p_i^{r_i}}(a)$, and thus $d_i^{\prime } \mid C_m(a)$. Otherwise if $p_i>2$ or $r_i \ne 2$, then $d_i^{\prime } \mid n_i$, and so $d_i^{\prime } \mid C_m(a)$. Hence, we have $d^{\prime } \mid C_m(a)$ for any integer a coprime to m.

Let $b = \gcd (m, m_1m_1^{\prime }n_1, \ldots , m_km_k^{\prime }n_k)$. Then, there exist integers $X_1,\ldots , X_k$ such that

$$\begin{aligned} b \equiv \sum _{i=1}^{k} m_i m_i^{\prime } n_i X_i \pmod {m}. \end{aligned}$$

(1.3)

If we denote $b_i = \gcd (p_i^{r_i}, m_i m_i^{\prime }n_i)$ for each $1 \le i \le k$, then $b = \prod _{i=1}^{k} b_i$; here, we remark that $b_i = \gcd (p_i^{r_i}, n_i)$. It is easy to see that for each $1 \le i \le k$, if $p_i>2$ or $r_i \ne 2$, we have $d_i^{\prime } = b_i$. Further, when $p_i=2$ and $r_i=2$, $d_i^{\prime } = 2b_i$ if $8 \not \mid \lambda (2p_1\ldots p_k)$, and $d_i^{\prime } = b_i$ otherwise.

We now have three cases for m:

• (i) There exists $1\le j \le k$ such that $p_j=2, r_j=2$ and

  $$\begin{aligned} 8 \not \mid \lambda (2p_1\ldots p_k). \end{aligned}$$

• (ii) There exists $1\le j \le k$ such that $p_j=2, r_j=2$ and

  $$\begin{aligned} 8 \mid \lambda (2p_1\ldots p_k). \end{aligned}$$

• (iii) All the other cases.

Clearly, in Cases (ii) and (iii) we have $d^{\prime }=b$, and in Case (i) $d^{\prime } = 2b$.

According to (I), there exist integers $a_i$ with $p_i \not \mid a_i$ for $1 \le i \le k$ defined by

$$\begin{aligned} C_{p_i^{r_i}}(a_{i}) \equiv \left\{ \begin{array}{ll} 2X_i &{} \text {in Case (i)}, \\ X_i &{} \text {in Case (iii)}, \\ &{} \qquad \qquad \qquad \qquad \qquad \pmod {p_i^{r_i}} \\ X_i &{} \text {in Case (ii)}\ \text {and}\ i\ne j, \\ 0 &{} \text {in Case (ii)}\ \text {and}\ i=j. \end{array} \right. \end{aligned}$$

By the Chinese Remainder Theorem, we can choose a positive integer a such that $a \equiv a_i \pmod {p_i^{2r_i}}$. So, by [3, Proposition 2.2 (2)] we have $C_{p_i^{r_i}}(a) \equiv C_{p_i^{r_i}}(a_i) \pmod {p_i^{r_i}}$. Then, combining with (1.3) and the relation between b and $d^{\prime }$, we obtain $m_im_i^{\prime }n_i C_{p_i^{r_i}}(a) \equiv d^{\prime } \pmod {p_i^{r_i}}$ for each $1\le i \le k$ in all the three cases. Finally, using (1.2) we have $C_m(a) \equiv d^{\prime } \pmod {m}$, which completes the proof. $\square$

Comparing (1.1) with Proposition 1.1, we have $d^{\prime } \mid d$. Moreover, by [3, Proposition 2.1] we get

$$\begin{aligned} \frac{\varphi (m)}{\lambda (m)}d^{\prime }{\mathbb Z}/m{\mathbb Z}= d{\mathbb Z}/m{\mathbb Z}, \end{aligned}$$

which implies that $\gcd (\frac{\varphi (m)}{\lambda (m)}d^{\prime },m)=d$.

3 Another error

We take this opportunity to correct another error. In the proof of [3, Lemma 3.4], the last identity “$\equiv \ell n^{-1} 2^{r-2}$” may be not true, and it should be deleted. Because by using $n^{2^{r-2}} \equiv 1 \pmod {2^r}$, we only know that $(n^{2^{r-2}}+1)/2$ is an odd integer, which may be not congruent to 1 modulo $2^r$. Clearly, this error does not change the result there.