A note on two orthogonal totally $C_4$-free one-factorizations of complete graphs

TL;DR

This paper proves the existence of pairs of orthogonal totally $C_4$-free one-factorizations for complete graphs of order $q+1$, where $q$ is a prime power congruent to 3 mod 4 and at least 11.

Contribution

It establishes the existence of such factorizations for a new class of complete graphs based on prime powers congruent to 3 mod 4.

Findings

Existence of orthogonal totally $C_4$-free one-factorizations for $K_{q+1}$ when $q ot o 4$ (mod 4) and $q extgreater 10$.

Construction methods for these factorizations based on properties of prime powers.

Extension of previous results on one-factorizations avoiding 4-cycles.

Abstract

A pair of orthogonal one-factorizations $\mathcal{F}$ and $\mathcal{G}$ of the complete graph $K_n$ is totally $C_4$-free, if the union $F\cup G$, for any $F,G\in\mathcal{F}\cup\mathcal{G}$, does not include a cycle of length four. In this note, we prove if $q\equiv3$ (mod 4) is a prime power with $q\geq11$, then there is a pair of orthogonal totally $C_4$-free one-factorizations of $K_{q+1}$.

Equations

$$S_{\beta }=\left\{\{x,x\beta \}:x\in QR(q)\right\},$$

$${\beta }^{t_2}+{\beta }^{t_1}-1=0,\text{mbox}where{t}_1,t_2\in \{-1,1\}$$

$$a({\beta }^{t_1+1}-{\beta }^{t_2+t_3})+i({\beta }^{t_2+t_3}-{\beta }^{t_3}-{\beta }^{t_1}+1)=0,$$

$$\beta^{t_{2}}+\beta^{t_{1}}-1=0,\mbox{where }t_{1},t_{2}\in\{-1,1\}\mbox{ with $t_{1}=t_{2}=t$},$$

$$a({\beta }^{t_1+1}-{\beta }^{t_2+t_3})+i({\beta }^{t_2+t_3}-{\beta }^{t_3}-{\beta }^{t_1}+1)=0,$$

$$M_1$$

$$M_2$$

Full text

In this note, we prove if $q\equiv 3$ (mod 4) is a prime power with $q\geq 11$, then there is a pair of orthogonal totally $C_{4}$-free one-factorizations of $K_{q+1}$.

Keywords. One-factorization, Strong starters, $C_{4}$-free.

1 Introduction

An one-factor of a graph $G$ is a regular spanning subgraph of degree one. An one-factorization of a graph G is a set $\mathcal{F}=\{F_{1},F_{2},\ldots,F_{n}\}$ of edge disjoint one-factors such that $E(G)=\displaystyle\cup_{i=1}^{n}E(F_{i})$. Two one-factorizations $\mathcal{F}=\{F_{1},F_{2},\ldots,F_{n}\}$ and $\mathcal{H}=\{H_{1},H_{2},\ldots,H_{n}\}$ of a graph $G$ are orthogonal if $|F\cap H|\leq 1$, for every $F\in\mathcal{F}$ and $H\in\mathcal{H}$. A one-factorization $\mathcal{F}=\{F_{1},F_{2},\ldots,F_{n}\}$ of a graph $G$ is said to be $k$-cycle free if $F_{i}\cup F_{j}$ does not contains a cycle of length $k$, $C_{k}$.

The existence of a $C_{4}$-free one-factorization of the complete graph $K_{n}$, for even $n\geq 6$, has already been observed by Phelps et. al. in [24], where they were used to give the existence of simple quadruple systems with index three.

Theorem 1.1**.**

[24]** A $C_{4}$-free one-factorization of complete graphs $K_{n}$ exists, if and only if, $n$ is even $n\geq 6$.

In general, Meszka in [20] proved for each even $n$ and each even $k\geq 4$ with $k\not=\frac{n}{2}$, the complete graph $K_{n}$ has a $C_{k}$-free one-factorization.

A pair of orthogonal one-factorizations $\mathcal{F}$ and $\mathcal{H}$ of the complete graph $K_{n}$ is $C_{4}$-free if $F\cup H$ does not contains a cycle of length four, for all $F\in\mathcal{F}$ and $H\in\mathcal{H}$. On the other hand, a pair of orthogonal one-factorizations $\mathcal{F}$ and $\mathcal{H}$ of complete the graph $K_{n}$ is totally $C_{4}$-free, if $F\cup H$ does not include a cycle of length four, for all $F,H\in\mathcal{F}\cup\mathcal{H}$ [4].

An interesting way for constructing one-factorizations of complete graphs is using starters of aditive Abelian groups of odd order: Let $\Gamma$ be a finite additive Abelian group of odd order $n=2k+1$, and let $\Gamma^{*}=\Gamma\setminus\{0\}$ be the set of non-zero elements of $\Gamma$. A starter for $\Gamma$ is a set $S=\{\{x_{1},y_{1}\},\ldots,\{x_{k},y_{k}\}\}$ such that $\left\{x_{1},\ldots,x_{k},y_{1},\ldots,y_{k}\right\}=\Gamma^{*}$ and $\left\{\pm(x_{i}-y_{i}):i=1,\ldots,k\right\}=\Gamma^{*}$. Moreover, if $\left\{x_{i}+y_{i}:i=1,\ldots,k\right\}\subseteq\Gamma^{*}$ and $\left|\left\{x_{i}+y_{i}:i=1,\ldots,k\right\}\right|=q$, then $S$ is called strong starter for $\Gamma$. There are some interesting results on strong starters for cyclic groups [1, 19], in particular for $\mathbb{Z}_{n}$ [23, 29, 31, 30] and $\mathbb{F}_{q}$ [1, 7, 8, 13, 21, 22], and for finite Abelian groups [12, 17].

Strong starters were first introduced by Mullin and Stanton in [27] in constructing of Room squares. Starters and strong starters have been useful to construct many combinatorial designs such as Room cubes [11], Howell designs [3], Kirkman triple systems [19, 25], Kirkman squares and cubes [26, 28], and factorizations of complete graphs [2, 10, 12, 16].

Let $\Gamma$ be a finite additive Abelian group of odd order $n=2k+1$. It is well known that $F_{\gamma}=\{\{\infty,\gamma\}\}\cup\{\{x_{i}+\gamma,y_{i}+\gamma\}:1\leq i\leq k\}$, for all $\gamma\in\Gamma$, forms a one-factorization of the complete graph on $\Gamma\cup\{\infty\}$. Hence, if $F_{0}=\{\{\infty,0\}\}\cup\{\{x_{i},y_{i}\}:1\leq i\leq k\}$, then $F_{\gamma}=F_{0}+\gamma$, for all $\gamma\in\Gamma^{*}$. On the other hand, let $S=\{\{x_{i},y_{i}\}:1\leq i\leq k\}$ and $T=\{\{u_{i},v_{i}\}:1\leq i\leq k\}$ be two starters for $\Gamma$. Without loss of generality, we assume $x_{i}-y_{i}=u_{i}-v_{i}$, for all $i=1,\ldots,k$. Then $S$ and $T$ are orthogonal starters if $u_{i}-x_{i}=u_{j}-x_{j}$ implies $i=j$, and if $u_{i}\neq x_{i}$, for all $i=1,\ldots,q$.

Let $S=\{\{x_{i},y_{i}\}:i=1,\ldots,k\}$ be a starter for a finite additive Abelian group $\Gamma$ of odd order $n=2k+1$. It is easy to see $-S=\{\{-x_{i},-y_{i}\}:i=1,\ldots,k\}$ is also a starter for $\Gamma$. Also, the set of pairs $P=\{\{x,-x\}:x\in\Gamma^{*}\}$ is a starter for $\Gamma$, called the patterned starter.

Theorem 1.2**.**

[15]** If there is a strong starter $S$ in an additive Abelian group of odd order, then $S$, $-S$ and $P$ are pairwise orthogonal starters.

Lemma 1.3**.**

[9]** For odd $n\geq 5$, the one-factorization of $K_{n+1}$ generated by the patterned starter is $C_{4}$-free, if and only if, $n\not\equiv 0$ (mod 3).

Let $q$ be an odd prime power. An element $x\in\mathbb{F}_{q}^{*}$ is called a quadratic residue if there exists an element $y\in\mathbb{F}_{q}^{*}$ such that $y^{2}=x$. If there is no such $y$, then $x$ is called a non-quadratic residue. The set of quadratic residues of $\mathbb{F}_{q}^{*}$ is denoted by $QR(q)$ and the set of non-quadratic residues is denoted by $NQR(q)$. It is well known $QR(q)$ is a cyclic subgroup of $\mathbb{F}_{q}^{*}$ of order $\frac{q-1}{2}$ (see for example [14]). As well as, it is well known, if either $x,y\in QR(q)$ or $x,y\in NQR(q)$, then $xy\in QR(q)$. Also, if $x\in QR(q)$ and $y\in NQR(q)$, then $xy\in NQR(q)$. For more details of this kind of results the reader may consult [5, 14].

An interesting (strong) starter for $\mathbb{F}_{q}$, the finte field of order prime power $q$. Horton in [15] proved (also proved by the author in [1]) the following

Proposition 1.4**.**

[15]** If $q\equiv 3$ (mod 4) is an odd prime power ($q\neq 3$) and $\beta\in NQR(q)\setminus\{-1\}$, then

[TABLE]

is a strong starter for $\mathbb{F}_{q}$.

In same paper,[15], Horton proved the following

Theorem 1.5**.**

[15]** Let $q\equiv 3$ (mod 4) be a prime power ($q\neq 3$). If $\beta_{1},\beta_{2}\in NQR(q)\setminus\{-1\}$, with $\beta_{1}\neq\beta_{2}$, then $S_{\beta_{1}}$ and $S_{\beta_{2}}$ are orthogonal.

In this note, we prove the following theorem given by Bao and Ji in [4].

Theorem 1.6**.**

[4]** Let $q\equiv$3 (mod 4) be an odd prime power with $q\geq 11$, then there is a pair of orthogonal totally $C_{4}$-free one-factorizations of $K_{q+1}$.

The method utilized for constructing such a pair of orthogonal totally $C_{4}$-free one-factorizations is different than used in [4]. Moreover, for the case when $q\equiv 3$ (mod 8), our result generalizes Theorem 7 given by Bao and Ji in [4].

2 Results

In this section, we present some results of orthogonal $C_{4}$-free one-factorizations of the complete graph. Also, we present the main result of this note.

Lemma 2.1**.**

Let $q\equiv 3$ (mod 4) be an odd prime power and $\beta\in NQR(q)\setminus\{-1\}$. If $\frac{\beta^{2}+1}{\beta-1}\in QR(q)$, then the one-factorization generated by the starter $S_{\beta}$ is $C_{4}$-free.

Proof.  Let $F_{i}=\{\{\infty,i\}\}\cup\{\{x+i,x\beta+i\}:x\in QR(q)\}$, for $i\in\mathbb{F}_{q}$. Then $\mathcal{F}=\{F_{i}:i\in\mathbb{F}_{q}\}$ is a one-factorization of the complete graph on $\mathbb{F}_{q}\cup\{\infty\}$. Assume this one-factorization is not $C_{4}$-free. Since $F_{i}=F_{0}+i$,then without loss of generality, it is sufficient to prove $F_{0}\cup F_{i}$, for some $i\in\mathbb{F}_{q}$, contain a cycle of length 4, $C_{4}$.

First, suppose that $\infty\in V(C_{4})$. Hence $\{\infty,0\},\{\infty,i\}\in E(C_{4})$, where $\{\infty,0\}\in F_{0}$ and $\{\infty,i\}\in F_{i}$. We have $\{-i+i,-i\beta^{t_{1}}+i\},\{i,i\beta^{t_{2}}\}\in E(C_{4})$, with $t_{1}=1$ if $-i\in QR(q)$, and $t_{1}=-1$ if $-i\in NQR(q)$, and $t_{2}=1$ if $i\in QR(q)$, and $t_{2}=-1$ if $i\in NQR(q)$. Since $F_{0}\cup F_{i}$ contain a cycle of length 4, then $\{\infty,0\}\{0,-i\beta^{t_{1}}+i\}\{i,i\beta^{t_{2}}\}\{\infty,i\}=C_{4}$. Therefore

[TABLE]

Since $-i^{2}\in NQR(q)$, then $t_{1}\neq t_{2}$, which implies that $\beta^{2}-\beta+1=0$. If $\beta^{2}+1\in NQR(q)$, then $\beta-1\in NQR(q)$. Therefore, $\beta^{2}-\beta+1=0$ implies $\beta(\beta-1)=-1$, which is a contradiction, since $\beta(\beta-1)\in QR(q)$. On the other hand, if $\beta^{2}+1\in QR(q)$ then $\beta-1\in QR(q)$, hence $\beta^{2}-\beta+1=0$ implies $\beta^{2}+1=\beta$, which is a contradiction, since $\beta\in NQR(q)$.

Now, suppose that $\infty\not\in V(C_{4})$. Since $E(C_{4})\cap F_{0}\neq\emptyset$, then without loss of generality there is $a\in V(C_{4})\cap QR(q)$ such that

$\{a,a\beta\}\in F_{0}$ and $\{a\beta,(a\beta-i)\beta^{t_{1}}+i\}\in F_{i}$,

where $t_{1}=1$ if $a\beta-i\in QR(q)$, and $t_{1}=-1$ if $a\beta-i\in NQR(q)$. On the other hand

$\{a,(a-i)\beta^{t_{2}}+i\}\in F_{i}$ and $\{(a-i)\beta^{t_{2}}+i,((a-i)\beta^{t_{2}}+i)\beta^{t_{3}}\}\in F_{0}$,

where $t_{2}=1$ if $a-i\in QR(q)$, and $t_{2}=-1$ if $a-i\in NQR(q)$, and $t_{3}=1$ if $(a-i)\beta^{t_{2}}+i\in QR(q)$, and $t_{3}=-1$ if $(a-i)\beta^{t_{2}}+i\in NQR(q)$. Therefore, we have

[TABLE]

for all $t_{1},t_{2},t_{3}\in\{-1,1\}$.

• case (i)

Suppose $t_{1}=t_{2}=t_{3}=1$. In this case $\beta^{2}-2\beta+1=0$, which is a contradiction.

• case (ii)

Suppose $t_{1}=t_{2}=1$ and $t_{3}=-1$. In this case $i=\frac{a\beta(\beta+1)}{\beta-1}$. Since $a-i\in QR(q)$, we have $a-i=-a\frac{\beta^{2}+1}{\beta-1}\in NQR(q)$, which is a contradiction, since $\frac{\beta^{2}+1}{\beta-1}\in QR(q)$ and $-a\in NQR(q)$.

• case (iii)

Suppose $t_{1}=t_{3}=1$ and $t_{2}=-1$. In this case $i=\frac{a(\beta+1)}{2}$. We have $\frac{(a-i)\beta^{-1}+i}{a\beta-i}\in QR(q)$, since $a\beta-i\in QR(q)$ and $(a-i)\beta^{-1}+i\in QR(q)$. However $\frac{(a-i)\beta^{-1}+i}{a\beta-i}=\beta^{-1}\frac{\beta^{2}+1}{\beta-1}\in NQR(q)$, a contradiction.

• case (iv)

Suppose $t_{2}=t_{3}=1$ and $t_{1}=-1$. In this case $i=a\beta\frac{(\beta+1)}{\beta^{2}+1}$. Since $a-i\in QR(q)$, we have $a-i=-a\frac{\beta-1}{\beta^{2}+1}\in NRQ(q)$, which is a contradiction.

• case (v)

Suppose $t_{1}=1$ and $t_{2}=t_{3}=-1$. In this case $i=a(\beta+1)$. Since $a-i\in NQR(q)$, we have $a-i=-a\beta\in QR(q)$, a contradiction.

• case (vi)

Suppose $t_{2}=1$ and $t_{1}=t_{3}=-1$. In this case $\beta-1=0$, which is a contradiction.

• case (vii)

Suppose $t_{3}=1$ and $t_{1}=t_{2}=-1$. In this case $\beta^{2}-2\beta+1=0$, which is a contradiction.

• case (viii)

Suppose $t_{1}=t_{2}=t_{3}=-1$. In this case $i=-a\frac{\beta+1}{\beta-1}$. Since $a\beta-i\in NQR(q)$, we have $a\beta-i=a\frac{\beta^{2}+1}{\beta-1}\in QR(q)$, a contradiction.

Example 1**.**

For $q=19$, if $\beta\in\{2,3,10,13\}$, then $\frac{\beta^{2}+1}{\beta-1}\in QR(q)$. It is not difficult to check the one-factorization generated by the starter $S_{\beta}$ is $C_{4}$-free.

Lemma 2.2**.**

Let $q\equiv 3$ (mod 4) be an odd prime power and $\beta\in NQR(q)\setminus\{-1\}$. If either $\frac{\beta^{2}+1}{2}\in QR(q)$ or $\frac{\beta-1}{2}\in QR(q)$, with $\beta\not\in\{2,\frac{1}{2}\}$, then the pair of orthogonal one-factorizations generated by the starter $S_{\beta}$ and $S_{-\beta}$ is $C_{4}$-free.

Proof.  Let $F_{i}=\{\{\infty,i\}\}\cup\{\{x+i,x\beta+i\}:x\in QR(q)\}$ and $G_{i}=\{\{\infty,i\}\}\cup\{\{-x+i,-x\beta+i\}:x\in QR(q)\}$, for $i\in\mathbb{F}_{q}$. Then $\mathcal{F}=\{F_{i}:i\in\mathbb{F}_{q}\}$ and $\mathcal{G}=\{G_{i}:i\in\mathbb{F}_{q}\}$ are orthogonal one-factorizations of the complete graph on $\mathbb{F}_{q}\cup\{\infty\}$. Assume this pair of one-factorizations is not $C_{4}$-free. Since $F_{i}=F_{0}+i$ and $G_{i}=G_{0}+i$, then without loss of generality we can suppose $F_{0}\cup G_{i}$, for some $i\in\mathbb{F}_{q}$, contain a cycle of length 4, $C_{4}$.

First, suppose that $\infty\in V(C_{4})$. Hence, $\{\infty,0\},\{\infty,i\}\in E(C_{4})$, where $\{\infty,0\}\in F_{0}$ and $\{\infty,i\}\in G_{i}$. Therefore, we have $\{0,-i\beta^{t_{1}}+i\},\{i,i\beta^{t_{2}}\}\in E(C_{4})$, with $t_{1}=t_{2}=1$ if $i\in QR(q)$, and $t_{1}=t_{2}=-1$ if $i\in NQR(q)$. Since $F_{0}\cup F_{i}$ contain a cycle of length 4, then $\{\infty,0\}\{0,-i\beta^{t_{1}}+i\}\{i,i\beta^{t_{2}}\}\{\infty,i\}=C_{4}$. Hence

[TABLE]

which implies that $2\beta^{t}=1$, which is a contradiction if $\beta\not\in\{2,\frac{1}{2}\}$.

Now, suppose $\infty\not\in V(C_{4})$. Since $-1\in NQR(q)$, then $NQR(q)=-QR(q)$. Therefore, $G_{i}=\{\{y+i,y\beta+i\}:y\in NQR(q)\}$. Since $E(C_{4})\cap F_{0}\neq\emptyset$, without loss of generality there is $a\in V(C_{4})\cap QR(q)$ such that

$\{a,a\beta\}\in F_{0}$ and $\{a\beta,(a\beta-i)\beta^{t_{1}}+i\}\in G_{i}$,

where $t_{1}=1$ if $a\beta-i\in NQR(q)$, and $t_{1}=-1$ if $a\beta-i\in QR(q)$.

On the other hand

$\{a,(a-i)\beta^{t_{2}}+i\}\in G_{i}$ and $\{(a-i)\beta^{t_{2}}+i,((a-i)\beta^{t_{2}}+i)\beta^{t_{3}}\}\in F_{0}$,

where $t_{2}=1$ if $a-i\in NQR(q)$, and $t_{2}=-1$ if $a-i\in QR(q)$, and $t_{3}=1$ if $(a-i)\beta^{t_{2}}+i\in QR(q)$, and $t_{3}=-1$ if $(a-i)\beta^{t_{2}}+i\in NQR(q)$.

Therefore, we have

[TABLE]

for all $t_{1},t_{2},t_{3}\in\{-1,1\}$, which is an analogous equation obtained in the proof of Lemma 2.1.

• case (i)

Suppose $t_{1}=t_{2}=t_{3}=1$. In this case $\beta^{2}-2\beta+1=0$, which is a contradiction.

• case (ii)

Suppose $t_{1}=t_{2}=1$ and $t_{3}=-1$. In this case $i=a\frac{\beta(\beta+1)}{\beta-1}$. We have that $a\beta-i\in NQR(q)$, but $a\beta-i=-a\beta\frac{2}{\beta-1}\in QR(q)$, since $-a\beta\in QR(q)$, which is a contradiction. On the other hand, since $a-i\in NQR(q)$ and $a\beta-i\in NQR(q)$, then $\frac{a-i}{a\beta-i}\in QR(q)$, but $\frac{a-i}{a\beta-i}=\frac{\beta^{2}+1}{2\beta}\in NQR(q)$, which is a contradiction.

• case (iii)

Suppose $t_{1}=t_{3}=1$ and $t_{2}=-1$. In this case $i=a\frac{\beta+1}{2}$. We have $a\beta-i\in NQR(q)$, but $a\beta-i=a\frac{\beta-1}{2}\in QR(q)$, which is a contradiction. On the other hand, we have $(a-i)\beta^{-1}+i\in QR(q)$, but $(a-i)\beta^{-1}+i=\frac{a(\beta^{2}+1)}{2\beta}\in NQR(q)$, which is a contradiction, since $\frac{a}{\beta}\in NQR(q)$.

• case (iv)

Suppose $t_{2}=t_{3}=1$ and $t_{1}=-1$. In this case $i=a\beta\frac{(\beta+1)}{\beta^{2}+1}$. Since $(a-i)\beta+i\in QR(q)$, we have $(a-i)\beta+i=a\beta\frac{2}{\beta^{2}+1}\in NRQ(q)$, which is a contradiction. On the other hand, since $a-i\in NQR(q)$ and $(a-i)\beta+i\in QR(q)$, then $\frac{a-i}{(a-i)\beta+i}\in NQR(q)$, but $\frac{a-i}{(a-i)\beta+i}=-\frac{\beta-1}{2\beta}\in QR(q)$, which is a contradiction, since $-\beta^{-1}\in QR(q)$.

• case (v)

Suppose $t_{1}=1$ and $t_{2}=t_{3}=-1$. In this case $i=a(\beta+1)$. If there is a cycle of lenght 4, then $a\neq(a\beta-i)\beta+i=((a-i)\beta^{-1}+i)\beta^{-1}$, but $(a\beta-i)\beta+i=a=((a-i)\beta^{-1}+i)\beta^{-1}$, which is a contradiction.

• case (vi)

Suppose $t_{2}=1$ and $t_{1}=t_{3}=-1$. In this case $\beta-1=0$, which is a contradiction.

• case (vii)

Suppose $t_{3}=1$ and $t_{1}=t_{2}=-1$. In this case $\beta^{2}-2\beta+1=0$, which is a contradiction.

• case (viii)

Suppose $t_{1}=t_{2}=t_{3}=-1$. In this case $i=-a\frac{\beta+1}{\beta-1}$. We have $a-i\in QR(q)$, but $a-i=a\beta\frac{2}{\beta-1}\in NQR(q)$, wich is a contradiction. On the other hand, since $a\beta-i\in QR(q)$ and $a-i\in QR(q)$, then $\frac{a\beta-i}{a-i}\in QR(q)$, but $\frac{a\beta-i}{a-i}=\frac{\beta^{2}+1}{2\beta}\in NQR(q)$, which is a contradiction.

2.1 Main result

Let $q=ef+1$ be a prime power and let $H$ be the subgroup of $\mathbb{F}_{q}^{*}$ of order $f$ with $\{H=C_{0},\ldots,C_{e-1}\}$ the set of (multiplicative) cosets of $H$ in $\mathbb{F}_{q}^{*}$ (that is, $C_{i}=g^{i}C_{0}$, where $g$ is the least primitive element of $\mathbb{F}_{q}$). The cyclotomic number $(i,j)$ is defined as $|\{x\in C_{i}:x+1\in C_{j}\}|$. In particular, if $e=2$ and $f$ is odd, then $(0,0)=\frac{f-1}{2}$, $(0,1)=\frac{f+1}{2}$, $(1,0)=\frac{f-1}{2}$ and $(1,1)=\frac{f-1}{2}$, see [6], Table VII.8.50. Hence, if $C_{0}=QR(q)$ and $C_{1}=NQR(q)$ are the cosets of $QR(q)$ in $\mathbb{F}_{q}^{*}$, then the following is satisfied:

Lemma 2.3**.**

Let $p\equiv 3$ (mod 4) be an odd prime power with $p\neq 3$. Then, there exists $\beta_{1},\beta_{2}\in NQR(q)$ such that

$(\beta_{1}+1)\in NQR(q)$* and $(\beta_{2}+1)\in QR(q)$.* 2. 2.

$(\beta_{1}-1)\in NQR(q)$* and $(\beta_{2}-1)\in QR(q)$.*

Which implies too that:

Lemma 2.4**.**

Let $p\equiv 3$ (mod 4) be an odd prime power with $p\neq 3$. Then there are $\alpha_{1},\alpha_{2}\in QR(q)$ such that

$(\alpha_{1}+1)\in NQR(q)$* and $(\alpha_{2}+1)\in QR(q)$.* 2. 2.

$(\alpha_{1}-1)\in NQR(q)$* and $(\alpha_{2}-1)\in QR(q)$.*

The following Lemma 2.5 is analogous then given in [1], and we give the proof given in the same paper.

Lemma 2.5**.**

Let $p\equiv 3$ (mod 4) be an odd prime power with $p\neq 3$. Then there is $\beta\in NQR(q)$ such that $(\beta+1)(\beta-1)\in NQR(q)$.

Proof.  For each $\beta\in NQR(q)$ define $A_{\beta}=\{a^{\beta}_{1},\ldots,a^{\beta}_{l_{\beta}}\}$, where $a^{\beta}_{i}\in NQR(q)$ and $a^{\beta}_{i+1}=a^{\beta}_{i}+1$, for all $i=1,\ldots,l_{\beta-1}$ with $\beta=a_{1}^{\beta}$. By Lemma 2.3, there is $\beta\in NQR(q)$ such that $|A_{\beta}|>1$. If $\beta^{*}=a^{\beta}_{l_{\beta}}$ then $(\beta^{*}+1)\in NQR(q)$ and $(\beta^{*}-1)\in QR(q)$. On the other hand, if $\beta^{*}=\beta_{1}$ then $(\beta^{*}+1)\in QR(q)$ and $(\beta^{*}-1)\in NQR(q)$. Hence $(\beta^{*}+1)(\beta^{*}-1)\in NQR(q)$.

Lemma 2.6**.**

Let $p\equiv 3$ (mod 4) be an odd prime power with $p\neq 3$. Then there is $\alpha\in QR(q)$ such that $(\alpha+1)(\alpha-1)\in NQR(q)$.

Proof.  For each $\alpha\in QR(q)$ define $A_{\alpha}=\{a^{\alpha}_{1},\ldots,a^{\alpha}_{l_{\alpha}}\}$, where $a^{\alpha}_{i}\in QR(q)$ and $a^{\alpha}_{i+1}=a^{\alpha}_{i}+1$, for all $i=1,\ldots,l_{\alpha-1}$ with $\alpha=a_{1}^{\alpha}$. By Lemma 2.3, there is $\alpha\in QR(q)$ such that $|A_{\alpha}|>1$. If $\alpha^{*}=a^{\alpha}_{l_{\alpha}}$ then $(\alpha^{*}+1)\in NQR(q)$ and $(\alpha^{*}-1)\in QR(q)$. On the other hand, if $\alpha^{*}=\alpha_{1}$ then $(\alpha^{*}+1)\in QR(q)$ and $(\alpha^{*}-1)\in NQR(q)$. Hence $(\alpha^{*}+1)(\alpha^{*}-1)\in NQR(q)$.

The following theorem is the main theorem of this note.

Theorem 2.7**.**

Let $q\equiv$3 (mod 4) be an odd prime power with $q\geq 11$, then there is a pair of orthogonal totally $C_{4}$-free one-factorizations of $K_{q+1}$.

Proof.  Let $M=M_{1}\cup M_{2}$, where

$M_{1}=\left\{\beta\in NQR(q)\setminus\{2,2^{-1},-1\}:\frac{\beta^{2}+1}{2}\in QR(q),\beta^{3}\neq-1\right\},$

$M_{2}=\left\{\beta\in NQR(q)\setminus\{2,2^{-1},-1\}:\frac{\beta-1}{2}\in QR(q),\beta^{3}\neq-1\right\}$,

By Lemma 2.1 and Lemma 2.2, we need only show that $M$ is not the empty set: Since $-1\in NQR(q)$ then $\beta\to\beta^{2}$, for all $\in\beta\in NQR(q)$, is a bijection between $NQR(q)$ and $QR(q)$. By Lemma 2.6 there is $\beta\in NQR(q)$ such that $(\beta^{2}+1)(\beta^{2}-1)\in NQR(q)$. Furthermore, there are $\beta_{1},\beta_{2}\in NQR(q)$ such that $\beta^{2}_{1}-1\in NQR(q)$ and $\beta^{2}_{1}+1\in QR(q)$, and $\beta^{2}_{2}-1\in QR(q)$ and $\beta^{2}_{2}+1\in NQR(q)$, see proof of Lemma 2.6.

If $2\in QR(q)$ then let $\beta\in NQR(q)$ such that $\beta^{2}+1\in QR(q)$ and $\beta^{2}-1\in NQR(q)$, by Lemma 2.6. We assume that $\beta^{3}\neq-1$, since if $\beta^{3}+1=0$ then $\beta^{2}-\beta+1=0$, which implies that $\beta^{2}+1=\beta$, a contradiction, since $\beta^{2}+1\in QR(q)$ and $\beta\in NQR(q)$. Hence $M_{1}\neq\emptyset$. Since $(\beta^{2}-1)=(\beta-1)(\beta+1)\in NQR(q)$, then either $\beta-1\in QR(q)$ and $\beta+1\in NQR(q)$, which implies that $M_{2}\neq\emptyset$, or $\beta-1\in NQR(q)$ and $\beta+1\in QR(q)$. On the other hand, if $2\in NQR(q)$ then let $\beta\in NQR(q)$ such that $\beta-1\in NQR(q)$ and $\beta+1\in QR(q)$, see proof of Lemma 2.5. We assume that $\beta^{3}\neq-1$, since if $\beta^{3}+1=0$ then $\beta^{2}-\beta+1=0$, which implies that $\beta(\beta-1)=-1$, which is a contradiction, since $\beta(\beta-1)\in QR(q)$. Since $\beta-1\in NQR(q)$ then $\beta\neq 2$. Hence $M_{2}\neq\emptyset$. Finally, if $\beta^{2}+1\in NQR(q)$ then $M_{1}\neq\emptyset$.∎

It is well known, if $q\equiv 3$ (mod 8), then $2\in NQR(q)$ (see for example [18]). Therefore

Corollary 2.8**.**

Let $q\equiv$3 (mod 8) be an odd prime power with $q\geq 11$. If $M=M_{1}\cup M_{2}$, where

[TABLE]

then the one-factorization generated by the starters $S_{\beta_{1}}$ and $S_{\beta_{2}}$, for any pair of different elements $\beta_{1},\beta_{2}\in M$, are totally $C_{4}$-free of $K_{q+1}$.

Hence, for the case when $q\equiv 3$ (mod 8), Corollary 2.8 generalizes Theorem 7 given by Bao and Ji in [4].

Theorem 2.9**.**

[4]** Let $q\equiv$3 (mod 8) be an odd prime power with $q\geq 11$. If

$M=\left\{\beta\in NQR(q)\setminus\{2,2^{-1},-1\}:\beta^{2}+1\in NQR(q),\beta^{3}\neq-1\right\},$**

then the one-factorization generated by the starters $S_{\beta_{1}}$ and $S_{\beta_{2}}$, for any pair of different elements $\beta_{1},\beta_{2}\in M$, are totally $C_{4}$-free of $K_{q+1}$.

Acknowledgment

Research was partially supported by SNI and CONACyT.