In this section, we give several lemmas, which are essential in the proofs of our theorems. Hereinafter, we are going to use some properties of the classical Gauss sums, which can be found in some analytic number theory books, such as [13]; so we will not repeat them here. For convenience, first, we give the definition of the classical Gauss sums \(\tau (\chi )\) as follows: for any integer \(q>1\), let χ be any Dirichlet character \(\bmod\ q\). Then the famous Gauss sum \(\tau ( \chi )\) is defined as
$$ \tau (\chi )=\sum_{a=1}^{q}\chi (a)e \biggl(\frac{a}{q} \biggr), $$
where \(e(y)=e^{2\pi iy}\). With this mark, we have the following:
Lemma 1
Given
p
be an odd prime with
\(p\equiv 1 \bmod\ 3\), and let
ψ
be any third-order character
\(\bmod \ p\). Then we have the identity
$$ \tau ^{3} (\psi )+ \tau ^{3} (\overline{\psi } )=dp, $$
where
d
is uniquely determined by
\(4p=d^{2}+27b^{2}\)
and
\(d\equiv 1 \bmod\ 3\).
Proof
See Lemma 3 of [9] or references [14] and [10]. □
Lemma 2
Given
p
be an odd prime with
\(p\equiv 1\bmod\ 3\), and for any integer
b
with
\((b, p)=1\), let
\(U(b, p)=\sum_{a=0}^{p-1}e (\frac{ba^{3}}{p} )\). Then we have the identity
$$ U^{3}(b,p)=dp+ 3p\cdot U(b, p), $$
where
d
is the same as in Lemma 1.
Proof
Let χ be any third-order character \(\bmod\ p\). Then \(\chi ^{2}=\overline{\chi }\), and from Lemma 1 and the properties of Gauss sums we have
$$\begin{aligned} U(b,p)&=\sum_{a=0}^{p-1}e \biggl( \frac{ba^{3}}{p} \biggr)=1+\sum_{a=1} ^{p-1}e \biggl(\frac{ba^{3}}{p} \biggr) \\ & =1+ \sum_{a=1}^{p-1} \bigl(1+\chi (a)+ \overline{\chi }(a) \bigr)e \biggl(\frac{ba}{p} \biggr) \\ &= \sum_{a=0}^{p-1}e \biggl(\frac{ba}{p} \biggr)+\sum_{a=1}^{p-1} \chi (a)e \biggl( \frac{ba}{p} \biggr)+ \sum_{a=1}^{p-1} \overline{ \chi }(a)e \biggl(\frac{ba}{p} \biggr) \\ &= \overline{\chi }(b)\tau (\chi )+ \chi (b)\tau (\overline{ \chi } ). \end{aligned}$$
(2)
Note that \(\chi ^{3}=\chi _{0}\) and \(\tau (\chi )\tau (\overline{ \chi } )=p\), so from (2) we immediately infer that
$$\begin{aligned} U^{3}(b,p)&= \bigl(\overline{\chi }(b)\tau (\chi )+ \chi (b)\tau ( \overline{\chi } ) \bigr)^{3} \\ &= \tau ^{3}(\chi )+ 3\tau (\chi )\tau (\overline{\chi } ) \bigl( \overline{\chi }(b)\tau (\chi )+ \chi (b)\tau (\overline{ \chi } ) \bigr)+ \tau ^{3} (\overline{\chi } ) \\ &= \tau ^{3}(\chi )+ \tau ^{3} (\overline{\chi } )+ 3p\cdot U(b, p)=dp+3p\cdot U(b,p). \end{aligned}$$
This proves Lemma 2. □
Lemma 3
Given
p
be an odd prime with
\(p\equiv 1\bmod\ 4\), and let
ψ
be any fourth-order character
\(\bmod\ p\). Then we have the identity
$$ \tau ^{2}(\psi )+ \tau ^{2} (\overline{\psi } )=\sqrt{p} \cdot \sum_{a=1}^{p-1} \biggl(\frac{a+\overline{a}}{p} \biggr)= 2 \sqrt{p}\cdot \alpha , $$
where
\(\alpha = \sum_{a=1}^{\frac{p-1}{2}} (\frac{a+\overline{a}}{p} )\), \((\frac{*}{p} )\)
denotes the Legendre symbol
\(\bmod\ p\), and
a̅
denotes the multiplicative inverse of
\(a\bmod\ p\), that is, \(a\overline{a}\equiv 1\bmod\ p\).
Proof
In fact, this is Lemma 2.2 in [15]. Therefore we omit its proof. □
Lemma 4
Let
p
be an odd prime with
\(p\equiv 1\bmod\ 4\), and for any integer
b
with
\((b, p)=1\), let
\(A(b, p)=\sum_{a=0}^{p-1}e (\frac{ba^{4}}{p} )\). Then we have the identities
$$ A^{4}(b,p)=2p \bigl(C(p)+2 \bigr)A^{2}(b,p)+8p\alpha A(b,p) +4p\alpha ^{2}-C^{2}(p)p^{2}, $$
where
\(C(p)=-3\)
if
\(p=8k+5\)
and
\(C(p)=1\)
if
\(p=8k+1\).
Proof
Let ψ be a fourth-order character \(\bmod\ p\). Then \(\psi ^{2}=\chi _{2}\) (the Legendre symbol \(\bmod\ p\)), and by the properties of Gauss sums we get
$$\begin{aligned} A(b,p)&=\sum_{a=0}^{p-1}e \biggl( \frac{ba^{4}}{p} \biggr)=1+\sum_{a=1} ^{p-1}e \biggl(\frac{ba^{4}}{p} \biggr) \\ & = 1+ \sum_{a=1}^{p-1} \bigl(1+\psi (a)+ \chi _{2}(a)+ \overline{ \psi }(a) \bigr)e \biggl(\frac{ba}{p} \biggr) \\ &= \sum_{a=0}^{p-1}e \biggl(\frac{ba}{p} \biggr)+\sum_{a=1}^{p-1} \psi (a)e \biggl( \frac{ba}{p} \biggr)+ \sum_{a=1}^{p-1}\chi _{2}(a)e \biggl(\frac{ba}{p} \biggr)+ \sum _{a=1}^{p-1}\overline{\psi }(a)e \biggl( \frac{ba}{p} \biggr) \\ &= \chi _{2}(b)\tau (\chi _{2})+ \overline{\psi }(b)\tau ( \psi )+ \psi (b)\tau (\overline{\psi } ). \end{aligned}$$
(3)
Note that \(\psi ^{2}=\overline{\psi }^{2}=\chi _{2}\), \(\tau (\psi ) \tau (\overline{\psi } )=\psi (-1)\tau (\psi )\overline{ \tau (\psi )}=\psi (-1)p\), and \(\tau (\chi _{2})= \sqrt{p}\). Combining (3) and Lemma 2, we have
$$\begin{aligned} A^{2}(b,p)={}& \bigl(\chi _{2}(b)\tau (\chi _{2})+ \overline{\psi }(b) \tau (\psi )+ \psi (b)\tau (\overline{\psi } ) \bigr) ^{2} \\ ={}& p + 2\chi _{2}(b)\sqrt{p} \bigl( \overline{\psi }(b)\tau (\psi )+ \psi (b)\tau (\overline{\psi } ) \bigr)+ \chi _{2}(b) \bigl(\tau ^{2}(\psi )+ \tau ^{2} (\overline{\psi } ) \bigr) \\ &{}+ 2\tau (\psi )\tau (\overline{\psi } ) \\ ={}& p +2\chi _{2}(b)\sqrt{p}A(b, p)-2p+ \chi _{2}(b)2\sqrt{p} \cdot \alpha + \psi (-1)2p \\ = {}&C(p)p +2\chi _{2}(b)\sqrt{p}A(b, p)+ \chi _{2}(b)2\sqrt{p} \cdot \alpha , \end{aligned}$$
(4)
where \(C(p)=-3\) if \(p=8k+5\) and \(C(p)=1\) if \(p=8k+1\).
Now, according to (4), we have
$$\begin{aligned} \bigl(A^{2}(b, p)-C(p)p \bigr)^{2}&=A^{4}(b, p)-2C(p)p A^{2}(b,p)+C ^{2}(p)p^{2} \\ &= \bigl(2\sqrt{p}A(b,p)+ 2\sqrt{p}\cdot \alpha \bigr)^{2}=4p \bigl(A^{2}(b,p)+2\alpha A(b,p)+\alpha ^{2} \bigr). \end{aligned}$$
(5)
Applying (5), we immediately deduce that
$$ A^{4}(b,p)=2p \bigl(C(p)+2 \bigr)A^{2}(b,p)+8p\alpha A(b,p) +4p\alpha ^{2}-C^{2}(p)p^{2}. $$
This proves Lemma 4. □
Lemma 5
Let
p
be an odd prime with
\(p\equiv 1\bmod\ 3\), and let
χ
be any third-order character
\(\bmod\ p\). Then we have the identities
$$\begin{aligned} M_{1}(3, \chi ; p)=p-1;\qquad M_{2}(3,\chi ;p)= \frac{p-1}{p} \cdot \tau ^{3}(\chi );\qquad M_{3}(3,\chi ; p)= 3p(p-1) \end{aligned}$$
and
$$\begin{aligned} M_{k}(3, \chi ; p) =3p\cdot M_{k-2}(3,\chi ; p)+ dp\cdot M_{k-3}(3, \chi ; p) \end{aligned}$$
for all integers
\(k\geq 4\), where
d
is the same as in Lemma 1.
Proof
It is obvious that \(M_{1}(3,\chi ; p)=p-1\). According to (2) and the properties of Gauss sums, we get
$$\begin{aligned} M_{2}(3, \chi ; p)&=\frac{1}{\tau (\overline{\chi } )} \sum _{b=1}^{p-1}\overline{\chi }(b)\sum _{c=0}^{p-1}\sum_{d=0}^{p-1} e \biggl(\frac{b(c^{3}+d^{3})}{p} \biggr) \\ &= \frac{1}{\tau (\overline{\chi } )}\sum_{b=1}^{p-1}\overline{ \chi }(b) \bigl(\overline{\chi }(b)\tau (\chi )+ \chi (b)\tau (\overline{ \chi } ) \bigr)^{2} \\ &= \frac{1}{\tau (\overline{\chi } )}\sum_{b=1}^{p-1}\overline{ \chi }(b) \bigl(\overline{\chi }^{2}(b)\tau ^{2}(\chi )+ 2p+ \chi ^{2}(b) \tau ^{2} (\overline{\chi } ) \bigr) \\ &= \frac{\tau ^{2}(\chi )}{\tau (\overline{\chi } )} \cdot (p-1)=\frac{p-1}{p}\cdot \tau ^{3}(\chi ), \end{aligned}$$
(6)
$$\begin{aligned} M_{3}(3, \chi ; p)&=\frac{1}{\tau (\overline{\chi } )} \sum _{b=1}^{p-1}\overline{\chi }(b)\sum _{c=0}^{p-1}\sum_{d=0}^{p-1} \sum_{e=0}^{p-1} e \biggl(\frac{b(c^{3}+d^{3}+e^{3})}{p} \biggr) \\ &= \frac{1}{\tau (\overline{\chi } )}\sum_{b=1}^{p-1}\overline{ \chi }(b) \bigl(\overline{\chi }(b)\tau (\chi )+ \chi (b)\tau (\overline{ \chi } ) \bigr)^{3} \\ &= \frac{1}{\tau (\overline{\chi } )}\sum_{b=1}^{p-1}\overline{ \chi }(b) \bigl(\tau ^{3}(\chi )+ 3pU(b,p)+ \tau ^{3} ( \overline{ \chi } ) \bigr) = 3p(p-1). \end{aligned}$$
(7)
For any integer \(k\geq 4\), according to Lemma 1, we have
$$\begin{aligned} M_{k}(3, \chi ; p)&=\frac{1}{\tau (\overline{\chi } )} \sum _{b=1}^{p-1}\overline{\chi }(b) \Biggl(\sum _{a=0}^{p-1}e \biggl(\frac{ba ^{3}}{p} \biggr) \Biggr)^{k} \\ &= \frac{1}{\tau (\overline{\chi } )}\sum_{b=1}^{p-1}\overline{ \chi }(b)U^{k-3}(b, p) \bigl(\tau ^{3}(\chi )+\tau ^{3} (\overline{ \chi } )+3p U(b,p) \bigr) \\ &= \frac{3p}{\tau (\overline{\chi } )}\sum_{b=1}^{p-1} \overline{ \chi }(b)U^{k-2}(b,p)+ \frac{dp}{\tau (\overline{\chi } )}\sum _{b=1}^{p-1}\overline{ \chi }(b)U^{k-3}(b,p) \\ &= 3pM_{k-2}(3,\chi ; p)+dp\cdot M_{k-3}(3, \chi ; p). \end{aligned}$$
(8)
Now Lemma 5 follows from (6), (7), and (8). □
Lemma 6
Let
\(p=8h+5\)
be a prime, and let
χ
be any fourth-order character
\(\bmod\ p\). Then we have the identities
$$\begin{aligned} &M_{1}(4, \chi ; p)=p-1; \\ &M_{2}(4, \chi ; p) =2(p-1)\sqrt{p}\cdot \frac{\tau (\chi )}{\tau (\overline{ \chi })}=- \frac{2(p-1)}{\sqrt{p}}\cdot \tau ^{2}(\chi ); \\ &M_{3}(4, \chi ; p)=(p-1) \frac{\tau ^{3}(\chi )}{\tau (\overline{\chi })}=-\frac{p-1}{p}\cdot \tau ^{4}(\chi ). \end{aligned}$$
For every integer
\(k\geq 0\), we obtain the fourth-order linear recurrence formula
$$\begin{aligned} M_{k+4}(4, \chi ; p)=-2pM_{k+2}(4, \chi ;p) + 8p\alpha M_{k+1}(4, \chi ; p) -p \bigl(9p-4\alpha ^{2} \bigr)M_{k}(4, \chi ; p), \end{aligned}$$
where
\(M_{0}(4, \chi ; p)=0\), and
α
is the same as in Lemma 3.
Proof
Let p be an odd prime with \(p=8h+5\), and let χ be any fourth-order character \(\bmod\ p\). Then this time we obtain \(C(p)=-3\). For any integer \(k\geq 0\), according to the properties of Gauss sums and Lemma 4, we get
$$\begin{aligned} M_{k+4}(4, \chi ; p)&=\frac{1}{\tau (\overline{\chi })}\sum _{x_{1}=0} ^{p-1}\sum_{x_{2}=0}^{p-1} \cdots \sum_{x_{k+4}=0}^{p-1}\sum _{b=1}^{p-1}\overline{ \chi }(b) e \biggl( \frac{b(x_{1}^{4}+x_{2}^{2}+\cdots +x_{k+4}^{4})}{p} \biggr) \\ &= \frac{1}{\tau (\overline{\chi })}\sum_{b=1}^{p-1} \overline{\chi }(b) \Biggl(\sum_{x=0}^{p-1}e \biggl(\frac{bx^{4}}{p} \biggr) \Biggr)^{k+4}=\frac{1}{ \tau (\overline{\chi })}\sum _{b=1}^{p-1}\overline{\chi }(b)A^{k+4}(b, p) \\ &= \frac{1}{\tau (\overline{\chi })}\sum_{b=1}^{p-1} \overline{\chi }(b)A ^{k}(b,p) \bigl(-2pA^{2}(b,p)+8p\alpha A(b,p)-p \bigl(9p-4\alpha ^{2} \bigr) \bigr) \\ &= -2pM_{k+2}(4, \chi ; p) + 8p\alpha M_{k+1}(4, \chi ; p)-p \bigl(9p-4 \alpha ^{2} \bigr)M_{k}(4, \chi ; p), \end{aligned}$$
(9)
where \(M_{0}(4, \chi ; p)=0\).
Suppose that χ is a fourth-order character \(\bmod \ p\). Then we get
$$\begin{aligned} M_{1}(4, \chi , p)=\sum_{a=0}^{p-1} \chi \bigl(a^{4} \bigr)=\sum_{a=0}^{p-1} \chi ^{4} (a )=p-1. \end{aligned}$$
(10)
Note that \(\overline{\chi }(b) (\frac{b}{p} )=\chi (b)\), so from (4) and (3) we get
$$\begin{aligned} M_{2}(4, \chi ; p)&= \frac{1}{\tau (\overline{\chi })}\sum _{b=1}^{p-1}\overline{ \chi }(b) \Biggl(\sum _{x=0}^{p-1}e \biggl(\frac{bx^{4}}{p} \biggr) \Biggr) ^{2} \\ &= \frac{1}{\tau (\overline{\chi })}\sum_{b=1}^{p-1} \overline{\chi }(b) \bigl(-3p+\chi _{2}(b)2\sqrt{p}A(b,p)+ \chi _{2}(b)2\sqrt{p}\alpha \bigr) \\ &= \frac{2\sqrt{p}}{\tau (\overline{\chi })}\sum_{b=1}^{p-1}\chi (b)A(b,p) \\ &= \frac{2\sqrt{p}}{\tau (\overline{\chi })}\sum_{b=1}^{p-1}\chi (b) \bigl(\chi _{2}(b)\sqrt{p}+ \overline{\chi }(b)\tau (\chi )+ \chi (b) \tau (\overline{\chi } ) \bigr) \\ &= -\frac{2(p-1)}{\sqrt{p}}\cdot \tau ^{2}(\chi ). \end{aligned}$$
(11)
In the same way, applying (3) and the orthogonality of the characters \(\bmod \ p\), we also have
$$\begin{aligned} M_{3}(4, \chi ; p)&= \frac{1}{\tau (\overline{\chi })}\sum _{b=1}^{p-1}\overline{ \chi }(b) \Biggl(\sum _{x=0}^{p-1}e \biggl(\frac{bx^{4}}{p} \biggr) \Biggr) ^{3} \\ &= \frac{1}{\tau (\overline{\chi })}\sum_{b=1}^{p-1} \overline{\chi }(b) \bigl(\chi _{2}(b)\sqrt{p}+ \overline{\chi }(b)\tau (\chi )+ \chi (b) \tau (\overline{\chi } ) \bigr)^{3} \\ &= \frac{3p}{\tau (\overline{\chi })}\sum_{b=1}^{p-1} \overline{\chi }(b) \chi (b)\tau (\overline{\chi } )+ \frac{1}{\tau (\overline{ \chi })}\sum _{b=1}^{p-1}\tau ^{3}(\chi )+ \frac{3}{\tau (\overline{ \chi })}\sum_{b=1}^{p-1}\tau (\chi ) \tau ^{2} (\overline{\chi } ) \\ &= -\frac{p-1}{p}\cdot \tau ^{4}(\chi ). \end{aligned}$$
(12)
Combining (9)–(12), we immediately obtain Lemma 6. □
Lemma 7
Let
\(p=8h+1\)
be a prime, and let
χ
be any fourth-order character
\(\bmod\ p\). Then we have the identities
$$\begin{aligned} &M_{1}(4, \chi ; p)=p-1; \\ &M_{2}(4, \chi ; p) =2(p-1)\sqrt{p}\cdot \frac{\tau (\chi )}{\tau (\overline{ \chi })}= \frac{2(p-1)}{\sqrt{p}}\cdot \tau ^{2}(\chi ); \\ &M_{3}(4, \chi ; p)=\frac{p-1}{p}\cdot \tau ^{4} (\chi ). \end{aligned}$$
For every integer
\(k\geq 0\), we have the fourth-order linear recurrence formula
$$\begin{aligned} M_{k+4}(4, \chi ; p) = 6pM_{k+2}(4, \chi ; p) + 8p\alpha M_{k+1}(4, \chi ; p)+p \bigl(4\alpha ^{2} -p \bigr)M_{k}(4, \chi ; p). \end{aligned}$$
Proof
If \(p=8h+1\), then from Lemma 4 we have
$$\begin{aligned} M_{k+4}(4, \chi ; p)&=\frac{1}{\tau (\overline{\chi })}\sum _{x_{1}=0} ^{p-1}\sum_{x_{2}=0}^{p-1} \cdots \sum_{x_{k+4}=0}^{p-1}\sum _{b=1}^{p-1}\overline{ \chi }(b) e \biggl( \frac{b(x_{1}^{4}+x_{2}^{2}+\cdots +x_{k+4}^{4})}{p} \biggr) \\ &= \frac{1}{\tau (\overline{\chi })}\sum_{b=1}^{p-1} \overline{\chi }(b)A ^{k}(b,p) \bigl(6pA^{2}(b, p)+8p\alpha A(b, p)+p \bigl(4\alpha ^{2}-p \bigr) \bigr) \\ &= 6pM_{k+2}(4, \chi ; p) + 8p\alpha M_{k+1}(4,\chi ; p)+p \bigl(4 \alpha ^{2}-p \bigr)M_{k}(4,\chi ; p). \end{aligned}$$
(13)
It is not complicated to prove that
$$\begin{aligned} M_{1}(4, \chi ; p)=\sum_{x=0}^{p-1} \chi \bigl(x^{4} \bigr)=\sum_{x=0}^{p-1} \chi ^{4} (x )=p-1. \end{aligned}$$
(14)
Note that \(\overline{\chi }(b) (\frac{b}{p} )=\chi (b)\), \(\tau (\chi )\tau (\overline{\chi })=p\). By the method of proving (11) we have
$$\begin{aligned} M_{2}(4, \chi ; p)&= \frac{1}{\tau (\overline{\chi })}\sum _{b=1}^{p-1}\overline{ \chi }(b) \Biggl(\sum _{x=0}^{p-1}e \biggl(\frac{bx^{4}}{p} \biggr) \Biggr) ^{2} \\ &= \frac{2\sqrt{p}\cdot \tau (\chi )}{\tau (\overline{\chi })}\sum_{a=1}^{p-1}1 = \frac{2(p-1)}{\sqrt{p}}\cdot \tau ^{2}(\chi ). \end{aligned}$$
(15)
Similarly, combined with the method of proving (12), we also get
$$\begin{aligned} M_{3}(4, \chi ; p)&= \frac{1}{\tau (\overline{\chi })}\sum _{b=1}^{p-1}\overline{ \chi }(b) \Biggl(\sum _{x=0}^{p-1}e \biggl(\frac{bx^{4}}{p} \biggr) \Biggr) ^{3} \\ &= (p-1)\cdot \frac{\tau ^{3}(\chi )}{\tau (\overline{\chi } )}= \frac{p-1}{p}\cdot \tau ^{4}( \chi ). \end{aligned}$$
(16)
Now Lemma 7 follows from (13)–(16). □