On Some Congruences Involving Generalized Central Trinomial Coefficients

Abstract

In this paper, we mainly prove some congruences involving generalized central trinomial coefficients

$$\begin{aligned} T_n(b,c):&=[x^n](x^2+bx+c)^n=[x^0](b+x+cx^{-1})^n\\&=\sum _{k=0}^{\lfloor n/2\rfloor }\left( {\begin{array}{c}n\\ 2k\end{array}}\right) \left( {\begin{array}{c}2k\\ k\end{array}}\right) b^{n-2k}c^k=\sum _{k=0}^{\lfloor n/2\rfloor }\left( {\begin{array}{c}n-k\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) b^{n-2k}c^k, \end{aligned}$$

where $b, c\in \mathbb {Z}$. For example, let $p>3$ be a prime, b, c be integers and $p\not \mid d=b^2-4c$. Then,

$$\begin{aligned}&\sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}\\&\quad \equiv {\left\{ \begin{array}{ll} -p^2\pmod {p^3} &{} \texttt{if}\ \textit{p}\mid \textit{c},\\ -p^2-\frac{2}{3}p^2\left( \frac{d}{p}\right) +p^2\left( \frac{7d}{6c}+\frac{d^2}{3c^2}\right) \left( \left( \frac{d}{p}\right) -1\right) \pmod {p^3} &{} \texttt{if} \ \textit{p}\not \mid \textit{c}, \end{array}\right. } \end{aligned}$$

where $\left( \frac{\cdot }{p}\right)$ denotes the Legendre symbol.

1 Introduction

It is well known that the nth central trinomial coefficient

$$\begin{aligned} T_n=[x^n](1+x+x^2)^n\ \ \ \ \ n\in \mathbb {N}=\{0,1,2,\ldots \} \end{aligned}$$

is the coefficient of $x^n$ in the expansion of $(1+x+x^2)^n$. By the multi-nomial theorem, we have

$$\begin{aligned} T_n=\sum _{k=0}^{\lfloor n/2\rfloor }\frac{n!}{k!k!(n-2k)!}=\sum _{k=0}^{\lfloor n/2\rfloor }\left( {\begin{array}{c}n\\ 2k\end{array}}\right) \left( {\begin{array}{c}2k\\ k\end{array}}\right) =\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}n-k\\ k\end{array}}\right) \end{aligned}$$

since $T_n$ is the constant term of $(x+1+x^{-1})^n$. Central trinomial coefficients arise naturally in enumerative combinatorics (see, [4]), e.g., $T_n$ is the number of lattice paths from the point (0, 0) to (n, 0) with only allowed steps (1, 0), (1, 1) and $(1,-1)$. Andrews [1] pointed out that central trinomial coefficients were first studied by Euler, and in 1987, he and Baxter [2] found that the q-analogues of central trinomial coefficients have applications in the hard hexagon model.

Given $b, c\in \mathbb {Z}, n\in \mathbb {N}$, Sun [6] defined the generalized central trinomial coefficients

$$\begin{aligned} T_n(b,c):&=[x^n](x^2+bx+c)^n=[x^0](b+x+cx^{-1})^n\\&=\sum _{k=0}^{\lfloor n/2\rfloor }\left( {\begin{array}{c}n\\ 2k\end{array}}\right) \left( {\begin{array}{c}2k\\ k\end{array}}\right) b^{n-2k}c^k=\sum _{k=0}^{\lfloor n/2\rfloor }\left( {\begin{array}{c}n-k\\ k\end{array}}\right) \left( {\begin{array}{c}n\\ k\end{array}}\right) b^{n-2k}c^k \end{aligned}$$

and he proved that for each $n=1,2,3, \ldots$,

$$\begin{aligned} \sum _{k=0}^{n-1}(2k+1)T_k(b,c)^2(b^2-4c)^{n-1-k}\equiv 0\pmod {n^2}. \end{aligned}$$

And in [7], sun proved that for any integer b, c, positive integer n and $d=b^2-4c$, we have

$$\begin{aligned} b\sum _{k=0}^{n-1}(2k+1)T_k(b,c)^2(-d)^{n-1-k}=nT_n(b,c)T_{n-1}(b,c), \end{aligned}$$

and for all nonnegative integer k,

$$\begin{aligned} T_k(b,c)^2=\sum _{j=0}^k\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) \left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2c^jd^{k-j}. \end{aligned}$$

(1.1)

Note that

$$\begin{aligned} T_n=T_n(1,1),\ \ \ T_n(2,1)=[x^n](x+1)^{2n}=\left( {\begin{array}{c}2n\\ n\end{array}}\right) . \end{aligned}$$

So $T_n(b,c)$ can be viewed a natural common extension of central binomial coefficients. Let $d=b^2-4c$. Wilf [8, p. 195] showed that

$$\begin{aligned} \sum _{n=0}^{\infty }T_n(b,c)x^n=\frac{1}{\sqrt{1-2bx+{\textrm{d}}x^2}} \end{aligned}$$

for $\mid x\mid <\varepsilon$ with $\varepsilon >0$ sufficiently small. This gives the recurrence (see also [3])

$$\begin{aligned} (n+1)T_{n+1}(b,c)=(2n+1)bT_n(b,c)-dnT_{n-1}(b,c)\ \ (n\in \mathbb {Z}^{+}). \end{aligned}$$

It is known that

$$\begin{aligned} \sum _{n=0}^{\infty }P_n(t)x^n=\frac{1}{\sqrt{1-2tx+x^2}}. \end{aligned}$$

So, if $b, c\in \mathbb {Z}$ and $d=b^2-4c\ne 0$, then

$$\begin{aligned} \sum _{n=0}^{\infty }T_n(b,c)\left( \frac{x}{\sqrt{d}}\right) ^n=\frac{1}{\sqrt{1-2bx/\sqrt{d}+d(x/\sqrt{d})^2}}=\sum _{n=0}^{\infty }P_n\left( \frac{b}{\sqrt{d}}\right) x^n, \end{aligned}$$

hence

$$\begin{aligned} T_n(b,c)=(\sqrt{d})^nP_n\left( \frac{d}{\sqrt{d}}\right) . \end{aligned}$$

It follows that $T_n(2x+1,x^2+x)=P_n(2x+1)$ for all $x\in \mathbb {Z}$.

Motivated by the above, we obtain the following results.

Theorem 1.1

Let $p>3$ be a prime, b, c be integers and $p\not \mid d=b^2-4c$. Then,

$$\begin{aligned}&\sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}\\&\quad \equiv {\left\{ \begin{array}{ll} -p^2\pmod {p^3} &{}\texttt{if}\ \textit{p}\mid \textit{c},\\ -p^2-\frac{2}{3}p^2\left( \frac{d}{p}\right) +p^2\left( \frac{7d}{6c}+\frac{d^2}{3c^2}\right) \left( \left( \frac{d}{p}\right) -1\right) \pmod {p^3} &{}\texttt{if} \ \textit{p}\not \mid \textit{c}, \end{array}\right. } \end{aligned}$$

where $\left( \frac{\cdot }{p}\right)$ denotes the Legendre symbol.

Remark 1.1

It is easy to see that if $c=0$,

$$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}=\sum _{k=0}^{p-1}(2k+1)^3=p^2(2p^2-1)\equiv -p^2\pmod {p^4}. \end{aligned}$$

Corollary 1.1

For any prime $p>3$, $x\in \mathbb {Z}$, we have

$$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)^3P_k(2x+1)^2\equiv {\left\{ \begin{array}{ll} -p^2\pmod {p^3} &{}\texttt{if}\ \textit{p}\mid \textit{x}(\textit{x}+1),\\ -\frac{5}{3}p^2\pmod {p^3} &{}\texttt{if} \ \textit{x}\not \equiv 0,-1\pmod {\textit{p}}. \end{array}\right. } \end{aligned}$$

$$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)^3\frac{T_k^2}{(-3)^k}\equiv -\frac{p^2}{2}-\frac{7p^2}{6}\left( \frac{p}{3}\right) \pmod {p^3}. \end{aligned}$$

Proof

Since $T_n(2x+1,x^2+x)=P_n(2x+1)$, so we have $b=2x+1, c=x^2+x$, $d=(2x+1)^2-4(x^2+x)=1$, hence by Theorem 1.1 we immediately obtain the first congruence. The second one can be deduced by taking $b=c=1$ in Theorem 1.1 and noting that $\left( \frac{-3}{p}\right) =\left( \frac{p}{3}\right)$. $\square$

We also obtain the following results.

Theorem 1.2

Let $p>3$ be a prime, b, c be integers and $p\not \mid d=b^2-4c$.

• (i). We have

  $$\begin{aligned} \sum _{k=0}^{p-1}(-1)^k(2k+1)\frac{T_k(b,c)^2}{d^k}\equiv p\left( \frac{d}{p}\right) \pmod {p^2} \end{aligned}$$

  (1.2)

  and

  $$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)\frac{T_k^2}{3^k}\equiv p\left( \frac{p}{3}\right) \pmod {p^3}. \end{aligned}$$

  (1.3)

• (ii). If $p\not \mid b$, then we have

  $$\begin{aligned} \sum _{k=0}^{p-1}\frac{(-1)^k(2k+1)^3T_k(b,c)^2}{d^k}\equiv \frac{8pc}{b^2}\left( \frac{d}{p}\right) -3p\left( \frac{d}{p}\right) \pmod {p^2} \end{aligned}$$

  (1.4)

  and

  $$\begin{aligned}&\sum _{k=0}^{p-1}\frac{(-1)^k(2k+1)^5T_k(b,c)^2}{d^k}\nonumber \\&\equiv 25p\left( \frac{d}{p}\right) -\frac{112pc}{b^2}\left( \frac{d}{p}\right) -\frac{64p(d-2c)}{b^4}\left( \frac{d}{p}\right) \pmod {p^2}. \end{aligned}$$

  (1.5)

Corollary 1.2

For any prime $p>3$, $x\in \mathbb {Z}$, we have

$$\begin{aligned} \sum _{k=0}^{p-1}(-1)^k(2k+1)P_k(2x+1)^2\equiv p\pmod {p^2}. \end{aligned}$$

Proof

Since $T_n(2x+1,x^2+x)=P_n(2x+1)$, so we have $b=2x+1, c=x^2+x$, $d=(2x+1)^2-4(x^2+x)=1$, hence by (1.2) we immediately obtain the first congruence. $\square$

We are going to prove Theorem 1.1 in Sect. 2. Section 3 is devoted to proving Theorem 1.2. Our proofs make use of some combinatorial identities which can be proved by integral, combinatorial properties and induction.

2 Proof of Theorem 1.1

Lemma 2.1

For any nonnegative integer n and rational number $x\ne 0$, we have

$$\begin{aligned}{} & {} \sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{x^k}{k+1}=\frac{(1+x)^{n+1}-1}{(n+1)x}, \end{aligned}$$

(2.1)

$$\begin{aligned}{} & {} \sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{x^k}{k+2}=\frac{1+(nx-1)(1+x)^n+(n+1)x^2(1+x)^n}{(n+1)(n+2)x^2}. \end{aligned}$$

(2.2)

Proof

It is easy to see that

$$\begin{aligned} \sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{x^k}{k+1}&=\int _0^1\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) x^ky^k {\textrm{d}}y=\int _0^1(1+xy)^n{\textrm{d}}y=\frac{1}{x}\int _1^{x+1}z^n{\textrm{d}}z\\&=\frac{1}{x}\left. \dfrac{z^{n+1}}{n+1}\right| _{1}^{x+1}=\frac{(1+x)^{n+1}-1}{(n+1)x}. \end{aligned}$$

And by the combinatorial property $\left( {\begin{array}{c}n+1\\ k\end{array}}\right) =\left( {\begin{array}{c}n\\ k\end{array}}\right) +\left( {\begin{array}{c}n\\ k-1\end{array}}\right)$, we have

$$\begin{aligned}&\sum _{k=0}^{n+1}\frac{\left( {\begin{array}{c}n+1\\ k\end{array}}\right) x^k}{k+1}-\sum _{k=0}^{n}\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) x^k}{k+1}=\frac{x^{n+1}}{n+2}+\sum _{k=1}^{n}\left( \left( {\begin{array}{c}n+1\\ k\end{array}}\right) -\left( {\begin{array}{c}n\\ k\end{array}}\right) \right) \frac{x^k}{k+1}\\&\quad =\frac{x^{n+1}}{n+2}+\sum _{k=1}^{n}\left( {\begin{array}{c}n\\ k-1\end{array}}\right) \frac{x^k}{k+1}=\frac{x^{n+1}}{n+2}+\sum _{k=0}^{n-1}\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{x^{k+1}}{k+2}\\&\quad =\sum _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{x^{k+1}}{k+2}. \end{aligned}$$

Hence, by (2.1), we have

$$\begin{aligned} \sum _{k=0}^{n}\left( {\begin{array}{c}n\\ k\end{array}}\right) \frac{x^{k}}{k+2}&=\frac{1}{x}\left( \sum _{k=0}^{n+1}\frac{\left( {\begin{array}{c}n+1\\ k\end{array}}\right) x^k}{k+1}-\sum _{k=0}^{n}\frac{\left( {\begin{array}{c}n\\ k\end{array}}\right) x^k}{k+1}\right) \\&=\frac{1+(nx-1)(1+x)^n+(n+1)x^2(1+x)^n}{(n+1)(n+2)x^2}. \end{aligned}$$

Now the proof of Lemma 2.1 is finished. $\square$

Lemma 2.2

For any prime $p>3$, integer $j\in \{0,1,\ldots ,p-1\}$, we have

$$\begin{aligned} (-1)^j(p+j)\left( {\begin{array}{c}2j\\ j\end{array}}\right) \left( {\begin{array}{c}p+j-1\\ 2j\end{array}}\right) \equiv p\pmod {p^3}. \end{aligned}$$

Proof

It is easy to check that

$$\begin{aligned} (-1)^j(p+j)\left( {\begin{array}{c}2j\\ j\end{array}}\right) \left( {\begin{array}{c}p+j-1\\ 2j\end{array}}\right)&=(-1)^j\frac{p(p^2-1^2)\cdots (p^2-j^2)}{j!^2}\\&\equiv p\pmod {p^3}. \end{aligned}$$

This proves Lemma 2.2. $\square$

Proof of Theorem 1.1

In view of (1.1), we have

$$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}&=\sum _{k=0}^{p-1}(2k+1)^3\sum _{j=0}^k\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) \left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\\&=\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\sum _{k=j}^{p-1}(2k+1)^3\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) . \end{aligned}$$

By a straight-forward induction on n, we can prove that, for all $0\le j\le n-1$,

$$\begin{aligned} \sum _{k=j}^{n-1}(2k+1)^3\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) =\frac{n(n+j)(4n^2j+4n^2-3j-2)}{(j+1)(j+2)}\left( {\begin{array}{c}n+j-1\\ 2j\end{array}}\right) . \end{aligned}$$

It is known that $\left( {\begin{array}{c}2k\\ k\end{array}}\right) \equiv 0\pmod {p}$ for each $(p+1)/2\le k\le p-1$, so when $k=p-1$ or $p-2$, we have $\left( {\begin{array}{c}2k\\ k\end{array}}\right) \equiv 0\pmod p$ since $p-1, p-2\ge (p+1)/2$ under $p>3$. Set $n=p$ in the above identity, then by Lemma 2.2, we have

$$\begin{aligned}&\sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}\\&\quad =\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\frac{p(p+j)(4p^2j+4p^2-3j-2)}{(j+1)(j+2)}\left( {\begin{array}{c}p+j-1\\ 2j\end{array}}\right) \\&\quad \equiv \sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) \left( \frac{-c}{d}\right) ^j\frac{p^2(-3j-2)}{(j+1)(j+2)}\\&\quad = \sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) \left( \frac{-c}{d}\right) ^j\left( \frac{p^2}{j+1}-\frac{4p^2}{j+2}\right) \pmod {p^3}. \end{aligned}$$

The case $j=p-1, p-2$ need to be considered alone, so

$$\begin{aligned}&\sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}\\&\quad \equiv p^2\sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\frac{1}{j+1}+p\left( {\begin{array}{c}2p-2\\ p-1\end{array}}\right) \left( -\frac{c}{d}\right) ^{p-1}\\&\qquad -4p^2\sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\frac{1}{j+2}-4p\left( {\begin{array}{c}2p-4\\ p-2\end{array}}\right) \left( -\frac{c}{d}\right) ^{p-2}\\&\quad \equiv p^2\sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\frac{1}{j+1}-4p^2\sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\frac{1}{j+2}\\&\qquad -p^2\left( -\frac{c}{d}\right) ^{p-1}+\frac{2}{3}p^2\left( -\frac{c}{d}\right) ^{p-2}\pmod {p^3}. \end{aligned}$$

If $p\mid c$, it is easy to see that

$$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}\equiv p^2-2p^2=-p^2\pmod {p^3}. \end{aligned}$$

If $p\not \mid c$. Set $n=(p-1)/2, x=4c/d$ in Lemma 2.1, we have the following modulo p,

$$\begin{aligned}{} & {} \sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\frac{1}{j+1}\equiv \frac{d}{2c}\left( \left( \frac{d}{p}\right) +\frac{4c}{d}\left( \frac{d}{p}\right) -1\right) , \end{aligned}$$

(2.3)

$$\begin{aligned}{} & {} \sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\frac{1}{j+2}\equiv \frac{d^2}{12c^2}\left( 1-\left( \frac{d}{p}\right) -\frac{2c}{d}\left( \frac{d}{p}\right) +\frac{8c^2}{d^2}\left( \frac{d}{p}\right) \right) . \nonumber \\ \end{aligned}$$

(2.4)

These, with Fermat’s little theorem yield that

$$\begin{aligned}&\sum _{k=0}^{p-1}(2k+1)^3\frac{T_k(b,c)^2}{d^k}\\&\quad \equiv -p^2-\frac{2}{3}p^2\left( \frac{d}{p}\right) +p^2\left( \frac{7d}{6c}+\frac{d^2}{3c^2}\right) \left( \left( \frac{d}{p}\right) -1\right) \pmod {p^3}. \end{aligned}$$

Now the proof of Theorem 1.1 is complete. $\square$

3 Proof of Theorem 1.2

Firstly, by straight-forward induction on n, we can prove that, for all $0\le j\le n-1$,

$$\begin{aligned}&\sum _{k=j}^{n-1}(-1)^k(2k+1)\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) =(n+j)\left( {\begin{array}{c}n+j-1\\ 2j\end{array}}\right) , \end{aligned}$$

(3.1)

$$\begin{aligned}&\sum _{k=j}^{n-1}(-1)^k(2k+1)^3\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) =(-1)^{n+1}(n+j)\delta _1\left( {\begin{array}{c}n+j-1\\ 2j\end{array}}\right) , \end{aligned}$$

(3.2)

$$\begin{aligned}&\sum _{k=j}^{n-1}(-1)^k(2k+1)^5\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) =(-1)^{n+1}(n+j)\delta _2\left( {\begin{array}{c}n+j-1\\ 2j\end{array}}\right) . \end{aligned}$$

(3.3)

where $\delta _1=4n^2-4j-3$, $\delta _2=25+56j+32j^2-40n^2-32n^2j+16n^4$.

Proof of (1.2)

In view of (1.1), we have

$$\begin{aligned} \sum _{k=0}^{p-1}(-1)^k(2k+1)\frac{T_k(b,c)^2}{d^k}&=\sum _{k=0}^{p-1}(-1)^k(2k+1)\sum _{j=0}^k\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) \left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\\&=\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\sum _{k=j}^{p-1}(-1)^k(2k+1)\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) . \end{aligned}$$

Substituting $n=p$ into (3.1), then by Lemma 2.2 and Fermat’s little theorem, we have

$$\begin{aligned} \sum _{k=0}^{p-1}(-1)^k(2k+1)\frac{T_k(b,c)^2}{d^k}&\equiv p\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) \left( -\frac{c}{d}\right) ^j\equiv p\sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\\&=p\left( 1+\frac{4c}{d}\right) ^{\frac{p-1}{2}}=p\frac{b^{p-1}}{d^{\frac{p-1}{2}}}\\&\equiv p\left( \frac{d}{p}\right) \pmod {p^2}. \\ \end{aligned}$$

$\square$

Proof of (1.3)

From above we have

$$\begin{aligned} \sum _{k=0}^{p-1}(2k+1)\frac{T_k^2}{3^k}&=\sum _{k=0}^{p-1}(-1)^k(2k+1)\frac{T_k(1,1)^2}{(-3)^k}\equiv p\sum _{j=0}^{p-1}\frac{\left( {\begin{array}{c}2j\\ j\end{array}}\right) }{3^j}\\&\equiv p\left( \frac{p}{3}\right) \pmod {p^3}, \end{aligned}$$

where the last equation we used $\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) /3^j\equiv \left( \frac{p}{3}\right) \pmod {p^2}$ from [5, Corollary 1.1]. $\square$

Proof of (1.4)

Similarly, in view of (1.1), we have

$$\begin{aligned}&\sum _{k=0}^{p-1}(-1)^k(2k+1)^3\frac{T_k(b,c)^2}{d^k}\\&\quad =\sum _{k=0}^{p-1}(-1)^k(2k+1)^3\sum _{j=0}^k\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) \left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\\&\quad =\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\sum _{k=j}^{p-1}(-1)^k(2k+1)^3\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) . \end{aligned}$$

Substituting $n=p$ into (3.2), then by Lemma 2.2 and Fermat’s little theorem, we have

$$\begin{aligned}&\sum _{k=0}^{p-1}(-1)^k(2k+1)^3\frac{T_k(b,c)^2}{d^k}\equiv p\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) \left( -\frac{c}{d}\right) ^j(-4j-3)\\&\quad \equiv -4p\sum _{j=0}^{\frac{p-1}{2}}j\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j-3p\left( \frac{d}{p}\right) \pmod {p^2}. \end{aligned}$$

It is easy to see that

$$\begin{aligned}&\sum _{j=0}^{\frac{p-1}{2}}j\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j=\sum _{j=1}^{\frac{p-1}{2}}j\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\equiv -\frac{1}{2}\sum _{j=1}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-3}{2}\\ j-1\end{array}}\right) \left( \frac{4c}{d}\right) ^j\\&\quad =-\frac{2c}{d}\sum _{j=0}^{\frac{p-3}{2}}\left( {\begin{array}{c}\frac{p-3}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j=-\frac{2c}{d}\left( \frac{b^2}{d}\right) ^{\frac{p-3}{2}}\equiv -\frac{2c}{b^2}\left( \frac{d}{p}\right) \pmod p. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{k=0}^{p-1}(-1)^k(2k+1)^3\frac{T_k(b,c)^2}{d^k}\equiv \frac{8pc}{b^2}\left( \frac{d}{p}\right) -3p\left( \frac{d}{p}\right) \pmod {p^2}.\\ \end{aligned}$$

$\square$

Proof of (1.5)

Similarly, in view of (1.1), we have

$$\begin{aligned}&\sum _{k=0}^{p-1}(-1)^k(2k+1)^5\frac{T_k(b,c)^2}{d^k}\\&\quad =\sum _{k=0}^{p-1}(-1)^k(2k+1)^5\sum _{j=0}^k\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) \left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\\&\quad =\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) ^2\frac{c^j}{d^j}\sum _{k=j}^{p-1}(-1)^k(2k+1)^5\left( {\begin{array}{c}k+j\\ 2j\end{array}}\right) . \end{aligned}$$

Substituting $n=p$ into (3.3), then by Lemma 2.2 and Fermat’s little theorem, we have

$$\begin{aligned}&\sum _{k=0}^{p-1}(-1)^k(2k+1)^5\frac{T_k(b,c)^2}{d^k}\equiv p\sum _{j=0}^{p-1}\left( {\begin{array}{c}2j\\ j\end{array}}\right) \left( -\frac{c}{d}\right) ^j(25+56j+32j^2)\\&\quad \equiv p\sum _{j=0}^{\frac{p-1}{2}}\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j(25+56j+32j^2)\\&\quad \equiv 25p\left( \frac{d}{p}\right) -\frac{112pc}{b^2}\left( \frac{d}{p}\right) +32p\sum _{j=0}^{\frac{p-1}{2}}j^2\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\pmod {p^2}. \end{aligned}$$

It is easy to see that modulo p, we have

$$\begin{aligned}&\sum _{j=0}^{\frac{p-1}{2}}j^2\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\\&\quad =\sum _{j=1}^{\frac{p-1}{2}}j^2\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\equiv -\frac{1}{2}\sum _{j=1}^{\frac{p-1}{2}}j\left( {\begin{array}{c}\frac{p-3}{2}\\ j-1\end{array}}\right) \left( \frac{4c}{d}\right) ^j\\&\quad =-\frac{2c}{d}\sum _{j=0}^{\frac{p-3}{2}}(j+1)\left( {\begin{array}{c}\frac{p-3}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j=-\frac{2c}{b^2}\left( \frac{d}{p}\right) -\frac{2c}{d}\sum _{j=0}^{\frac{p-3}{2}}j\left( {\begin{array}{c}\frac{p-3}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j \end{aligned}$$

and

$$\begin{aligned}&\sum _{j=0}^{\frac{p-3}{2}}j\left( {\begin{array}{c}\frac{p-3}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j=\sum _{j=1}^{\frac{p-3}{2}}j\left( {\begin{array}{c}\frac{p-3}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\equiv -\frac{3}{2}\sum _{j=1}^{\frac{p-3}{2}}\left( {\begin{array}{c}\frac{p-5}{2}\\ j-1\end{array}}\right) \left( \frac{4c}{d}\right) ^j\\&\quad =-\frac{6c}{d}\sum _{j=0}^{\frac{p-5}{2}}\left( {\begin{array}{c}\frac{p-5}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j=-\frac{6c}{d}\left( \frac{b^2}{d}\right) ^{\frac{p-5}{2}}\equiv -\frac{6cd}{b^4}\left( \frac{d}{p}\right) \pmod p. \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{j=0}^{\frac{p-1}{2}}j^2\left( {\begin{array}{c}\frac{p-1}{2}\\ j\end{array}}\right) \left( \frac{4c}{d}\right) ^j\equiv -\frac{2c}{b^2}\left( \frac{d}{p}\right) +\frac{12c^2}{b^4}\left( \frac{d}{p}\right) \pmod p. \end{aligned}$$

Therefore,

$$\begin{aligned}&\sum _{k=0}^{p-1}(-1)^k(2k+1)^5\frac{T_k(b,c)^2}{d^k}\\&\quad \equiv 25p\left( \frac{d}{p}\right) -\frac{112pc}{b^2}\left( \frac{d}{p}\right) -\frac{64pc(b^2-6c)}{b^4}\left( \frac{d}{p}\right) \\&\quad \equiv 25p\left( \frac{d}{p}\right) -\frac{112pc}{b^2}\left( \frac{d}{p}\right) -\frac{64pc(d-2c)}{b^4}\left( \frac{d}{p}\right) \pmod {p^2}. \end{aligned}$$

Now the proof of Theorem 1.2 is complete. $\square$