*Article* **On a Generalization of a Lucas' Result and an Application to the 4-Pascal's Triangle †**

## **Atsushi Yamagami \* and Kazuki Taniguchi**

Department of Information Systems Science, Soka University, Tokyo 192-8577, Japan; e1658229@soka-u.jp **\*** Correspondence: yamagami@soka.ac.jp

† This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

Received: 29 January 2020; Accepted: 10 February 2020; Published: 16 February 2020

**Abstract:** The Pascal's triangle is generalized to "the *k*-Pascal's triangle" with any integer *k* ≥ 2. Let *p* be any prime number. In this article, we prove that for any positive integers *n* and *e*, the *n*-th row in the *p<sup>e</sup>* -Pascal's triangle consists of integers which are congruent to 1 modulo *p* if and only if *n* is of the form *<sup>p</sup>em* <sup>−</sup> <sup>1</sup> *<sup>p</sup><sup>e</sup>* <sup>−</sup> <sup>1</sup> with some integer *<sup>m</sup>* <sup>≥</sup> 1. This is a generalization of a Lucas' result asserting that the *n*-th row in the (2-)Pascal's triangle consists of odd integers if and only if *n* is a Mersenne number. As an application, we then see that there exists no row in the 4-Pascal's triangle consisting of integers which are congruent to 1 modulo 4 except the first row. In this application, we use the congruence (*x* + 1)*p<sup>e</sup>* <sup>≡</sup> (*x<sup>p</sup>* <sup>+</sup> <sup>1</sup>)*pe*−<sup>1</sup> (mod *pe*) of binomial expansions which we could prove for any prime number *p* and any positive integer *e*. We think that this article is fit for the Special Issue "Number Theory and Symmetry," since we prove a symmetric property on the 4-Pascal's triangle by means of a number-theoretical property of binomial expansions.

**Keywords:** the *p<sup>e</sup>* -Pascal's triangle; Lucas' result on the Pascal's triangle; congruences of binomial expansions

**MSC:** 11A99.

## **1. Introduction**

As it is known, Pascal's triangle is constructed in the following way: Write the first row "1 1". Then each member of each subsequent row is given by taking the sum of the just above two members, regarding any blank as 0.

**Example 1.** *Here is the Pascal's triangle from the first row to the* 7*-th row:*

```
1 1
       121
     13 31
    14 6 41
  1 5 10 10 5 1
 1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
```
**Remark 1.** *For any integers n* ≥ 1 *and r* ≥ 0*, we put*

$$\,\_n\mathbb{C}\_r := \frac{n!}{r!(n-r)!} = \frac{n(n-1)\cdots(n-r+1)}{r\cdots 1},$$

*where we put* 0! = 1*. Then it is well-known that the n-th row in the Pascal's triangle is equal to the sequence*

$$\dots \mathbb{C}\_{0 \prime n} \mathbb{C}\_{1 \prime \cdots \prime n} \mathbb{C}\_{n-1 \prime n} \mathbb{C}\_n$$

*consisting of n* + 1 *terms.*

In ([1], Section 1.4), the construction above is generalized as follows:

**Definition 1.** *Let k* ≥ 2 *be any integer. The k-Pascal's triangle is constructed in the following way: Write the first row " k* \* +( ) 1 1 ··· <sup>1</sup>*". Then each member of each subsequent row is given by taking the sum of the just above k members regarding the blank as* 0*.*

**Example 2.** *In the case where k* = 4*, the* 4*-Pascal's triangle from the first row to the* 5*-th row is the following:*

1111 12 3 4 32 1 1 3 6 10 12 12 10 6 3 1 1 4 10 20 31 40 44 40 31 20 10 4 1 1 5 15 35 65 101 135 155 155 135 101 65 35 15 5 1

**Remark 2.** (1) *In ([1], Section* 1.4*), for any integers k* ≥ 2 *and n* ≥ 1*, it is mentioned that the n-th row in the k-Pascal's triangle consists of n*(*k* − 1) + 1 *integers*

$$\dots\_n\mathbb{C}\_0^{(k)}, \,\_n\mathbb{C}\_1^{(k)}, \dots, \,\_n\mathbb{C}\_{n(k-1)-1'}^{(k)}\mathbb{C}\_{n(k-1)}^{(k)}$$

*satisfying the equation*

$$\begin{aligned} & \left( \mathbf{x}^{k-1} + \mathbf{x}^{k-2} + \dots + \mathbf{x} + 1 \right)^n \\ &= \mathbf{\_nC\_0^{(k)}x^{n(k-1)} + \_nC\_1^{(k)}x^{n(k-1)-1} + \dots + \_nC\_{n(k-1)-1}^{(k)}\mathbf{x} + \_nC\_{n(k-1)}^{(k)} \end{aligned}$$

*of polynomials with indeterminate x and integral coefficients. A detailed proof of this fact is described in ([2], Lemma* 1.1*).*

(2) *In ([1], Section* 9.10*), the following formula for nC*(*k*) *<sup>i</sup> is described:*

$$\prescript{}{n}{\mathbf{C}}\_{i}^{(k)} = \sum\_{j=0}^{\lfloor \frac{j}{k} \rfloor} (-1)^{j} \prescript{}{n+i-jk-1}{\mathbf{C}}\_{n-1} \cdot \prescript{}{n}{\mathbf{C}}\_{j'} $$

*where* <sup>7</sup> *<sup>i</sup> k* 8 *is the greatest integer that is less than or equal to <sup>i</sup> k .*

In Example 1, we can see that the *n*-th row consists of odd integers when *n* is equal to the Mersenne number 1, 3 or 7. Actually, Lucas showed the following

**Theorem 1** ([3], Exemple I in Section 228)**.** *Let n* ≥ 1 *be any integer. Then nCr is odd for any* 0 ≤ *r* ≤ *n if and only if n is a Mersenne number, i.e., n is of the form* <sup>2</sup>*<sup>m</sup>* − <sup>1</sup> *with some integer m* ≥ <sup>1</sup>*.*

In Section 2 in this article, we generalize the Lucas' result above as the following

**Theorem 2.** *Let p be any prime number and e any positive integer. For any integer n* ≥ 1*, the n-th row in the p<sup>e</sup> -Pascal's triangle consists of integers which are congruent to* 1 *modulo p if and only if n is of the form <sup>p</sup>em* − <sup>1</sup> *<sup>p</sup><sup>e</sup>* <sup>−</sup> <sup>1</sup> *with some integer m* <sup>≥</sup> <sup>1</sup>*.*

**Remark 3.** (1) *Theorem 2 is a generalization of ([2], Theorem* 0.2*) which is in the case where e* = 1*.*

(2) *We can see that Example 2 gives a partial example of Theorem 2 in the case where p* = 2*, e* = 2 *and m* = 1, 2*.*

As an application of Theorem 2, we can prove that ([2], Conjecture 0.3) holds for *k* = 4, i.e., there exists no row in the 4-Pascal's triangle consisting of integers which are congruent to 1 modulo 4 except the first row as follows:

By Theorem 2, in the case where *k* = 4, we see that for any integer *n* ≥ 1, the *n*-th row in the 4-Pascal's triangle consists of odd integers if and only if *<sup>n</sup>* is of the form <sup>4</sup>*<sup>m</sup>* <sup>−</sup> <sup>1</sup> <sup>3</sup> with some integer *m* ≥ 1.

Moreover, we can see an essential property of the <sup>4</sup>*<sup>m</sup>* <sup>−</sup> <sup>1</sup> <sup>3</sup> -th row in the 4-Pascal's triangle for any integer *m* ≥ 2 as in the following theorem proved in Section 3.2:

## **Theorem 3.** *For any integer <sup>m</sup>* <sup>≥</sup> <sup>2</sup>*, the* <sup>4</sup>*<sup>m</sup>* <sup>−</sup> <sup>1</sup> <sup>3</sup> *-th row in the* <sup>4</sup>*-Pascal's triangle is congruent to the sequence*

$$\overbrace{1133\cdots1133}^{2^{2m}-3}\overbrace{3311\cdots3311}^{2^{2m}-3}$$

*modulo* 4*, which consists of the repeated* 1133*'s and* 3311*'s whose numbers are the same* 22*m*−3*.*

Therefore we can obtain the following

**Corollary 1.** *([2], Conjecture* 0.3*) holds for k* = 4*, i.e., there exists no row in the* 4*-Pascal's triangle consisting of integers which are congruent to* 1 *modulo* 4 *except the first row.*

**Remark 4.** (1) *By Example 2, in the case where m* = 2*, we can see that the* 5*-th row in the* 4*-Pascal's triangle is congruent to the sequence*

1133113333113311

*modulo* 4*, which matches the assertion of Theorem 3.*

(2) *It seems that one could obtain the forms of the sequenece to which the* -<sup>4</sup>*<sup>m</sup>* − <sup>1</sup> <sup>3</sup> <sup>±</sup> - *-th row in the* 4*-Pascal's triangle is congruent modulo* 4 *for some positive integers by means of Theorem 3. We would like to do these calculations in the future.*

In the proof of Theorem 3 in Section 3.2, we shall use the following lemma proved in Section 3.1:

**Lemma 1.** *For any prime number p and any positive integer e, we have the following coefficient-wise congruence*

$$(x+1)^{p^{\epsilon}} \equiv (x^p+1)^{p^{\epsilon-1}} \pmod{p^{\epsilon}}$$

*of binomial expansions with indetermiate x.*

#### **2. A Proof of Theorem 2**

Although Theorem 2 can be proved by the same argument as the proof of ([2], Theorem 0.2), we shall describe its detailed proof here to make this article self-contained.

Let *n* and *e* be any positive integers and *p* be any prime number.

Firstly, we assume that *<sup>n</sup>* is of the form *<sup>n</sup>* <sup>=</sup> *<sup>p</sup>em* <sup>−</sup> <sup>1</sup> *<sup>p</sup><sup>e</sup>* <sup>−</sup> <sup>1</sup> with some integer *<sup>m</sup>* <sup>≥</sup> 1. In the algebra F*p*[*x*] of polynomials of one varible *x* with coefficients in the finite field F*<sup>p</sup>* = Z/*p*Z of *p* elements, we see that for any positive integer -,

$$\begin{aligned} (\mathbf{x} - \mathbf{1})^{p^\ell - 1} &= \frac{(\mathbf{x} - \mathbf{1})^{p^\ell}}{\mathbf{x} - \mathbf{1}} = \frac{\mathbf{x}^{p^\ell} - \mathbf{1}}{\mathbf{x} - \mathbf{1}} \\ &= \mathbf{x}^{p^\ell - 1} + \mathbf{x}^{p^\ell - 2} + \dots + \mathbf{x} + \mathbf{1}. \end{aligned}$$

Therefore we see that

$$\begin{aligned} (\mathbf{x}^{p^e-1} + \mathbf{x}^{p^e-2} + \dots + \mathbf{x} + 1)^n &= (\mathbf{x}^{p^e-1} + \mathbf{x}^{p^e-2} + \dots + \mathbf{x} + 1)^{\frac{p^m-1}{p^e-1}} \\ &= ((\mathbf{x} - 1)^{p^e-1})^{\frac{p^m-1}{p^e-1}} \\ &= (\mathbf{x} - 1)^{p^{em}-1} \\ &= \mathbf{x}^{p^{em}-1} + \mathbf{x}^{p^{em}-2} + \dots + \mathbf{x} + 1 \\ &= \mathbf{x}^{n(p^e-1)} + \mathbf{x}^{n(p^e-1)-1} + \dots + \mathbf{x} + 1 \end{aligned}$$

in F*p*[*x*]. By Remark 2 (1), this implies that the *n*-th row in the *p<sup>e</sup>* -Pascal's triangle consists of integers which are congruent to 1 modulo *p* as desired.

Conversely, we now assume that *n* is of the form

$$n = 1 + p^c + \dots + p^{c(m-1)} + k$$

with some integers *<sup>m</sup>* ≥ 1 and 1 ≤ *<sup>k</sup>* ≤ *<sup>p</sup>em* − 1. Moreover, we assume that we have

$$(\mathbf{x}^{p^e-1} + \mathbf{x}^{p^e-2} + \dots + \mathbf{x} + 1)^n = \mathbf{x}^{n(p^e-1)} + \mathbf{x}^{n(p^e-1)-1} + \dots + \mathbf{x} + 1$$

in <sup>F</sup>*p*[*x*] to obtain some contradiction. Since the left hand side is equal to (*<sup>x</sup>* <sup>−</sup> <sup>1</sup>)*n*(*pe*−1) and the right hand side is equal to *<sup>x</sup>n*(*pe*−1)+<sup>1</sup> <sup>−</sup> <sup>1</sup> *<sup>x</sup>* <sup>−</sup> <sup>1</sup> , we then have the equality

$$(\mathfrak{x} - \mathbf{1})^{\mathfrak{n}(p^e - 1) + 1} = \mathfrak{x}^{\mathfrak{n}(p^e - 1) + 1} - 1$$

in <sup>F</sup>*p*[*x*]. Since *<sup>n</sup>* <sup>=</sup> *<sup>p</sup>em* <sup>−</sup> <sup>1</sup> *<sup>p</sup><sup>e</sup>* <sup>−</sup> <sup>1</sup> <sup>+</sup> *<sup>k</sup>*, this implies that

$$(\mathfrak{x} - 1)^{p^{em} + k(p^e - 1)} = \mathfrak{x}^{p^{em} + k(p^e - 1)} - 1.$$

Let *vp*(*a*) be the *<sup>p</sup>*-adic valuation of any non-zero integer *<sup>a</sup>*, i.e., *<sup>p</sup>vp*(*a*) <sup>|</sup> *<sup>a</sup>* and *<sup>p</sup>vp*(*a*)+<sup>1</sup> *<sup>a</sup>*. Since <sup>1</sup> ≤ *<sup>k</sup>* ≤ *<sup>p</sup>em* − 1, we see that *vp*(*k*) < *em* and then

$$v\_p(p^{\epsilon m} + k(p^{\epsilon} - 1)) = v\_p(k).$$

Therefore we can put

$$p^{\varepsilon m} + k(p^{\varepsilon} - 1) = p^{v\_P(k)}t^k$$

with some positive integer *t* which is prime to *p*. Then we have

$$(\mathfrak{x} - \mathfrak{1})^{p^{\overline{\mathfrak{p}}p(k)}t} = \mathfrak{x}^{p^{\overline{\mathfrak{p}}p(k)}t} - \mathfrak{1} = (\mathfrak{x}^t - \mathfrak{1})^{p^{\overline{\mathfrak{p}}p(k)}t}$$

which implies that

$$(\mathbf{x} - \mathbf{1})^{p^{v\_p(k)}(t-1)} = (\mathbf{x}^{t-1} + \mathbf{x}^{t-2} + \dots + \mathbf{x} + \mathbf{1})^{p^{v\_p(k)}}.$$

since <sup>F</sup>*p*[*x*] is an integral domain. Since *<sup>p</sup>vp*(*k*) <sup>&</sup>lt; *<sup>p</sup>em*, we see that *<sup>t</sup>* <sup>≥</sup> 2. Therefore substituting *<sup>x</sup>* <sup>=</sup> <sup>1</sup> leads a contradiction *t* = 0 in F*<sup>p</sup>* as desired, and Theorem 2 is proved.

#### **3. An Application to the 4-Pascal's Triangle**

By Theorem 2, in the case where *p* = 2 and *e* = 2, we see that for any integer *n* ≥ 1, the *n*-th row in the 4-Pascal triangle consists of odd integers if and only if *<sup>n</sup>* is of the form <sup>4</sup>*<sup>m</sup>* <sup>−</sup> <sup>1</sup> <sup>3</sup> with some integer *m* ≥ 1.

In this section, we shall prove Theorem <sup>3</sup> asserting that for any integer *<sup>m</sup>* <sup>≥</sup> 2, the <sup>4</sup>*<sup>m</sup>* <sup>−</sup> <sup>1</sup> <sup>3</sup> -th row in the 4-Pascal's triangle is congruent to the sequence

$$\overbrace{1133\cdots1133}^{2^{2m}-3}\overbrace{3311\cdots3311}^{2^{2m}-3}$$

modulo 4. Here we should note that 22*m*−<sup>3</sup> is the number of 1133's and 3311's, respectively.

Then Theorems 2 and 3 imply that ([2], Conjecture 0.3) holds in the case where *k* = 4, i.e., there exists no row in the 4-Pascal's triangle consisting of integers which are congruent to 1 modulo 4 except the first row as we have seen in Corollary 1.

## *3.1. On a Congruence of Binomial Expansions*

Before proving Theorem 3, we shall prove Lemma 1 on a congruence of binomial expansions in this subsection.

Let *p* be any prime number and *e* any positive integer. In order to prove the congruence

$$(x+1)^{p^{\epsilon}} \equiv (x^p+1)^{p^{\epsilon-1}} \pmod{p^{\epsilon}}$$

of binomial expansions with indeterminate *x*, it suffices to see the following two congruences hold:

(1) For any integer 1 ≤ -≤ *<sup>p</sup><sup>e</sup>* − 1 which is prime to *<sup>p</sup>*,

$${p^\ell} C\_\ell \equiv 0 \pmod{p^\ell} .$$

(2) In the case where *<sup>e</sup>* ≥ 2, for any integers 0 ≤ *<sup>f</sup>* ≤ *<sup>e</sup>* − 2 and *<sup>i</sup>* such that 1 ≤ *ip<sup>f</sup>* ≤ *<sup>p</sup>e*−<sup>1</sup> − 1 and (*i*, *p*) = 1,

$$\_{p^{\ell}}\mathbb{C}\_{ip^{f+1}} \equiv \_{p^{\ell-1}}\mathbb{C}\_{ip^{f}} \pmod{p^{\ell}} .$$

Firstly, we shall prove the part (1). In the case where -= 1, we see that

$$\_{p^\varepsilon}\mathbb{C}\_1 = p^\varepsilon \equiv 0 \pmod{p^\varepsilon}.$$

Moreover, in the case where 2 ≤ -≤ *<sup>p</sup><sup>e</sup>* − 1, we see that

$$\_{p^{\ell}}\mathbb{C}\_{\ell} = \frac{p^{\ell}}{\ell} \prod\_{j=1}^{\ell-1} \frac{p^{\ell}-j}{j}.$$

Since *vp*(*p<sup>e</sup>* − *<sup>j</sup>*) = *vp*(*j*) for any 1 ≤ *<sup>j</sup>* ≤ - − <sup>1</sup> < *<sup>p</sup><sup>e</sup>* and is prime to *p*, we then see that

$$\begin{aligned} \upsilon(\_{p^\epsilon}\mathcal{C}\_\ell) &= \mathfrak{e} - \upsilon\_p(\ell) + \sum\_{j=1}^{\ell-1} \upsilon\_p\left(\frac{p^\epsilon - j}{j}\right) \\ &= \mathfrak{e} + \sum\_{j=1}^{\ell-1} (\upsilon\_p(p^\epsilon - j) - \upsilon\_p(j)) \\ &= \mathfrak{e}. \end{aligned}$$

Therefore *<sup>p</sup>eC*-≡ <sup>0</sup> (mod *<sup>p</sup>e*), and part (1) is proved.

Secondly, we shall prove part (2). We see that

$$\begin{split} &p^{\mathcal{C}}\mathbf{C}\_{ip^{f+1}} - p^{e^{-1}}\mathbf{C}\_{ip^{f}} \\ &= \frac{p^{e}}{ip^{f+1}} \cdot \frac{\prod\_{j=0}^{i-1} \left( \prod\_{1 \le k \le p^{f} \atop 1 \le k \le p^{f+1} - 1, \ (k,p)=1} (k + ip^{f+1} + (p^{e} - ip^{f+1})) \right)}{\prod\_{j=0}^{i-1} \left( \prod\_{1 \le k \le p^{f} \nmid -1, \ (k,p)=1} (k + ip^{f+1}) \right)} \cdot & \cdot p^{e-1} \cdot \mathbf{C}\_{ip^{f}-1} \\ & - \frac{p^{e-1}}{ip^{f}} \cdot \prescript{}{p^{e-1}-1}{\cdot p^{e-1}-1} \mathbf{C}\_{ip^{f}-1} \\ &= \frac{p^{e-f-1}}{i} \cdot \prescript{}{p^{e-1}-1}{\cdot p^{e-1}-1} \mathbf{C}\_{ip^{f}-1} \left( \frac{\prod\_{j=0}^{i-1} \left( \prod\_{1 \le k \le p^{f+1}-1, (k,p)=1} (k + ip^{f+1} + (p^e - ip^{f+1})) \right)}{\prod\_{j=0}^{i-1} \left( \prod\_{1 \le k \le p^{f+1}-1, (k,p)=1} (k + ip^{f+1}) \right)} - 1 \right) \end{split}$$

and that

$$\begin{aligned} &\prod\_{j=0}^{i-1} \left( \prod\_{\substack{1 \le k \le p^{f+1}-1,\ (k,p)=1}} (k+jp^{f+1}+(p^e-ip^{f+1})) \right) \\ &\equiv \prod\_{j=0}^{i-1} \left( \prod\_{\substack{1 \le k \le p^{f+1}-1,\ (k,p)=1}} (k+jp^{f+1}) \right) \\ &\equiv \left( \prod\_{\substack{1 \le k \le p^{f+1}-1,\ (k,p)=1}} k \right)^i (\text{mod } p^{f+1}). \end{aligned}$$

Since (*i*, *<sup>p</sup>*) = 1, we then see that *<sup>p</sup>eCipf*<sup>+</sup><sup>1</sup> <sup>−</sup> *<sup>p</sup>e*−<sup>1</sup>*Cipf* is divisible by *<sup>p</sup>e*−*f*−<sup>1</sup> · *<sup>p</sup>f*+<sup>1</sup> <sup>=</sup> *<sup>p</sup><sup>e</sup>* as desired.

## *3.2. A Proof of Theorem 3*

Now we shall prove Theorem 3 by means of Lemma 1 with *p* = 2 and *e* = 2, i.e., the congruence of binomial expansions

$$(\mathfrak{x}+1)^4 \equiv (\mathfrak{x}^2+1)^2 \pmod{4} . \dots (\ast : (\*))$$

By Remark 2 (1), proving Theorem 3 is equivalent to proving that for any integer *m* ≥ 2, the coefficient-wise congruence

$$\begin{aligned} & \left(\mathbf{x}^3 + \mathbf{x}^2 + \mathbf{x} + 1\right)^{\frac{4^m - 1}{3}} \\ & \equiv \mathbf{x}^{4^m - 1} + \mathbf{x}^{4^m - 2} - \mathbf{x}^{4^m - 3} - \mathbf{x}^{4^m - 4} + \dots + \mathbf{x}^{\frac{4^m}{2} + 3} + \mathbf{x}^{\frac{4^m}{2} + 2} - \mathbf{x}^{\frac{4^m}{2} + 1} - \mathbf{x}^{\frac{4^m}{2}} \\ & - \mathbf{x}^{\frac{4^m}{2} - 1} - \mathbf{x}^{\frac{4^m}{2} - 2} + \mathbf{x}^{\frac{4^m}{2} - 3} + \mathbf{x}^{\frac{4^m}{2} - 4} - \dots - \mathbf{x}^3 - \mathbf{x}^2 + \mathbf{x} + 1 \pmod{4} \quad \dots \end{aligned}$$

holds with indeterminate *x* by the induction on *m*.

Before doing this, we see the following

**Lemma 2.** *The polynomial in the right hand side of the congruence relation* (∗∗) *can be decomposed as*

$$((\mathbf{x}+1)(\mathbf{x}^2-1)(\mathbf{x}^4+1)\cdots(\mathbf{x}^{2^{2m-2}}+1)(\mathbf{x}^{2^{2m-1}}-1)\dots)$$

**Proof.** By a direct calculation, we can see that there exists some positive integer such that the polynomial in the right hand side of the congruence relation (∗∗) can be decomposed as

$$\begin{split} &((x+1)(\mathbf{x}^{4m-2}-\mathbf{x}^{4m-4}+\mathbf{x}^{4m-6}-\mathbf{x}^{4m-8}+\cdots+\mathbf{x}^{4\frac{m}{2}+6}-\mathbf{x}^{\frac{4m}{2}+4}+\mathbf{x}^{\frac{4m}{2}+2}-\mathbf{x}^{\frac{4m}{2}}) \\ & \qquad \qquad -\mathbf{x}^{\frac{4m}{2}-2}+\mathbf{x}^{\frac{4m}{2}-4}-\mathbf{x}^{\frac{4m}{2}-6}+\mathbf{x}^{\frac{4m}{2}-8}-\cdots-\mathbf{x}^{6}+\mathbf{x}^{4}-\mathbf{x}^{2}+1) \\ & \qquad \qquad = (\mathbf{x}+1)(\mathbf{x}^{2}-1)(\mathbf{x}^{4m-4}+\mathbf{x}^{4m-8}+\cdots+\mathbf{x}^{\frac{4m}{2}+4}+\mathbf{x}^{\frac{4m}{2}}) \\ & \qquad \qquad \qquad -\mathbf{x}^{\frac{4m}{2}-4}-\mathbf{x}^{\frac{4m}{2}-8}-\cdots-\mathbf{x}^{4}-1) \\ & \qquad \qquad = \cdots \\ &= (\mathbf{x}+1)(\mathbf{x}^{2}-1)(\mathbf{x}^{4}+1)\cdots(\mathbf{x}^{2^{\ell}}+1)(\mathbf{x}^{3\cdot2^{\ell+1}}+\mathbf{x}^{\cdot2\cdot2^{\ell+1}}-\mathbf{x}^{\prime\ell+1}-1) \\ & \qquad = (\mathbf{x}+1)(\mathbf{x}^{2}-1)(\mathbf{x}^{4}+1)\cdots(\mathbf{x}^{2^{\ell}}+1)(\mathbf{x}^{2^{\ell+1}}+1)(\mathbf{x}^{2^{\ell+2}}-1). \end{split}$$

Since the degree of the polynomial in the right hand side of the congruence relation (∗∗) is equal to 4*<sup>m</sup>* − 1, we then see that

$$\begin{aligned} 4^m - 1 &= 1 + 2 + 2^2 + \dots + 2^\ell + 2^{\ell+1} + 2^{\ell+2} \\ &= 2^{\ell+3} - 1, \end{aligned}$$

which implies that -= 2*m* − 3 as desired.

Let us start to prove Theorem 3 by the induction on *m* ≥ 2. Firstly, in the case where *m* = 2, since

$$(\mathbf{x}^2 + 1)^4 \equiv (\mathbf{x}^4 + 1)^2 \pmod{4}$$

and

$$\begin{aligned} (\mathbf{x} + \mathbf{1})^4 &\equiv (\mathbf{x}^2 + \mathbf{1})^2 \equiv \mathbf{x}^4 + 2\mathbf{x}^2 + \mathbf{1} \equiv \mathbf{x}^4 - 2\mathbf{x}^2 + \mathbf{1} \\ &\equiv (\mathbf{x}^2 - \mathbf{1})^2 \pmod{4} \end{aligned}$$

by the congruence relation (∗), we see that

$$\begin{aligned} (\mathbf{x}^3 + \mathbf{x}^2 + \mathbf{x} + 1)^5 &\equiv (\mathbf{x} + 1)(\mathbf{x}^2 + 1)(\mathbf{x} + 1)^4(\mathbf{x}^2 + 1)^4 \\ &\equiv (\mathbf{x} + 1)(\mathbf{x}^2 + 1)(\mathbf{x}^2 - 1)^2(\mathbf{x}^4 + 1)^2 \\ &\equiv (\mathbf{x} + 1)(\mathbf{x}^2 - 1)(\mathbf{x}^4 + 1)(\mathbf{x}^8 - 1) \pmod{4}. \end{aligned}$$

Therefore the congruence relation (∗∗) holds for *m* = 2 by Lemma 2.

Secondly, we assume that the congruence relation (∗∗) holds for some *m* ≥ 2. By the congruence relation (∗), we see that

$$\begin{aligned} \left(\mathbf{x} + 1\right)^{4^m} &\equiv \left(\mathbf{x} + 1\right)^{2^{2m}} \equiv \left((\mathbf{x} + 1)^4\right)^{2^{2m-2}} \\ &\equiv \left(\mathbf{x}^2 + 1\right)^{2^{2m-1}} \equiv \left((\mathbf{x}^2 + 1)^4\right)^{2^{2m-3}} \\ &\equiv \left(\mathbf{x}^{2^2} + 1\right)^{2^{2m-2}} \\ &\equiv \dots \\ &\equiv \left(\mathbf{x}^{2^{2m-1}} + 1\right)^2 \\ &\equiv \left(\mathbf{x}^{\frac{4^m}{2}} + 1\right)^2 \pmod{4}. \end{aligned}$$

By Lemma 2, we then see that

$$\begin{split} & \left( \mathbf{x}^3 + \mathbf{x}^2 + \mathbf{x} + 1 \right) \xrightarrow[3]{4^{m+1}-1} \\ & \equiv \left( \mathbf{x}^3 + \mathbf{x}^2 + \mathbf{x} + 1 \right) \xrightarrow[3]{4^{m}-1} + 4^{\pi} \\ & \equiv \left( \mathbf{x} + 1 \right) (\mathbf{x}^2 - 1) (\mathbf{x}^4 + 1) \cdot \dots \cdot (\mathbf{x}^{2^{2m-2}} + 1) (\mathbf{x}^{2^{2m-1}} - 1) (\mathbf{x}^2 + 1)^{4^{\pi}} (\mathbf{x} + 1)^{4^{\pi}} \\ & \equiv \left( \mathbf{x} + 1 \right) (\mathbf{x}^2 - 1) (\mathbf{x}^4 + 1) \cdot \dots \cdot (\mathbf{x}^{2^{2m-2}} + 1) (\mathbf{x}^{\frac{4^{\pi}}{2}} - 1) (\mathbf{x}^{4^{\pi}} + 1)^2 (\mathbf{x}^{\frac{4^{\pi}}{2}} + 1)^2 \\ & \equiv \left( \mathbf{x} + 1 \right) (\mathbf{x}^2 - 1) (\mathbf{x}^4 + 1) \cdot \dots \cdot (\mathbf{x}^{2^{2m-2}} + 1) (\mathbf{x}^{2^{2m-1}} + 1) (\mathbf{x}^{2^{2m}} + 1) (\mathbf{x}^{2^{4^{\pi}}} - 1) \\ & \equiv \left( \mathbf{x} + 1 \right) (\mathbf{x}^2 - 1) (\mathbf{x}^4 + 1) \cdot \dots \cdot (\mathbf{x}^{2^{2m}} + 1) (\mathbf{x}^{2^{2m+1}} - 1) \text{ (mod 4)}, \end{split}$$

i.e., the congruence relation (∗∗) also holds for *m* + 1 as desired. This proves Theorem 3.

**Author Contributions:** Conceptualization, A.Y.; Investigation, A.Y. and K.T.; Writing—original draft, A.Y. All authors have read and agreed to the published version of the manuscript.

**Acknowledgments:** The first author is very grateful to the second author, who is one of his students at Soka University, for giving some interesting talks regarding his calculations of the *k*-Pascal's triangles with some specified composite numbers *k* in seminars held in 2019 at Soka University.

**Conflicts of Interest:** The authors declare no conflict of interest.

## **References**


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