The concrete theory of numbers: initial numbers and wonderful properties of numbers repunit

TL;DR

This paper explores properties of initial numbers and repunit numbers, proving key gcd and divisibility properties, and analyzing their prime divisors, contributing to the theoretical understanding of these special numbers.

Contribution

It introduces new proofs of gcd and divisibility properties of repunit numbers and examines their prime divisor structure, advancing the theoretical framework of number theory.

Findings

gcd(R_a, R_b) = R_{gcd(a,b)}

R_{ab}/(R_a R_b) is an integer iff gcd(a,b)=1

Divisors of repunit numbers relate to prime number degrees

Abstract

In this work initial numbers and repunit numbers have been studied. All numbers have been considered in a decimal notation. The problem of simplicity of initial numbers has been studied. Interesting properties of numbers repunit are proved: $gcd(R_a, R_b) = R_{gcd(a,b)}$; $R_{ab}/(R_aR_b)$ is an integer only if $gcd(a,b) = 1$, where $a\geq1$, $b\geq1$ are integers. Dividers of numbers repunit, are researched by a degree of prime number.

Equations

$$N=\{x{\}}_n=x\dots x,n>0.$$

$$E_{n,k}=\{1\{0{\}}_k{\}}_{n}1,$$

$$R_n=E_{n-1,0},$$

$$R_n=(10^n-1)/9,$$

$$E_{n,k}=\frac{R_{(k+1)(n+1)}}{R_{k+1}}=\frac{10^{(k+1)(n+1)}-1}{10^{k+1}-1}.$$

$$R_a=R_d\cdot \{10^{d(a_1-1)}+\dots +10^d+1\},$$

$$R_b=R_d\cdot \{10^{d(b_1-1)}+\dots +10^d+1\}.$$

$$A=10^{d(a_1-1)}+\dots +10^d+1,$$

$$B=10^{d(b_1-1)}+\dots +10^d+1.$$

$$A\equiv 0(modq),B\equiv 0(modq).$$

$$(10^d{)}^{d_{min}}\equiv 1(modq),$$

$$gcd(R_{p^k}/R_{p^t},R_{p^s})=1.$$

$$A=R_{p^k}/R_{p^t}=(10^{p^t}{)}^{p^{k-t}-1}+(10^{p^t}{)}^{p^{k-t}-2}+\dots +10^{p^t}+1.$$

$$gcd(R_{ab},R_a{R}_b)=R_a{R}_b.$$

$$R_a{R}_b/R_{(a,b)}\le gcd(R_{ab},R_a{R}_b)<R_a{R}_b.$$

$$R_a=R_{(a,b)}X,R_b=R_{(a,b)}Y,$$

$$R_{ab}=R_{a}A=R_{b}B,$$

$$A=10^{a(b-1)}+10^{a(b-2)}+\dots +10^a+1,$$

$$B=10^{b(a-1)}+10^{b(a-1)}+\dots +10^b+1.$$

$$(10^x-1)/9>x,$$

$$R_a\equiv 0(mod{3}^n),but{R}_a\not\equiv 0(mod{3}^{(n+1)}).$$

$$r_n=10^{11^n}+1\equiv 0(mod11^{n+1}),but{r}_n\not\equiv 0(mod11^{n+2}).$$

$$\begin{gathered} A=10^{11^{k-1}10}-10^{11^{k-1}9}+10^{11^{k-1}8}-10^{11^{k-1}7}+10^{11^{k-1}6}- \\ -10^{11^{k-1}5}+10^{11^{k-1}4}-10^{11^{k-1}3}+10^{11^{k-1}2}-10^{11^{k-1}}+1. \end{gathered}$$

$$R_a\equiv 0(mod11^{n+1}),but{R}_a\not\equiv 0(mod11^{n+2}).$$

$$gcd(R_{ab},R_a{R}_b)=((R_a{R}_b)/R_{(a,b)})\cdot 3^L,$$

$$gcd(R_{ab},R_a{R}_b)=((R_a{R}_b)/R_{(a,b)})\cdot 3^L\cdot 11^S.$$

$$E_{n,k}=E_{n,t-1}\cdot ((10^{(n+1)t}+1)/(10^t+1)),$$

$$E_{n,k}=R_{(n+1)(k+1)}/R_{(k+1)}=R_{(n+1)}\cdot (R_{(n+1)(k+1)}/(R_{k+1}\cdot R_{n+1})).$$

Full text

**Devoted to the tercentenary from the date of birth (4/15/1707)

of Leonhard Euler

**

1 Introduction

Let $x\geq 0$, $n\geq 0$ be integers. An integer $N$, which record consists from $n$ records of number $x$, we shall designate by

[TABLE]

For $n=0$ it is received $\{x\}_{0}=\varnothing$ an empty record. For example, $\{10\}_{3}1=1010101$, $\{10\}_{0}1=1$, etc.

Palindromic numbers of a kind

[TABLE]

where $n\geq 0$, $k\geq 0$ we will name initial numbers. We will notice that $E_{0,k}=1$ at any $k\geq 0$.

Numbers repunit(see[2, 3, 4]) are natural numbers, which records consist of units only, i.e. by definition

[TABLE]

where $n\geq 1$.

In decimal notation the general formula for numbers repunit is

[TABLE]

where $n=1,2,3,\ldots$ .

There are known only five prime repunit for $n=$2,19, 23, 317, 1031.

Known problem** ((Prime repunit numbers[3])).**

Whether exists infinite number of prime numbers repunit ?

Will we use designations further :

$(a,b)=gcd(a,b)$ the greatest common divider of integers $a>0$, $b>0$.

$p,q$ odd prime numbers.

If it is not stipulated specially, the integer positive numbers are considered.

2 Initial numbers

Let’s consider the trivial properties of initial numbers.

Theorem 1**.**

Following trivial statements are fair :

(1) General formula of initial numbers is

[TABLE]

*(2) For $k\geq 0$, $n\geq m\geq 1$ if $n+1\equiv 0(mod\,(m+1))$,

then $(E_{n,k},E_{m,k})=E_{m,k}$.*

*(3) For $k\geq 0$, $n>m\geq 1$ if integer $s\geq 1$, exists such

that $n+1\equiv 0(mod\,(s+1))$, $m+1\equiv 0(mod\,(s+1))$, then

$(E_{n,k},E_{m,k})\geq E_{s,k}>1$.*

*(4) For $k\geq 0$, $n>m\geq 1$ $(E_{n,k},E_{m,k})=1$ when and only then,

$(n+1,m+1)=1$.*

Proof.

1. Properties (1)—(3) are obvious.

2. The Proof of property (4). Necessity. Let

$(E_{n,k},E_{m,k})=~{}1$ and $(n+1,m+1)=s>1$, $s-1\geq 1$. From property (3) of the theorem follows that $(E_{n,k},E_{m,k})\geq E_{s-1,k}=\{1\{0\}_{k}\}_{s-1}1>1$. Appears the contradiction .

Sufficiency of property (4). Let $(n+1,m+1)=1$, then will be integers $a>0$, $b>0$, such that either $a(n+1)=b(m+1)+1$

or $b(m+1)=a(n+1)+1$. Let’s assume, that $(E_{n,k},E_{m,k})=d>1$.

a) Let $a(n+1)=b(m+1)+1$, then $E_{b(m+1),k}=E_{a(n+1)-1,k}=(10^{a(n+1)(k+1)}-1)/(10^{k+1}-1)\equiv 0(mod\,E_{n,k})\equiv 0(mod\,d)$.

On the other hand $E_{b(m+1),k}=(10^{(k+1)\{b(m+1)+1\}}-1)/(10^{k+1}-1)=((10^{b(m+1)(k+1)}-1)/(10^{k+1}-1))\cdot 10^{k+1}+1\equiv$

$\equiv 1(mod\,E_{m,k})\equiv 1(mod\,d)$. Appears the contradiction.

b) Let $b(m+1)=a(n+1)+1$, then $E_{a(n+1),k}=E_{b(m+1)-1,k}=(10^{b(m+1)(k+1)}-1)/(10^{k+1}-1)\equiv 0(mod\,E_{m,k})\equiv 0(mod\,d)$.

On the other hand $E_{a(n+1),k}=(10^{(k+1)\{a(n+1)+1\}}-1)/(10^{k+1}-1)=((10^{a(n+1)(k+1)}-1)/(10^{k+1}-1))\cdot 10^{k+1}+1\equiv$

$\equiv 1(mod\,E_{n,k})\equiv 1(mod\,d)$. Have received the contradiction. ∎

3 Numbers repunit

Let’s consider trivial properties of numbers repunit.

Theorem 2**.**

Following trivial statements are fair :

(1) The number $R_{n}$ is prime only if $n$ number is prime.

(2) If $p>3$ all prime dividers of number $R_{p}$ look like $1+2px$ where $x\geq 1$ is integer.

(3) $(R_{a},R_{b})=1$ if and only if $(a,b)=1$.

Proof.

Property (1) of theorem is proved in ([2, 3]), property (2) is proved in ([1]), as exercise. Property (3) is the corollary of the theorem 1. ∎

Theorem 3**.**

$(R_{a},R_{b})=R_{(a,b)}$, where $a\geq 1$, $b\geq 1$ are integers.

Proof.

Validity of the theorem for $(a,b)=1$ follows from property (3) of theorem 2. Let $(a,b)=d>1$, where $a=a_{1}d$, $b=b_{1}d$, $(a_{1},b_{1})=1$. Let’s consider equations

[TABLE]

[TABLE]

Let

[TABLE]

[TABLE]

Let’s assume, that $(A,B)>1$, and $q$ is a prime odd number such that

[TABLE]

If $q=3$, then $10^{t}\equiv 1(mod\,q)$ for any integer $t\geq 1$. Then from (6) it follows that $a_{1}\equiv b_{1}\equiv 0(mod\,q)$. Have received the contradiction.

Thus, $q>3$. Then there exists an index $d_{min}$, to which the number $10^{d}$ belongs on the module $q$.

[TABLE]

where $d_{min}\geq 1$.

If $d_{min}=1$, then it follows from (6) that $a_{1}\equiv b_{1}\equiv 0(mod\,q)$. Have received the contradiction. Hence $d_{min}>1$. As $R_{a}\equiv R_{b}\equiv~{}0(mod\,q)$, then $(10^{d})^{a_{1}}\equiv 1(mod\,q)$ and $(10^{d})^{b_{1}}\equiv 1(mod\,q)$.

Then $a_{1}\equiv b_{1}\equiv 0(mod\,d_{min})$. Have received the contradiction. ∎

Theorem 4**.**

Let $p>3$ be a prime number, $k\geq t\geq 1$, $t\geq s\geq 1$ integer numbers. Then

[TABLE]

Proof.

Let’s consider expression

[TABLE]

If $(A,R_{p^{s}})>1$, then the prime number $q$ exists such that

$A\equiv~{}0(mod\,q)$ $R_{p^{s}}\equiv~{}0(mod\,q)$. Hence $10^{p^{t}}\equiv~{}1(mod\,q)$, then $A\equiv~{}p^{k-t}\equiv~{}0(mod\,q)$, $p=q=3$. Have received the contradiction, because $p>3$. ∎

Theorem 5**.**

Let $a\geq 1$, $b\geq 1$ are integers, then the following statements are true :

(1) If $(a,b)=1$, then

[TABLE]

(2) If $(a,b)>1$, then

[TABLE]

Proof.

1. Let $(a,b)=1$, then $(R_{a},R_{b})=R_{(a,b)}=1$,

$R_{ab}=R_{a}X=R_{b}Y$, $X=cR_{b}$, where $c\geq 1$ is integer. $R_{ab}=cR_{a}R_{b}$.

2. Let $(a,b)=d>1$, $a=a_{1}d$, $b=b_{1}d$, $(a_{1},b_{1})=1$, $a_{1}\geq 1$, $b_{1}\geq 1$. As $gcd(R_{a},R_{b})=R_{(a,b)}$, we receive equality

[TABLE]

where $(X,Y)=1$.

Further, $R_{ab}=R_{a}A=R_{b}B=XAR_{(a,b)}=YBR_{(a,b)}$, $XA=YB$,

$A=Yz$, $B=Xz$, $z\geq 1$ is integer. Then $R_{ab}=XYR_{(a,b)}z$,

$R_{ab}=zR_{a}R_{b}/R_{(a,b)}$. We have proved, that

$R_{a}R_{b}/R_{(a,b)}\leq gcd(R_{ab},R_{a}R_{b})$.

Let’s assume, that $gcd(R_{ab},R_{a}R_{b})=R_{a}R_{b}$, then $R_{ab}=zR_{a}R_{b}$, where $z\geq 1$ is integer. Let’s consider equalities

[TABLE]

where

[TABLE]

[TABLE]

Since $A=R_{b}z$, $B=R_{a}z$, $10^{a}\equiv 1(mod\,R_{(a,b)})$,

$10^{b}\equiv 1(mod\,R_{(a,b)})$, then $A\equiv B\equiv 0(mod\,R_{(a,b)})$, hence

$a\equiv b\equiv 0(mod\,R_{(a,b)})$.

Thus, comparison $(a,b)\equiv 0(mod\,R_{(a,b)})$ or

$d\equiv 0(mod\,R_{d})$ is fair, that contradicts an obvious inequality

[TABLE]

where $x>1$ is real. ∎

{$\bigstar$} The Important corollary of the theorem 5.

Number $R_{ab}/(R_{a}R_{b})$ is integer when and only when $(a,b)=1$, where $a\geq 1$, $b\geq 1$ are integers.

Let’s quote some trivial statements for numbers repunit.

Lemma 1**.**

If $a=3^{n}b$, $(b,3)=1$, then

[TABLE]

Proof.

If $n=1$, then $R_{a}=R_{3}B$, where $B=10^{3(b-1)}+\,\ldots\,+10^{3}+1$, $R_{3}\equiv 0(mod\,3)$, $B\equiv b\not\equiv 0(mod\,3)$. Thus,

$R_{a}\equiv 0(mod\,3)$, but $R_{a}\not\equiv 0(mod\,3^{2})$.

Let comparisons (12) be proved for $n\leq k-1$. We shall consider $a=3^{k}b$, $(b,3)=1$. Then $R_{a}=R_{3^{k-1}b}A$, where $A=10^{3^{k-1}b2}+10^{3^{k-1}b}+1$.

$R_{3^{k-1}b}\equiv 0(mod\,3^{k-1})$, but $R_{3^{k-1}b}\not\equiv 0(mod\,3^{k})$, $A\equiv 0(mod\,3)$, but $A\not\equiv 0(mod\,3^{2})$. ∎

Lemma 2**.**

If $n\geq 0$ is integer, then

[TABLE]

Proof.

$r_{0}=11\equiv 0(mod\,11)$, but $r_{0}=11\not\equiv 0(mod\,11^{2})$.

$r_{1}=10^{11}+1\equiv 0(mod\,11^{2})$, but $r_{1}\not\equiv 0(mod\,11^{3})$.

Let’s make the inductive assumption, that formulas (13) are proved for

$n\leq k-1$, where $k-1\geq 1$, $k\geq 2$. Let $n=k$, then

$r_{k}=10^{11^{k}}+1=(10^{11^{k-1}})^{11}+1=r_{k-1}A$, where

[TABLE]

Since, due to the inductive assumption $10^{11^{k-1}}\equiv-1(mod\,11^{k})$, where

$k\geq 2$, then $A\equiv 11(mod\,11^{k})$. Then $A\equiv 0(mod\,11)$, but

$A\not\equiv 0(mod\,11^{2})$. Thus, we receive, that $r_{k}\equiv 0(mod\,11^{k+1})$, but

$r_{k}\not\equiv 0(mod\,11^{k+2})$. ∎

Lemma 3**.**

For an integer $a\geq 1$, the following statements are true :

(1) If $a$ is odd, then $R_{a}\not\equiv 0(mod\,11)$.

(2) If $a=2(11^{n})b$, $(b,11)=1$, then

[TABLE]

Proof.

If $a$ is odd, then $R_{a}\equiv 1(mod\,11)$. If $a=2(11^{n})b$, $(b,11)=1$, then $R_{a}=((10^{2(11)^{n}})^{b}-1)/9=R_{11^{n}}\cdot r_{n}\cdot A$, where $r_{n}=10^{11^{n}}+1$,

$A=10^{2(11^{n})(b-1)}+\,\ldots\,+10^{2(11^{n})}+1$. $R_{11^{n}}\not\equiv 0(mod\,11)$,

$A\equiv~{}b\not\equiv~{}0(mod\,11)$. Then validity of the statement (2) of lemma 3 follows from lemma 2. ∎

{$\bigstar$} The assumption: the general formula for $gcd(R_{ab},R_{a}R_{b})$.

If $a\geq 1$, $b\geq 1$ are integers, $d=(a,b)$, where $d=3^{L}\cdot 11^{S}\cdot c$, $(c,3)=1$, $(c,11)=1$, $L\geq 0$, $S\geq 0$, then equalities are true :

— if $c$ is an odd number, then

[TABLE]

— if $c$ is an even number, then

[TABLE]

Let’s give another two obvious statements in which divisors of numbers

repunit are studied, as degrees of prime number.

Lemma 4**.**

*If $p,q$ are prime numbers and $R_{p}\equiv 0(mod\,q)$, but

$R_{p}\not\equiv 0(mod\,q^{2})$, then statements are true :*

(1) For any integer $r$, $0<r<q$, $R_{pr}\not\equiv 0(mod\,q^{2})$.

(2) For any integer $n$, $n\geq 1$, $R_{p^{n}}\not\equiv 0(mod\,q^{2})$.

Proof.

1. $R_{pr}=R_{p}\cdot\widehat{R}_{pr}$, where $\widehat{R}_{pr}=10^{p(r-1)}+10^{p(r-2)}+\,\\ +\,\ldots\,+10^{p}+1$. If $R_{pr}\equiv 0(mod\,q^{2})$, then $\widehat{R}_{pr}\equiv 0(mod\,q)$,

$r\equiv~{}0(mod\,q)$. Have received the contradiction.

2. If $n>1$ found such that $R_{p^{n}}\equiv 0(mod\,q^{2})$, then from (7) follows $(R_{p^{n}}/R_{p},R_{p})=1$. Have received the contradiction. ∎

Lemma 5**.**

If $p,q$ are prime numbers and $R_{p}\equiv 0(mod\,q)$, then

$R_{pq^{n}}\equiv 0(mod\,q^{n+1})$.

Proof.

Since $R_{pq}=R_{p}\cdot\widehat{R}_{pq}$, where $\widehat{R}_{pq}=10^{p(q-1)}+\\ +\,10^{p(q-2)}+\,\ldots\,+10^{p}+1$, then $\widehat{R}_{pq}\equiv 0(mod\,q)$, $R_{pq}\equiv 0(mod\,q^{2})$.

Let’s assume that $R_{pq^{n-1}}\equiv 0(mod\,q^{n})$. Then

$R_{pq^{n}}=R_{pq^{n-1}\cdot q}=R_{pq^{n-1}}\cdot\widehat{R}_{pq^{n-1}\cdot q}$, where

$\widehat{R}_{pq^{n-1}\cdot q}=10^{pq^{n-1}\cdot(q-1)}+10^{pq^{n-1}\cdot(q-2)}+\,\ldots\,+10^{pq^{n-1}}+1\equiv~{}0(mod\,q)$,

$R_{pq^{n}}\equiv 0(mod\,q^{n+1})$. ∎

4 Problem of simplicity of initial numbers

Let’s consider the problem of simplicity of initial numbers $E_{n,k}$, where

$k\geq 0$, $n\geq 0$.

If $k=0$, then $E_{n,0}=R_{n+1}$. Thus, simplicity of numbers $E_{n,0}$ – is known problem of prime numbers repunit $R_{p}$, where $p$ is prime number.

If $n=1$, then $E_{1,k}=1\{0\}_{k}1=10^{k+1}+1$. As number $E_{1,k}$ can be prime only when $k+1=2^{m}$, $m\geq 0$ is integer, then we come to the known problem of simplicity of the generalized Fermat numbers $f_{m}(a)=a^{2^{m}}+1$ for $a=10$. Generalized Fermat numbers nave been define by Ribenboim [5] in 1996, as numbers of the form $f_{n}(a)=a^{2^{n}}+1$, where $a>2$ is even.

The generalized Fermat numbers $f_{n}(10)=10^{2^{n}}+1$ for $n\leq 14$ are prime only if $n=0,1$. $f_{0}(10)=11$, $f_{1}(10)=101$.

Theorem 6**.**

Let $n>1$, $k>0$. If any of conditions

(1) $n$ number is odd,

(2) $k$ number is odd,

(3) $n+1\equiv 0(mod\,3)$,

*(4) $(n+1,k+1)=1$,

is true, then number $E_{n,k}$ is compound.*

Proof.

1. $n+1=2t$, $t>1$. Then $E_{n,k}=E_{t-1,k}\cdot(10^{t(k+1)}+1)$, where $t>1$, $t-1\geq 1$. As $E_{t-1,k}>1$, then $E_{n,k}$ is compound number.

2. Let $k$ be an odd number. Due to the proved condition (1) we count that number $(n+1)$ is odd. $k+1=2t\geq 2$, $t\geq 1$. Further,

[TABLE]

where $n>1$, $t-1\geq 0$, $E_{n,t-1}>1$, number $(10^{(n+1)t}+1)/(10^{t}+1)>1$ is integer.

3. If $n+1\equiv 0(mod\,3)$, then $E_{n,k}\equiv 0(mod\,3)$, $E_{n,k}>11$.

4. Let $n>1$, $k\geq 1$, $(n+1,k+1)=1$, then

[TABLE]

Due to the theorem 5 number $z=R_{(n+1)(k+1)}/(R_{k+1}\cdot R_{n+1})$ is integer. Further,

$z>(10^{(n+1)(k+1)}-1)/(10^{n+k+2})=10^{nk-1}-1/(10^{n+k+2})$, $nk-1\geq 1$, thus, $z>1$. ∎

Question of simplicity of initial numbers under conditions, when

$(n+1,k+1)>1$, $(n+1)$ number is odd, $(k+1)$ number is odd,

$n+1\not\equiv 0(mod\,3)$, remains open.

In particular, it is interesting to considerate numbers $E_{p-1,p-1}=R_{p^{2}}/R_{p}$, where $p$ is prime number. For $p<100$ numbers $E_{p-1,p-1}$ are compound.

5 The open problems of numbers repunit

The known problem of numbers repunit remains open.

Problem 1** ((Prime repunit numbers[3])).**

Whether there exists infinite number of prime numbers $R_{p}$, $p$–prime number ?

Problem 2**.**

Whether all numbers $R_{p}$, $p$–prime number, are numbers free from squares ?

The author has checked up for $p<97$, that numbers $R_{p}$ are free from squares. Another following open questions are interesting :

Problem 3**.**

If number $R_{p}$ is free from squares, where $p>3$ is prime number, whether will number $n$, be found such what number $R_{p^{n}}$ contains a square ?

Problem 4**.**

$p$* is prime number, whether there are simple numbers of a kind

$E_{p-1,p-1}=R_{p^{2}}/R_{p}$ ?*

The author has checked up to $p\leq 127$, that numbers $E_{p-1,p-1}$ is compound. It is known, that $R_{p}$ divide by number $(2p+1)$ for prime numbers $p=41,53$, $R_{p}$ divide by number $(4p+1)$ for prime numbers $p=13,43,79$.

There appears a question :

Problem 5**.**

Whether there is infinite number of prime numbers $p$, such that $R_{p}$ divide by number $(2p+1)$ or is number $(4p+1)$ ?

The remark.

If the number $p>5$ Sophie Germain prime (i.e. number $2p+1$ is prime too), then either $R_{p}$ or $R_{p}^{+}=(10^{p}+1)/11$ divide by number $(2p+1)$.

6 The conclusion

Leonhard Euler, professor of the Russian Academy of sciences since 1731, has paid mathematics forever ! Euler’s invisible hand directs the development of concrete mathematics for more than 200 years.

Euler’s titanic work which has opened a way to freedom to mathematical community, admires. The pleasure caused by Euler’s works warms hearts.