1.Introduction

Repunit is a number like 11, 111 that contains only the digit 1.

The repunit is defined as R(n) = (10^n-1)/9.

The number of digits must be a prime number so that repunit may become a prime number.

The only known repunit primes are R(2), R(19), R(23), R(317), and R(1031).
R(2)=11
R(19)=1111111111111111111
R(23)=11111111111111111111111
R(317)=1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
       1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
       1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
       11111111111111111111111111111111111111111111111111111

And known probable prime repunits are R(49081), R(86453), R(109297),and R(270343).


I show a theorem about repunit.


Euler showed the following theorem about Mersenne number.

Theorem(Euler)

Let p = 3 (mod 4) be prime and q=2p+1 is also prime.
Then q divides Mersenne number M(p)=2^p-1.

The theorem of Euler is related to Quadratic residue.


I searched the similar theorem that is related to Quadratic residue.

In the same way as Euler,we can prove the following theorem.
Numerical example:

(10^41-1)/9 = 11111111111111111111111111111111111111111
            = (83)*(1231)*(201763709900322803748657942361)*(538987)

2. Theorem
          

Theorem 1.

Let p = 1, 13, and 19  (mod 20) be prime and q = 2p+1 is also prime.
Then q divides repunit R(p) = (10^p-1)/9.


Proof.
    p,q: prime number.
    n: integer.
(n/q): Legendre's symbol.

(10/q)=10^{(q-1)/2} mod q (Euler's criterion).

(10/q)=(2/q)*(5/q) (quadratic reciprocity law).

Case 1: (2/q)=1 and (5/q)=1
        Prime number q must be 1, 9, 31,and 39 mod 40 to satisfy the condition.
        As prime number p= (q-1)/2,p must be =19 mod 20.

Case 2: (2/q)=-1 and (5/q)=-1
        In the same way,q must be 3, 13, 27,and 37 mod 40.
        Then,p must be =1, 13 mod 20.


Therefore,if p = 1, 13, and 19 mod 20 and q = 2p+1 is also prime,then 10^p=1 mod q.


Example: p < 1000

 

         p    q     factors of (10^p-1)/9 (from Torbjorn Granlund's page) 

        41    83    83 1231 538987 201763709900322803748657942361
        53   107    107 1659431 1325815267337711173 47198858799491425660200071
       113   227    227 908191467191 p98

       173   347    347 21528169344472027 46194618816084982100679234312974236345786346173 p107

       179   359    359 36558961 4201521652717 352543640588653 p141

       233   467    467 47533 115081654907 89412401116867 42983757528467293 10857980200444401822643 20131492120828919814484857298874674155298711142397769181347 p103

       239   479    479 142847911 383155477843726029783939406113226468701730728790004161 128780300340244872385688233345188210841783983757299260103530718169486826135819357 p94

       281   563    563 2597610354323 380431988961690791 133355295524380732608674846305692470329 c210

       293   587    587 9476124263231731100867 72753092406224481018194292737207864791672573037959 c218

       359   719    719 73237 799853 33569502677 18200054748466867994849 8745075387933004096394385246656921347 c275

       419   839    839 82963 481154717070862987053019025329 12627408558266078891364887027326567507 c344

       593  1187    1187 c589

       641  1283    1283 49999 105138885091009723 223886701054873016774410942923722863481 c577

       653  1307    1307 3919 c646

       659  1319    1319 5283793003008413 37362749172404658991 c620

       719  1439    1439 1153277 6699643 3713656876665286297046096029015677459199 c663

       761  1523    1523 6089 1095841 15286969 c740

       953  1907    1907 6318391 c942
                    


Theorem 2.

Let p and q be primes. If q^2 divides R(p), then 10^{(q-1)/2} = 1 mod q^2,
so, q is a generalized Wieferich prime. 
There doesn't exist a prime q such that q^2 divides R(p) where q < 10^9.


Proof.

Let a is an integer and gcd(q,a)=1.

We define that the generalized Wieferich primes are the primes q for which a^(q-1) = 1 (mod q^2).

If q divides R(p),then q = 2*k*p+1 for some integer k.

10^p=10^{(q-1)/(2*k)} = 1 mod q^2.

Hence, 10^(q-1) = 1 mod q^2 shows that q is a generalized Wieferich prime.

I searched and found q=487 and 56598313 as the solutions of 10^(q-1) = 1 mod q^2 where q < 10^9.

Actually,any R(p) is not divisible by q=487 and 56598313.

So, there doesn't exist a prime q such that q^2 divides repunit where q < 10^9.