Activity 1-8: Repunits

1. www.carom-maths.co.uk Activity 1-8: Repunits

2. 111, 11111, 11111111111, 11111111111111 are all repunits. They have received a lot of attention down the years. In particular, when are they prime? R1 = 1, no, R2 = 11, yes, R3 = 111, no...

3. 111111111111111= 111111111111111= 111  1001001001001 Task: prove that any repunit having a composite (non-prime) number of digits must be composite. A common move with repunits is to consider their value in bases other than 10. The above argument is exactly the same in bases other than 10.

5. So for a repunit to be a prime in base 10 is rare. Conjecture: there are infinitely many repunitprimes in base 10.

6. When is a repunitsquare? Well, 1 is a square – but if we search for others, they seem hard to find. Conjecture: 1 is the only squarerepunit. Task: how could we prove this?

7. How to approach this? Firstly, what remainders can a square have if you divide by 4? (2n)2 = 4n2, and so has remainder 0, while (2n+1)2 = 4n2 + 4n + 1, and so has remainder 1. Conclusion: a square can never have remainder 2 or 3 when divided by 4.

8. Now consider 111...111. 1 is a square. 11 is not a square. For repunits bigger than these, We have 111...111 = 111...11100 + 11. 4 goes into111...11100, and leaves a remainder 3 when it divides 11. So 111...111 cannot be a square, and 1 is the only square repunit.

9. Theorem: 1 is the only square repunit. And so we have... Another theorem: if a number is not divisible by either 2 or 5, then some multiple of this number must be a repunit. 7  15873 = 111111 3  37 = 111 13  8547 = 111111 11  1 = 11 17  65359477124183 = 1111111111111111 19  5847953216374269 = 111111111111111111 215291=111111

10. How to prove this? We can use a theorem due to Euler. Leonhard Euler, Swiss (1707-1783)

11. The numbers a and b are COPRIME if gcd(a, b) = 1, where gcd = ‘greatest common divisor’. Define (n) to be the number of numbers in {1, 2, 3…, n - 1} that are coprime with n. So (2) = 1, (3) = 2, (4) = 2, (5) = 4, (20) = 8. Task: find(5), (9), (45). What do you notice?

12. It turns out that (n) (the totient function)is what we call multiplicative. That is to say, if a and b are coprime, then (ab) = (a) (b). Thus (45) = (9) (5). Now Euler’s Theorem tells us: if a and b are coprime, then a divides b(a)– 1. So, for example, since 11 and 13 are coprime, 11 divides 13(11)  1, 137858491848 = 11 x 12532590168 and 13 divides 11(13) 1, 3138428376720 = 13 x 241417567440

13. Note that a repunit is of the form (10n–1)/9. Now supposeaand 10 have no common factor. Then 9a and 10 have no common factor. So by Euler’s Theorem, 9a divides 10(9a) 1. So 10(9a) 1 = 9ak, for some k. So a  k= (10(9a) 1)/9, which is a repunit. Thus some multiple of ais a repunit.

14. Are there any numbers that are repunits in more than one base? 31 = 111 (base 5) = 11111 (base 2) 8191 = 111 (base 90) = 1111111111111 (base 2) Goormaghtigh Conjecture: these are the only two.

15. With thanks to:Shaun Stevens, for his help and advice. Wikipedia, for another excellent article. Carom is written by Jonny Griffiths, hello@jonny-griffiths.net