Activity 1-8: Repunits

111, 11111, , are all repunits. They have received a lot of attention down the years. In particular, when are they prime? R 1 = 1, no, R 2 = 11, yes, R 3 = 111, no...

= = 111  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.

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

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

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

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

Another theorem: if a number is not divisible by either 2 or 5, then some multiple of this number must be a repunit. 3  37 =  =  1 =  8547 =  =  =  5291= Theorem: 1 is the only square repunit. And so we have...

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

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, Task: find  (5),  (9),  (45). What do you notice? The numbers a and b are COPRIME if gcd(a, b) = 1, where gcd = ‘greatest common divisor’.  (20) = 8.

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). Now Euler’s Theorem tells us: if a and b are coprime, then a divides b  (a) – 1. Thus  (45) =  (9)  (5). So, for example, since 11 and 13 are coprime, 11 divides 13  (11)  1, = 11 x and 13 divides 11  (13)  1, = 13 x

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

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

With thanks to: Shaun Stevens, for his help and advice. Wikipedia, for another excellent article. Carom is written by Jonny Griffiths,