Activity 1-8: Repunits

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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...

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.

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 = 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.

Now consider 111...111.

1 is a square.

For repunits bigger than these, We have 111...111 = 111...11100 + 11.

4 goes into 111...11100, and leaves a remainder 3 when it divides 11.

So 111...111 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 = 111 7 15873 = 111111

11 1 = 11 13 8547 = 111111

17 65359477124183 = 1111111111111111

19 5847953216374269 = 111111111111111111

215291=111111

Theorem: 1 is the only square repunit.

And so we have...

We can use a theorem due to Euler.

Leonhard Euler,Swiss

(1707-1783)

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,

137858491848 = 11 x 12532590168 and

13 divides 11(13) 1, 3138428376720 = 13 x 241417567440

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 (10n – 1)/9.

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.

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