Design a site like this with WordPress.com

Twins!!

Today is my twin brother’s birthday, so I decided that’s a good enough excuse to talk a bit about twins. Whilst “twin brothers” are my favourite variety of twin, I’m actually going to be writing about my second favourite twins: Twin Primes.

Twin Primes refer to any pairs $(p, p+2)$ where both $p$ and $p + 2$ are prime. For example $(3,5)$, and $(5,7)$, and $(11,13)$. There are bigger twin primes too, like $(137,139)$, and $(1446089, 1446091)$, and $(18407771, 18407773)$.

As of September $2018$, the largest known pair of twin primes was $2996863034895 \times {2}^{1290000} \pm 1$. But what about twin primes bigger than that? How many are there? What are they? Well the truth is, we don’t know.It is conjectured there there are infinitely many twin primes, but this remains unproven.

So what do we know about twin primes? Well, aside from $5$, every prime number has at most one twin. This is because every third odd number is a multiple of $3$, and the only prime multiple of $3$ is $3$ itself. Using modular arithmetic, we can extend this idea further to say that, aside from $3$, if $(p, p+2)$ are twin primes, we must have $p, p_2 \not \equiv 0$ (mod $3$). The only way this is possible is if $p \equiv -1$ (mod $3$). Further, $2$ is the only even prime number, so $p$ and $p+2$ must both be odd. Combining this with the previous relation, we get that $p \equiv -1$ (mod $6$).

There is a cool theorem in number theory called Wilson’s Theorem. This states that for odd $p$, $p$ is prime if and only if $(p-1)! \equiv -1$ (mod $p$). This is the same as saying $(p-1)! +1 \equiv 0$ (mod $p$). As $p$ is odd, $p \not = 4$, so $4[(p-1)! +1] \equiv 0$ (mod $p$). Finally, $p \equiv 0$ (mod $p$) so we can say $4[(p-1)! +1] \equiv -p$ (mod $p$).

Now consider twin primes $(p, p+2)$. We established earlier that they must both be odd so the above relation must hold, i.e. $(p-1)! \equiv -1$ (mod $p$), and therefore $4[(p-1)! +1] \equiv -p$ (mod $p$).
Using Wilson’s Theorem on $p+2$ yields $(p+1)! \equiv -1$ (mod $p+2$).
Note that $(p+1)! = (p^2+p)(p-1)! = [(p+2)(p-1) +2](p-1)!$.
Hence we can write $[(p+2)(p-1) +2](p-1)! \equiv -1$ (mod $p+2$), i.e. $[(p+2)(p-1) +2](p-1)! + 1 \equiv 0$ (mod $p+2$).
This therefore means that $2(p-1)! + 1 \equiv 0$ (mod $p+2$).
We then do the following algebraic manipulation:

$2(p-1)! + 1 \equiv 0$ (mod $p+2$)
$2[2(p-1)! + 1] \equiv 0$ (mod $p+2$)
$2[2(p-1)! + 1] + (p+2) \equiv 0$ (mod $p+2$)
$4[(p-1)! + 1] + p \equiv 0$ (mod $p+2$)
$4[(p-1)! + 1] \equiv -p$ (mod $p+2$)

So now we have that $4[(p-1)! +1] \equiv -p$ (mod $p$) and $4[(p-1)! + 1] \equiv -p$ (mod $p+2$). Because $p$ and $p+2$ are coprime, we can combine these two equations and get $4[(p-1)! +1] \equiv -p$ (mod $p(p+2)$) for all twin primes $(p, p+2)$

If this has got you intrigued, see what you can work out (or find out) about cousin primes (where both $p$ and $p + 4$ are prime) and sexy primes (where both $p$ and $p + 6$ are prime)!

One final type of number I’ll mention is “Isolated Primes”. This refers to any number that isn’t a twin prime. It’s quite a fitting name really, as I know I’m a lot happier because of my twin.

Happy Birthday Jack!