Ulams Prime Spiral |

It's the 157th day of the year, and I pointed out on my daily post that 157 is not only prime, but is also part of a sexy triple; a string of three prime numbers each successively differing by 6. In this case the three are 151, 157, and 163 (for younger students, the Latin name for the number six was sex, thus Roman names that have "sextus", the sixth, in them.). In fact, it's reverse, 751 is also a sexy prime paired with 757.

With the mention of sexy pairs and sexy triplets, a curious young person might wonder can there be sexy quadruplets, sexy quintuplets, or even sexy sextuplets. To which I would answer, Yes..... and Yes.... and NO!

Since the ideas are all easy to explain and understand for even elementary school students with a little nod to modular, or clock arithmetic (or simple division) I will remind teachers (and inform some young readers) of what I think might be an approach at the most basic level.

First, it's easy to prove there are sexy quadruplets by simply searching and finding one, not too difficult since the first such four primes are 5, 11, 17, 23. and to prove there are four more sets below 100.

It's almost as easy to prove that there is a set of five in a row; since 29 is also prime, the 5, 11, 17, 23, 29 string is your proof there is at least one sexy quintuplet. But if you go on searching, you will not find another, ever..... and to prove that, we have to do a little very primitive number theory.

So let's prove, all in one go, that there is only one sexy quintuplet, and that there are no sexy n-tuples greater that n=5.

**a brief introduction to a property of divisibility and remainders:**

If you divide any number m by some other number, say three, you may get a reminder of zero, one, or two.

Now if you add 4 to m, and divide m+4 by three, you will get a remainder that is one more than the remainder for m (unless m has a reminder of 2, in which case m+4 will have remainder zero).

Another way to say this is if you have three numbers all differing by four, say 13, 17, 21... ONE of them has to be divisible by three (remainder zero) another must have a remainder of one, and another a remainder of two.

When you divide m by 15, you can get 15 different reminders depending on m... and whatever you get for a remainder, when you divide m+16 you will get a remainder that is one more than the last remainder.

**And now to the important application**, if you have five numbers in a row that differ by 6 (this is called an arithmetic sequence....) for example I will pick out of the far reaches of my mind, 83, 89, 95, 101, 107. We don't care if they are prime or not right now, we just want to show that one of them is divisible by 5, and it's pretty obvious which one is; but check the reminders of all the others. Notice that they appear in order.... R3, R4, R0, R1, R2 So if you have an arithmetic sequence of five numbers differing by 6, ONE of them must be divisible by five. Notice that in our one Quintuplet 5, 11, 17, 23, 29, one of them is divisible by five, but it is five itself which is prime. For any larger starting number, the multiple of five can not be five itself, and so no more sequences of sexy quintuplets exist beyond this one.

BUT what about sexy sextuplets or heptuplets or octuplets?????? ahh.... if they all differ by six, then one of them is divisible by five. So there can't be any sexy n-tuplets with n greater than five.