# A Surprising Fact about Prime Numbers

## Background

Basically, a **prime number** is an integer that's divisible only by itself and by 1. The second part of the definition is the convention that **1** is not counted as a prime number. Apparently this is a useful convention.

The **Larry Primes** are prime numbers that are greater than 3. Here are the Larry Primes that are less than 100: 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.

Let **m** and **n** be two Larry Primes. And let's suppose that **n** is greater than **m**.

**Larry's Prime Theorem (LPT): The difference between the squares of any two Larry Primes (**written in mathematical shorthand as** n^2 - m^2) is always divisible by 6. **

The 'hat' notation means "to the power of. Thus

**2^3 = 2*2*2 = 8 **

As an example of **LPT**, let's consider the primes 5 and 13.

13^2 - 5^2 = 169 - 25 = 144 which is divisible by 6. (144/6 = 24.)

Whadayaknow, I'm right so far! If you're still skeptical, please feel free to test-drive some other prime pairs on this list, before moving on to the next section.

## Why?

By definition, Larry Primes greater than 3 are *not* divisible by 2, or by 3.

6 is divisible by both 2 and 3.

When we divide a Larry Prime **p** by 6, the only possible remainders are 1 and 5.

Why?

By definition, zero cannot be a remainder.

2 cannot be a remainder, because that would make **p** an even number.

Ditto for 4 as a remainder.

3 cannot be a remainder, because that would make **p** divisible by 3.

Thus any Larry Prime can be written as either **6a + 1** or **6b + 5**

where **a** and **b** are both integers.

Now let's square both expressions.

**Equation 1: (****6a + 1)^2 =** **36a^2 + 12a + 1 = 6j + 1**

where **j** is an integer.

On the other hand,

**Equation 2: ****(****6b + 5)^2 =** **36b^2 + 60b + 25 = 6k + 1**

where **k** is an integer.

Note that the right hand sides of Equations 1 and 2 have the same form.

Finding the difference between the two right hand sides,

**(6j + 1) - (6k + 1) = 6(j - k)**

which is divisible by 6.

## A second surprising fact

Let **a, b, c, k, m, **and **n** be any six Larry Primes. Then

**a^2 + b^2 + c^2 + k^2 + m^2 + n^2 is divisible by 6. **

In other words, the sum of the squares of any 6 Larry Primes is always divisible by 6.

To see why, use an approach similar to that of the previous section.

## Hat-tip

The inspiration for this hub is a conversation from many years ago, with mathematician extraordinaire, Mamikon Mnatsakanian. Mamikon showed me an educational graphic of his design. Spider Math should help students visualize the distribution of prime numbers. However I have not been able to find it online yet.

Here's a LINK to Mamikon's website.

Copyright 2012 by Larry Fields

## Comments

Very Nice one

Hmmm. It took me a minute to see the whole thing. Very interesting.

Very interesting. Is there a reason why it is always divisible by 6?

Clear as mud Larry! lol! on a dark day in a tunnel full of mush! but thats not a reflection on your math, its a reflection on my mind and maths! me and numbers just do not get on, fascinating though! lol!

Thanks Larry

good knowledge based article.

I liked your mathematics behind prime numbers. It is very interesting indeed.

At one time, as an engineering student, I was very much interested in secrets of mathematical numbers, but then after graduating, I never practiced engineering. However, I do try to keep connected with mathematics in one way or the other.

I will shortly be reading a book titled 'Mathematical Nature Walk'. It seems to show mathematics behind many of the nature's phenomenon. Mathematics still intrigues and, in a way, intimidates me.

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