Copyright C Brian Warburton August 2017

The Determination of Prime Numbers

By means of an Inequality

# Introduction

There is a great current interest in the possibility of the determination of prime numbers as these are used

in the transmission of secretised information e.g. commercial credit card numbers.

However, over the centuries, it has proved virtually impossible to provide a simple algorithm to

achieve this purpose. The following procedure will confirm or otherwise the identity of a given candidate

for prime number.

# Procedure

A prime number is a real, positive, integer whose only factors are itself and unity.

Consider the following inequality:-

2q+1 <> (2r+1)(2p+1)                                                          (1)

Where q, r and p are any real, positive and definite integers.

The L.H.S represents a prime number, which (apart from 2) must be odd.

The R.H.S represents a non-prime odd integer of factors 2r+1 and 2p+1.

As will be shown later, and in order to simplify and inject some symmetry into the R.H.S,

we make the substitution:-

p = r+t                                                                                     (2)

Perhaps a more satisfactory explanation of (2) above is as follows. In ordinary algebra

and in arithmetic, the order of multiplication of two variables is unimportant. Hence the

simplification enabled in (2). It is also helpful to visualize the elements r,p as members

of a rectangular matrix : Apr . Note that when t=0, the R.H.S is a perfect square.

After (2) above and further simplification:-

q <> 2r(r+1) + t(2r+1)                                                          (3)

Let a new variable, I be 0.5r(r+1)

We note that r(r+1) is always even, which invites simplification of division by four:-

0.25q <> I + 0.5t(r+0.5)                                                       (4)

Rearranging:-

q-4I <> 2t(r+0.5) and q-4I <> t(2r+1)

It will be useful at this stage to split the R.H.S ito two functions:_

F(r) =  r(r+1)  = 2I                                                                 (5)

And

F(r,t) = 0.5t(r+0.5)                                                                (6)

Finally:-

t <> (4I  q)/(2r +1)                                                              (7)

If we consider a scanning process, where r is in the range 1,rmax , for values of F(  r),

and also a scanning process for t in the function F(t,r) , we can classify F( r) as a fast

function whereas F(t,r) as a slow. This is because q is a constant and so t is only

a function of r as can be seen in (7).

Let us suppose that the prime candidate is in fact not a prime e.g. 102 (3x34) , then by definition

all values for t must be whole numbers. However, if the prime candidate is a prime e.g. 101,

then logically t must be a fraction.