**Copyright C Brian Warburton
August 2017**

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**The Determination of Prime
Numbers**

**By means of an Inequality**

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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.

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 : A_{pr} . 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,r_{max} , 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.