conjectured

that

For each prime number p, it exists x pairs (x>0) of smaller prime numbers q

r such that

p-q=q-r , which can be expressed by: p is symmetric from r through q.

At first I thought this not significant because this is a property of odd

integers

For any two odd integers p, q, where q is less than p,

r = 2 q - p is odd,

and p - q = q - p.

Marc Persuy conjectures that his property of odd integers partially extends

to

the prime number subset of the odd integers.

He conjectures that if p is prime then

there exist primes r and q with r < q such that p - q = q - r.

It's easily proven that if p is an odd prime, then

there exist odd numbers r and q with r < q such that p - q = q - r.

For take q any odd number between (p+1)/2 and p; define r = 2 q - p.

So now we can almost prove Marc's conjecture.

We can prove that if p is an odd prime then there exist a prime q and and

odd positive integer r < q

such that q - r = p - q.

It has been proven that between every integer and its double, there is at

least one prime.

Thus if p is an odd prime > 5

Thus there is at least one odd prime between (p+1)/2 and p.

Let q be such a prime.

Then define r = 2 q - p.

Then q - r = p - q.

Now can we go one more step and prove Marc's conjecture?

We've selected the prime p, and a prime q between (p+1)/2 and p.

How can we find which odd numbers r = 2 q - p are prime?

We proceed as follows:

We seek primes r, q, p such that

q - r = p - q.

Define k = q - r.

Then this relationship becomes

q = r + k or r = q - k.

p = q + k = r + 2 k

or, in terms of p,

q = p - k,

r = p - 2 k = q - k

Thus marc's conjecture is equivalent to the conjecture that

for every prime p, it's possible to find primes q and r such that

for some positive integer k,

q = p - k and r = q - k.

The following has a trivial proof.

If q is any odd positive integer, and k is any positive integer,

and

if the equation

m n + k m + n = q has only the solutions

{m = 0, n = q } and { m = q+k-1, n = 1 - k } in integral m and n,

and

m n + k m - n = q

also has only the solutions

{ m = 2, n = q - 2 k} and { m = q - k + 1, n = 1 - k}

then both

p = q + k

and

r = q - k

are prime.

So to complete the proof of Marc's conjecture, we need only prove that

given a prime q between (p+1)/2 and p,

there exist an integer k such that

that each of the equations

m n + k m + n = q

and

m n + k m - n = q

have exactly two integral solutions.

In fact, since we have already specified p as prime, we need only look at

the

second equation.

If p and q are prime, with p > q,

and

if the positive integer k = p - q has the property that

the equation

m n + k m - n has exactly two integral solutions,

then r = p - q is prime.