- I've just noticed that if p and p+2 are prime then

(p+2)^p % p = 2

p^(p+2) % (p+2) = p

If the first condition is met then p is prime and if the second condition is met then p+2 is prime.

I've tested this for twin primes up to 101/103; but of course it needs to be proved. Any pointers on going about proving this?

Cheers,

David

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[Non-text portions of this message have been removed] - David Litchfield wrote:

> I've just noticed that if p and p+2 are prime then

(p+2)^p % p = 2^p % p, for all numbers.

>

> (p+2)^p % p = 2

> p^(p+2) % (p+2) = p

>

> If the first condition is met then p is prime and if the second condition is

> met then p+2 is prime.

>

> I've tested this for twin primes up to 101/103; but of course it needs to be

> proved. Any pointers on going about proving this?

Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp.

Fermat's little theorem says all primes are 2-prp.

However some 2-prp are composites called pseudoprimes, specifically 2-psp in

this case.

There are infinitely many 2-psp: 341, 561, 645, 1105, 1387, ...

All Fermat numbers 2^(2^n)+1 are 2-prp.

All 2-psp below 10^13 have been computed by Richard Pinch:

http://www.chalcedon.demon.co.uk/carpsp.html

Setting q = p+2 makes the second condition:

(q-2)^q % q = q-2

q satisfying this are called (q-2)-prp, and are exactly the same numbers as

2-prp.

--

Jens Kruse Andersen - Thanks, Jens!

Jens Kruse Andersen <jens.k.a@...> wrote:David Litchfield wrote:

> I've just noticed that if p and p+2 are prime then

(p+2)^p % p = 2^p % p, for all numbers.

>

> (p+2)^p % p = 2

> p^(p+2) % (p+2) = p

>

> If the first condition is met then p is prime and if the second condition is

> met then p+2 is prime.

>

> I've tested this for twin primes up to 101/103; but of course it needs to be

> proved. Any pointers on going about proving this?

Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp.

Fermat's little theorem says all primes are 2-prp.

However some 2-prp are composites called pseudoprimes, specifically 2-psp in

this case.

There are infinitely many 2-psp: 341, 561, 645, 1105, 1387, ...

All Fermat numbers 2^(2^n)+1 are 2-prp.

All 2-psp below 10^13 have been computed by Richard Pinch:

http://www.chalcedon.demon.co.uk/carpsp.html

Setting q = p+2 makes the second condition:

(q-2)^q % q = q-2

q satisfying this are called (q-2)-prp, and are exactly the same numbers as

2-prp.

--

Jens Kruse Andersen

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