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## More twin primes

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• 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
Message 1 of 3 , Oct 3, 2004
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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• ... (p+2)^p % p = 2^p % p, for all numbers. Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp. Fermat s little theorem says all primes
Message 2 of 3 , Oct 4, 2004
David Litchfield wrote:

> 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?

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

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! ... (p+2)^p % p = 2^p % p, for all numbers. Numbers satisfying 2^p % p = 2 are called base 2 probable primes, or 2-prp. Fermat s little theorem
Message 3 of 3 , Oct 4, 2004
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+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?

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

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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