Is this (prime numbers) formula known ?
 

Good morning,

Is this formula known? does it work or does it not work?

The formula is :

[TEX]((p^1*p^2)+(p^2 or p^1))-1 = prime number[/TEX]

Thank you.


.

[QUOTE=Godzilla;496799]Good morning,

Is this formula known? does it work or does it not work?

The formula is :

[TEX]((p^1*p^2)+(p^2 or p^1))-1 = prime number[/TEX]

Thank you.


.[/QUOTE]Yes, it works perfectly.

p1=2, p2=13 ---> winner

[QUOTE=retina;496800]Yes, it works perfectly.

p1=2, p2=13 ---> winner[/QUOTE]




[B]and while p > 2 ?[/B]

Thanks

.

[QUOTE=Godzilla;496801][B]and while p > 2 ?[/B]

Thanks

.[/QUOTE]You could also test it for yourself and find your own counter examples. Just a thought.

[QUOTE=retina;496804]You could also test it for yourself and find your own counter examples. Just a thought.[/QUOTE]



You forum members do you know or not if it works?


.

I found a counterexample that does not work.

With the primes up to 10^5 this works 10240881 out of 91996872 times (about 11%).

[QUOTE=Godzilla;496808]I found a counterexample that does not work.[/QUOTE]
If you find a counterexample that works, you can get the Nobel prize...

559526371/6161857506 for primes up to a million. (I'm done.)

I'm not sure what the [i]or[/i] means [possibilities include the bitwise OR of p1 and p2, which in Pari-GP would be bitor(p1,p2)].

For the sake of discussion I will assume it means

p1*p2 + p1 - 1 or p1*p2 + p2 - 1.

It occurred to me that one could make both expressions divisible by the same prime q. This would require that p1 == p2 (mod q). Calling the common residue class x, we have

x^2 + x - 1 == 0 (mod q).

This quadratic has two solutions (mod q) when 5 is a quadratic residue (mod q), i.e. when q == 1 or 9 (mod 10). For q = 11, the two values of x are 3 (mod 11) and 7 (mod 11).

Thus, if p1 and p2 are both congruent to 3 (mod 11) or both are congruent to 7 (mod 11), both p1*p2 + p1 - 1 and p1*p2 + p2 - 1 will be divisible by 11.

For example, we could take p1 = 3 and p2 = 47, or p1 = 7 and p2 = 29.

I was thinking about the possibility that "p1 or p2" meant bitor(p1, p2), and noticed that, if 2 < p1 < p2, and 2^(r-1) < p1 < 2^r, then

bitor(p1,p2) = p2 when p2 == p1 (mod 2^r).

This would make the expression equal to

p2*(p1 + 1) - 1.

If q is a prime which does not divide p1 + 1, p1 + 1 has a multiplicative inverse (mod q), and the expression is divisible by q when p2 == (p1 + 1)[sup]-1[/sup] (mod q).

OK, so it's easy to construct counterexamples in which

bitor(p1,p2) = p2 and q divides p1*p2 + p2 - 1,

provided q does not divide p1 + 1.

Then, I noticed something curious. The first 2 odd primes are 3 and 5. The congruence classes 3 (mod 4) and 5 (mod 8) cover the congruence classes 3, 5, and 7 (mod 8). That only leaves the class 1 (mod 8) uncovered. So it occurred to me to wonder whether it ever does get covered by classes p1 (mod 2^r).

Since the first prime congruent to 1 (mod 8) is 17 and the next is 41, things start off badly. You need to cover 8 residue classes (mod 64) and only 3 of them (17, 49, and 41) are covered. That leaves 5 classes (mod 64). Make that 10 classes (mod 128).

You have to cover 1/4 of the odd residue classes (mod 2^r) for some r (those congruent to 1 (mod 8)), and the primes congruent to 1 (mod 8) are about a quarter of the primes, and the primes keep getting thinner on the ground. So I sort of doubt the class 1 (mod 8) ever gets covered.