- Hello All:

you can see this conjecture at Primepuzzles.net:

Problems & Puzzles: Conjectures

Conjecture 44. A property of prime twins, only?

Sebastián Martín Ruiz sends the following conjecture

from his own invention:

Let p, q be consecutive prime numbers, p<q.

Let z=sqrt[(p^2+q^2)/2-1]

Conjecture: p&q are prime twins iff z is integer.

Question: Prove it or show it false.

Sincerely

Sebastian Martin Ruiz

http://perso.wanadoo.es/smaranda

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http://es.voice.yahoo.com > Let p, q be consecutive prime numbers, p<q.

Since p and q are twin primes q=p+2

> Let z=sqrt[(p^2+q^2)/2-1]

>

> Conjecture: p&q are prime twins iff z is integer.

So z becomes

sqrt( (p^2 + (p+2)^2)/2 -1) =

sqrt( (p^2 + p^2 + 4p +4)/2 -1) =

sqrt( (2p^2+4p+4)/2 -1 ) =

sqrt(p^2+2p+2 - 1) =

sqrt(p^2+2p+1) =

sqrt( (p+1)^2 ) =

p+1

Now let p and q be any (non-twin!) consecutive primes with 2 < p < q

Now there exist an n>1 such that q=p+2*n

Doing the same as above leads to:

z = sqrt((p^2 + p^2+4np+ 4n^2 )/2 -1)

= sqrt(p^2+2np+2n^2-1)

Its roots are -n +/- sqrt( (-n)^2 - 2n^2+1 ) = -n +/- sqrt( n^2 - 2n^2+1 )

= -n +/- sqrt( -n^2+1 )

which has solutions in integers only for n=0,1

So the conjecture is indeed true.

Ciao,

Ronny> Let p, q be consecutive prime numbers, p<q.

Let p = a and q = a+2.

> Let z=sqrt[(p^2+q^2)/2-1]

>

> Conjecture: p&q are prime twins iff z is integer.

>

> Question: Prove it or show it false.

Let a = any integer. Then:

(a^2 + (a+2)^2)/2 - 1 =

(a^2 + a^2 + 4a + 4)/2 - 1 =

(2a^2 + 4a + 4)/2 - 1 =

(a^2 + 2a + 2) - 1 =

a^2 + 2a + 1 =

(a+1)*(a+1) =

(a+1)^2

So, the the sqrt[(a+1)^2] = (a+1) an integer.

There is no requirement for (a, a+2) to be prime. They can be any two

integers that differ by 2.

The equation does hold for twin primes and none of the other primes, but

it also holds for non-prime integers as well. So, the equation, by

itself, tells us nothing about the character of the numbers.

Also, its not a very helpful test since you start with the statement:> Let p, q be consecutive prime numbers, p<q.

At this point, since you already know that p and q are prime numbers,

all you have to do is take their difference to see if they are twins.

However, if you remove the constraint of knowing beforehand that they

are primes, then the test itself doesn't reveal anything about the

prime/non-prime character of the numbers. It will just tell you if the

numbers differ by 2.

-David C.