## A property of prime twins, only?

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• Hello All: you can see this conjecture at Primepuzzles.net: Problems & Puzzles: Conjectures Conjecture 44. A property of prime twins, only? Sebastián Martín
Message 1 of 3 , Jul 1, 2006
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

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• ... Since p and q are twin primes q=p+2 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 -
Message 2 of 3 , Jul 1, 2006
> 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.

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

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 = a and q = a+2. 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 +
Message 3 of 3 , Jul 1, 2006
> 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.

Let p = a and q = a+2.
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.