> 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