--- On Thu, 9/23/10, Sebastian Martin Ruiz <s_m_ruiz@...
> OEIS A086381 is:
> Numbers n such that n^2+2 and n^2+4 are primes.
> I have found that this secuence is the same:
> Numbers n such that n=((p^2-q)/(p+1))^(1/2) is integer.
> p<q consecutive primes.
Throw away the obfuscation of the squareroot - you want (p^2-q)/(p+1) to be an integer first, and and a square second.
Trivially p^2-q = p^2+p-(p+q)
So (p^2-q)/(p+1) = p - ((p+nextprime(p))/(p+1))
We know nextprime(p) is near p, so, for p greater than not very much, that subtrahend is only ever going to be an integer when it's 2. I.e. q=p+2.
And you've set p=n^2+2, so q=n^2+4.
Q.E.D. (given that 'not very much' is bugger all)
Please try harder.