--- On Thu, 9/23/10, Sebastian Martin Ruiz <

s_m_ruiz@...> wrote:

> Hello

>

> 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)

Too easy.

Please try harder.

Or don't.

Phil