--- In

primenumbers@yahoogroups.com,

Kevin Acres <research@...> wrote:

> Was the hint that generous or did I miss the point entirely?

The hint was indeed very generous and my point was this:

> I conjecture that this generous hint gives the only answer.

Explanation: I had solved Aldrich's "issquare problem"

(5*r^2 - F)/11 = a^2, with F = 20*z^2 - 11,

by setting r = 3*p, a = 2*p, where

p = 63304192701203884454712276208334615126937940298786013924413

is, most crucially, a prime.

Then F = p^2 is the square of a prime. Thus I conjecture that

(5*r^2 - F)/11 is never a square for 2*z < r < 3*p,

though it is certainly a square for r = 2*z and for r = 3*p.

As Kevin noticed, 3*p is very close to 6*sqrt(5)*z, since

z^2 = (p^2 + 11)/20.

Puzzle: (p^2 + 11)/20 is a square for the primes

p = 3, 13, 67, 4253, 21587, 6950947 ...

Find the 17th prime in this sequence.

Comment: Extra credit will be gained for proving

that your answer is correct.

David