21355Re: F = 20z^2 - n and (5r^2 -F)/n
- Feb 28, 2010--- In firstname.lastname@example.org,
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.
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