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## 21353Re: F = 20z^2 - n and (5r^2 -F)/n

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• Feb 27, 2010
"aldrich617" <aldrich617@...> conjectured:

> Let F = 20*z^2 - 7, for integer z > 0.
> If there is at least one integer r where 2*z < r <= z*sqrt(14)
> such that (5*r^2 - F)/7 = a perfect square, then F is composite.

Thanks for writing sqrt(14), which indeed looks like
the correct limit, in this case. You may write
r < z*sqrt(14), since the issue of equality cannot arise.

> Its companion conjecture that " otherwise F is prime " would
> also probably be true.

No. Here you need the caveat that for z = 0 mod 7
it is F/7, and not F, that is conjectured to be prime.

Your list of "n" values for which you have a conjecture was

> n = 1,6,10,12,13,14,15,19,21,22, and 23

Now consider the case n = 11, about which you were strangely
silent. Here, the closely parallel conjecture is as follows:

Let F = 20*z^2 - 11, for integer z > 0, and let
G = F/11, if z = 0 mod 11, and G = F, otherwise.
Then I conjecture that G is prime if and only if
there is no integer r such that 2 < r/z < 6*sqrt(5)
and (5*r^2 - F)/11 is the square of an integer.

Exercise: Let
z = 14155247814063791862875274459749906082392395640576933548527
and let F = 20*z^2 - 11. Find an integer r such that
2 < r/z < 6*sqrt(5) and (5*r^2 - F)/11 is the square of an integer.

Hint: An answer can be found very easily, if you pay good
attention to my carefully chosen upper limit for r/z. Moreover,
I conjecture that this generous hint gives the only answer.

David
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