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

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• ... An examination of a few of the neighboring equations to F where other integers n are substituted for 3 reveals that for many possible n, Propositions
Message 1 of 7 , Feb 27, 2010

> Proposition: Let F = 20*z^2 - 3, for integer z > 0.
> If there are integers r > a > 1 such that
> (5*r^2 - F)/3 = a^2, then F is composite.

An examination of a few of the neighboring equations
to F where other integers n are substituted for 3 reveals
that for many possible n, Propositions similar to that above
can also be made that are likely true. For example,
if n = 7 then the Proposition could read:

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.

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

A quick look at some of the other equations shows that
for n = 1,6,10,12,13,14,15,19,21,22, and 23 a similar treatment
could probably be applied. In at least one case, at n = 4,
there are subsequences that can be handled the same way.

No general rule for sorting out the different patterns
in advance nor for estimating their true extent now suggests
itself, but variants of the ISSQUARE functions seem to be
proliferating. If anyone knows the origin and history of the

Aldrich
• ... Thanks for writing sqrt(14), which indeed looks like the correct limit, in this case. You may write r
Message 2 of 7 , 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
• ... Maybe: r=189912578103611653364136828625003845380813820896358041773239 where (5*r^2 - F)/11 = 126608385402407768909424552416669230253875880597572027848826^2
Message 3 of 7 , Feb 27, 2010
At 04:17 PM 28/02/2010, djbroadhurst wrote:

>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.

Maybe:

r=189912578103611653364136828625003845380813820896358041773239

where (5*r^2 - F)/11 =
126608385402407768909424552416669230253875880597572027848826^2

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

Kevin.
• ... 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 =
Message 4 of 7 , Feb 28, 2010
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

David
• ... Kevin sent me a correct answer, in decimal format. But for that he needed 454 characters. The solution can be written in less than 20 characters, using a
Message 5 of 7 , Mar 2, 2010

> 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.

Kevin sent me a correct answer, in decimal format. But for
that he needed 454 characters. The solution can be written
in less than 20 characters, using a form that OpenPFGW
readily understands. So the puzzle is still open, for a

Meanwhile, congratulations to Kevin, for a valid result

David
• ... Kevin has now found a reasonably compact answer as a linear combination of 3 Fibonacci numbers. In fact the answer can be written in terms of only 2:
Message 6 of 7 , Mar 3, 2010

> 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.

Kevin has now found a reasonably compact answer
as a linear combination of 3 Fibonacci numbers.
In fact the answer can be written in terms of only 2:

Solution [in 19 characters]:

F(2166)/2+3*F(2168)

Proof [by exhaustion, with primality proving]:

P(n,s) = fibonacci(6*n)/2 + 3*fibonacci(6*n+2*s);
{c=0;for(n=0,361,forstep(s=-1,1,2,p=P(n,s);
if(issquare((p^2+11)/20)&&isprime(p),c++;
print1(s*n" "))));print([c]);}

0 -1 1 -3 3 5 -11 15 -45 -47 -65 75 105 253 -303 -359 361 [17]

may be used to prove that my Fibonacci method is exhaustive.
A probable continuation of the sequence of primes is as follows:

F(3966)/2+3*F(3964), 829
F(5334)/2+3*F(5336), 1116
F(8682)/2+3*F(8684), 1816
F(15066)/2+3*F(15068), 3150
F(15438)/2+3*F(15436), 3227
F(41166)/2+3*F(41168), 8604
F(50874)/2+3*F(50876), 10633
F(114702)/2+3*F(114704), 23972
F(117978)/2+3*F(117980), 24657

with the number of digits of each PRP given, after the comma.
The last 3 have been consigned to Henri's repository of PRPs.

David
• ... My solution, in 29 characters, is: 3/2*F(2169)+5*F(2164)+F(2161) David s sequence, including the non-prime values, starting with the 13,67 pair can be
Message 7 of 7 , Mar 4, 2010
At 03:15 PM 4/03/2010, djbroadhurst wrote:

> > 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.
>
>Kevin has now found a reasonably compact answer
>as a linear combination of 3 Fibonacci numbers.
>In fact the answer can be written in terms of only 2:
>
>Solution [in 19 characters]:
>
>F(2166)/2+3*F(2168)

My solution, in 29 characters, is:

3/2*F(2169)+5*F(2164)+F(2161)

David's sequence, including the non-prime values, starting with the
13,67 pair can be reproduced by the following:

Let f1=F(12*n+9), f2=F(12*n+4), f3=F(12*n+1), a=f1+2*f2 and o=1/2*f1+3*f2+f3

Then the lower of each pair is a-o and the higher is a+o or:

1/2*f1-f2-f3 and 3/2*f1+5*f2+f3

In the case of the p454 just substitute n for 180 in the above assignments.

The pari/gp script to generate the sequence follows:

for(i=0,10,
f1=fibonacci(12*i+9);
f2=fibonacci(12*i+4);
f3=fibonacci(12*i+1);
o=1/2*f1+3*f2+f3;
a=f1+2*f2;
print([a-o,a+o]);
);

I strayed a little from the puzzle by solving for 180z^2-p^2-11
before trying to re-create the sequence from 13/67 onwards.

Once again, Dario's page at: http://www.alpertron.com.ar/QUAD.HTM was
of great help.

Kevin.
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