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

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• ... 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 1 of 7 , Feb 28, 2010
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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 2 of 7 , Mar 2, 2010
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> 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 3 of 7 , Mar 3, 2010
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> 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 4 of 7 , Mar 4, 2010
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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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