- --- 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 - --- In primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

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

Kevin sent me a correct answer, in decimal format. But for

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

> Find the 17th prime in this sequence.

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

compact answer.

Meanwhile, congratulations to Kevin, for a valid result

David - --- In primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

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

Kevin has now found a reasonably compact answer

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

> Find the 17th prime in this sequence.

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]

Comments: Dario Alpern's http://www.alpertron.com.ar/QUAD.HTM

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 - At 03:15 PM 4/03/2010, djbroadhurst wrote:

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

My solution, in 29 characters, is:

> > 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)

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.