Inaugural puzzle

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• p1 is the smallest titanic prime of the form sqrt(3*x^2-2) with integer x. p2 is the smallest titanic prime of the form sqrt(5*y^2-4) with integer y. p3 is the
Message 1 of 8 , Jan 15, 2009
p1 is the smallest titanic prime of the form
sqrt(3*x^2-2) with integer x.

p2 is the smallest titanic prime of the form
sqrt(5*y^2-4) with integer y.

p3 is the smallest prime such that
2*p1*p2*p3-1 is prime.

Find a metrical connection between p3 and liberty.

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• From: David Broadhurst ... Is sqrt the exact square root defined over only squares, or merely the integer square root with truncation defined over the
Message 2 of 8 , Jan 15, 2009
> p1 is the smallest titanic prime of the form
> sqrt(3*x^2-2) with integer x.

Is 'sqrt' the exact square root defined over only squares, or merely the integer square root with truncation defined over the naturals?

Phil
• ... It is indeed. David Broadhurst ... The Open University is incorporated by Royal Charter (RC 000391), an exempt charity in England and Wales and a charity
Message 3 of 8 , Jan 16, 2009
<thefatphil@...> wrote:
>
> > p1 is the smallest titanic prime of the form
> > sqrt(3*x^2-2) with integer x.
>
> Is 'sqrt' the exact square root defined over only squares

It is indeed.

----------------------------------------------------------------
The Open University is incorporated by Royal Charter (RC 000391),
an exempt charity in England and Wales and
a charity registered in Scotland (SC 038302).
• Spoiler warning. ... I suck at diophantine equations so I didn t even try but just bruteforced the first few x values which gave integer square root. A quick
Message 4 of 8 , Jan 16, 2009
Spoiler warning.

> p1 is the smallest titanic prime of the form
> sqrt(3*x^2-2) with integer x.
>
> p2 is the smallest titanic prime of the form
> sqrt(5*y^2-4) with integer y.
>
> p3 is the smallest prime such that
> 2*p1*p2*p3-1 is prime.
>
> Find a metrical connection between p3 and liberty.

I suck at diophantine equations so I didn't even try but just
bruteforced the first few x values which gave integer square root.
A quick lookup of the x sequences in OEIS and there they were.
http://www.research.att.com/~njas/sequences/A001835 has comment:
Terms are the solutions to: 3x^2-2 is a square.
http://www.research.att.com/~njas/sequences/A001519 has comment:
Terms for n>1 are the solutions to : 5x^2-4 is a square.

A little PARI+PFGW later and I get a 1673-digit p1,
a 1223-digit p2, and p3 = 12241.
The metrical connection is left as an exercise for the reader
(actually, I don't know and didn't look hard).
I haven't run Primo on p1 and p2. Knowing David there might be
easy proofs but I'm not looking. OK, I'm lazy.
Assuming they are prime, PFGW proved 2*p1*p2*p3-1.

--
Jens Kruse Andersen
• ... Well done, Jens! Primo timing: Total=2h 11mn 27s DecimalSize=1673 ... No, that s too large. Try re-reading the OEIS page :-) David Broadhurst ... The Open
Message 5 of 8 , Jan 16, 2009
--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:

> A little PARI+PFGW later and I get a 1673-digit p1

Well done, Jens!

Primo timing:
Total=2h 11mn 27s
DecimalSize=1673

> a 1223-digit p2

No, that's too large. Try re-reading the OEIS page :-)

----------------------------------------------------------------
The Open University is incorporated by Royal Charter (RC 000391),
an exempt charity in England and Wales and
a charity registered in Scotland (SC 038302).
• ... Oops. My PARI/GP script to find p2 was correct but it output decimal expansions and didn t stop after the first. I had a bunch of open windows and failed
Message 6 of 8 , Jan 16, 2009
> --- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
> <jens.k.a@...> wrote:
>> a 1223-digit p2
>
> No, that's too large. Try re-reading the OEIS page :-)

Oops. My PARI/GP script to find p2 was correct but it output decimal
expansions and didn't stop after the first. I had a bunch of open windows
and failed to scroll up to see two earlier prp's when I copied the result
later. How sloppy. The old PARI window is still open and I just checked the
output. The first 3 titanic prp's of p2 type has 1001, 1002, 1223 digits.

I now get 1673-digit p1, 1001-digit p2, and p3=151. Primo is running on p2.
PFGW has proved p3 where p1 was enough help for an easy proof.
I still don't know what the metric connection is.

--
Jens Kruse Andersen
• ... Congratulations on completing the maths. ... Solution: 1) p1 = 3*lucasU(4,1,2925) - lucasV(4,1,2925)/2 is the smallest titanic prime in the integer
Message 7 of 8 , Jan 16, 2009
--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:

> I now get 1673-digit p1, 1001-digit p2, and p3=151.

Congratulations on completing the maths.

> I still don't know what the metric connection is.

Inaugural puzzle:

> p1 is the smallest titanic prime of the form
> sqrt(3*x^2-2) with integer x.
>
> p2 is the smallest titanic prime of the form
> sqrt(5*y^2-4) with integer y.
>
> p3 is the smallest prime such that
> 2*p1*p2*p3-1 is prime.
>
> Find a metrical connection between p3 and liberty.

Solution:

1) p1 = 3*lucasU(4,1,2925) - lucasV(4,1,2925)/2
is the smallest titanic prime in the integer sequence
http://www.research.att.com/~njas/sequences/A001834
with a[1] = 1, a[2] = 5, a[n] = 4*a[n-1] - a[n-2],
ensuring that a[n]^2 = 3*x^2 - 2, with integer x.
Then a[n] = 3*lucasU(4,1,n) - lucasV(4,1,n)/2 solves
the recursion, in terms of generalized Lucas numbers.

PFGW test:
Calling N-1 BLS with factored part 4.90%
and helper 1.80% (16.51% proof)
3*lucasU(4,1,2925)-lucasV(4,1,2925)/2
is Fermat and Lucas PRP!

Primo timing:
Total=2h 11mn 27s
DecimalSize=1673

2) p2 = (5*F(2*2394) - L(2*2394))/2
is the smallest titanic prime in the integer sequence
http://www.research.att.com/~njas/sequences/A002878
with a[1] = 1, a[2] = 4, a[n] = 3*a[n-1] - a[n-2],
ensuring that a[n]^2 = 5*y^2 - 4, with integer y.
Then a[n] = (5*F(2*n) - L(2*n))/2 solves the recursion,
in terms of the Fibonacci and classical Lucas numbers.

PFGW test:
Calling N+1 BLS with factored part 7.61%
and helper 3.79% (26.69% proof)
(5*F(2*2394)-L(2*2394))/2
is Fermat and Lucas PRP!

Primo timing:
Total=19mn 41s
DecimalSize=1001

3) p3 = 151
is the smallest prime such that 2*p1*p2*p3-1 is prime.

PFGW test with helpers p1 and p2:
Calling N+1 BLS with factored part 100.00%
and helper 0.65% (300.68% proof)
151*(3*lucasU(4,1,2925)-lucasV(4,1,2925)/2)*(5*F(2*2394)-L(2*2394))-1
is prime!

4) Then by googling
> 151 prime
we are directed, at very first choice, to Chris Caldwell's page
http://primes.utm.edu/curios/page.php/151.html
reminding us of the Statue of Liberty.

I reflected on the poem inscribed within its pedestal
http://www.libertystatepark.com/emma.htm
which served to remind me of a time when the
United States of America were a better nation
than that belittled so wantonly on 20/03/03
by the ignominious Bush and his shameless toady, Blair: