## Re: primes of the form x^3 - y^2

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• ... for primes p such that x^3 - p is conjecturally never a square for integer x. To prove that there are no integral points on the elliptic curves x^3 = y^2 +
Message 1 of 22 , Jun 17, 2009
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cino hilliard <hillcino368@...> wrote:

> 3, 5, 17, 29, 31, 37, 41, 43, 59, 73, 97,
> 101, 103, 113, 131, 137, 149, 157, 163 ...

for primes p such that x^3 - p is conjecturally never
a square for integer x.

To prove that there are no integral points on the
elliptic curves x^3 = y^2 + p, for those primes,
you might need to compute their Mordell-Weil groups:
www.dpmms.cam.ac.uk/Algebraic-Number-Theory/0086/paper.ps

There are 21 integral points with y > 0 on the elliptic curve
x^3 = y^2 + 28279
with a Mordell-Weil group of rank 5, namely those with
x = 32, 34, 40, 50, 67, 70, 122, 260, 295, 359, 515, 592,
952, 2284, 2327, 2410, 7330, 7580, 11702, 130184, 26507590.

David
• ... http://magma.maths.usyd.edu.au/calc/ enabled me to show that there are no more: Input: E := EllipticCurve([0, -28279]); Q, reps := IntegralPoints(E); Q;
Message 2 of 22 , Jun 17, 2009
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> There are 21 integral points with y > 0 on the elliptic curve
> x^3 = y^2 + 28279
> with a Mordell-Weil group of rank 5, namely those with
> x = 32, 34, 40, 50, 67, 70, 122, 260, 295, 359, 515, 592,
> 952, 2284, 2327, 2410, 7330, 7580, 11702, 130184, 26507590.

http://magma.maths.usyd.edu.au/calc/
enabled me to show that there are no more:

Input:

E := EllipticCurve([0, -28279]);
Q, reps := IntegralPoints(E);
Q;

Output:

[ (34 : -105 : 1), (40 : 189 : 1), (70 : 561 : 1),
(32 : 67 : 1), (50 : -311 : 1), (67 : -522 : 1),
(122 : 1337 : 1), (260 : -4189 : 1), (295 : 5064 : 1),
(592 : -14403 : 1), (952 : -29373 : 1), (359 : 6800 : 1),
(515 : 11686 : 1), (2284 : 109155 : 1), (2410 : 118311 : 1),
(2327 : -112252 : 1), (7330 : 627561 : 1), (7580 : -659939 : 1),
(11702 : -1265873 : 1), (130184 : -46971715 : 1),
(26507590 : 136475711439 : 1) ]

The CPU-time was comfortably below the 20 second limit:

> Total time: 7.219 seconds, Total memory usage: 137.04MB

David
• ... E := EllipticCurve([0, -101]); Q, reps := IntegralPoints(E); Q; [] Total time: 0.370 seconds, Total memory usage: 137.04MB But for x^3-y^2=149, Magma
Message 3 of 22 , Jun 17, 2009
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cino hilliard <hillcino368@...> wrote:

> My gut is x^3-y^2 = 101,103 and others are out there but we
> will not have the time to find them in this BB universe.

E := EllipticCurve([0, -101]);
Q, reps := IntegralPoints(E);
Q;

[]
Total time: 0.370 seconds, Total memory usage: 137.04MB

But for x^3-y^2=149, Magma issues a warning:

E := EllipticCurve([0, -149]);
Q, reps := IntegralPoints(E);
Q;

Warning: rank computed (0) is only a lower bound
(It may still be correct, though)
[]
Total time: 0.340 seconds, Total memory usage: 61.95MB

PS: It seems that Sm*r*nd*ch* once conjectured that there
are no rational points on x^3 = y^2 + 7, somehow
overlooking the fact that 2^3 = 1^2 + 7:
www.articlearchives.com/asia/northern-asia-china-south/951072-1.html

David
• Off list, Tony Noe kindly provided this link: http://www.research.att.com/~njas/sequences/A081120 From the table attached thereto, one may conclude that Cino s
Message 4 of 22 , Jun 18, 2009
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Off list, Tony Noe kindly provided this link:
http://www.research.att.com/~njas/sequences/A081120

From the table attached thereto, one may conclude that
Cino's sequence, "Primes not of the form x^3 - y^2", begins

3, 5, 17, 29, 31, 37, 41, 43, 59, 73, 97, 101, 103, 113, 131,
137, 149, 157, 163, 173, 179, 181, 197, 211, 227, 229, 241,
257, 263, 269, 281, 283, 311, 313, 317, 331, 337, 347, 349,
353, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 443,
449, 457, 461, 467, 479, 491, 509, 521, 523, 541, 563, 569,
571, 577, 601, 607, 613, 617, 619, 641, 643, 653, 659, 661,
677, 691, 701, 709, 733, 739, 743, 751, 757, 761, 773, 787,
809, 811, 821, 823, 827, 829, 839, 853, 857, 863, 877, 881,
883, 907, 911, 929, 937, 941, 947, 953, 967, 977, 983, 997 ...

This paper appears to have open access:
http://tinyurl.com/mtvsn5
See Section 5.1 for Cino's record holder, p = 28279,
with 21 integral points.

David
• Hi David, The elliptic functions in Pari are forboding at least to me. I contacted the Pari group and got a workaround for eliptic curves from Karim. He gave
Message 5 of 22 , Jun 18, 2009
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Hi David,

The elliptic functions in Pari are forboding at least to me.

I contacted the Pari group and got a workaround for eliptic curves from
Karim. He gave the following to find solutions for x^3-y^2 = p.

diffcubes(n,p)=local(x,y);setintersect(vector(n,x,x^3-p),vector(n,y,y^2))
which we found out will work in the next version of Pari. :-)

So Karim wrote:

> For "older" versions than that, use
>
> setintersect(Set(vector(n,x,x^3-p)), Set(vector(n,y,y^2)))
>
> ( slower but not *much* slower... )

Ok, I did it my way with this and asked a question at the end.
The timing aint bad. Magma and Sage would probably be faster offline.

We need a way of knowing upper bounds a priori though.

> diffcubes(n,p) =
> {
> local(j,x,y,c);
> a=eval(setintersect(Set(vector(n,x,x^3-p)), Set(vector(n,y,y^2))));
> c=length(a);
> a=vecsort(a);
> for(j=1,c,
> y=round(sqrt(a[j])); \\ this could be iffy
> x=round((a[j]+p)^(1/3));
> print(j": "x"^3 - "y"^2 = "p); \\ Too fancy? Change it.
> );
> c;
> }
>
>
>(14:51:50) gp > diffcubes(300000,431)
>1: 8^3 - 9^2 = 431
>2: 11^3 - 30^2 = 431
>3: 20^3 - 87^2 = 431
>4: 30^3 - 163^2 = 431
>5: 36^3 - 215^2 = 431
>6: 138^3 - 1621^2 = 431
>7: 150^3 - 1837^2 = 431
>8: 575^3 - 13788^2 = 431
>9: 3903^3 - 243836^2 = 431
>(15:02:35) gp > ##
> *** last result computed in 2,250 ms.
>(15:02:39) gp >
>
>This finds all instances because I new a prori 243836 was the big kahuna.
>I guess it will still be trial and error?
>
>Oh well, we are at least 2 quanta over what I had.
>
> Thank you >>>
>Cino

Cino

Date: Thu, 18 Jun 2009 16:26:36 +0000
Subject: [PrimeNumbers] Re: primes of the form x^3 - y^2

Off list, Tony Noe kindly provided this link:
http://www.research.att.com/~njas/sequences/A081120

From the table attached thereto, one may conclude that
Cino's sequence, "Primes not of the form x^3 - y^2", begins

3, 5, 17, 29, 31, 37, 41, 43, 59, 73, 97, 101, 103, 113, 131,
137, 149, 157, 163, 173, 179, 181, 197, 211, 227, 229, 241,
257, 263, 269, 281, 283, 311, 313, 317, 331, 337, 347, 349,
353, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 443,
449, 457, 461, 467, 479, 491, 509, 521, 523, 541, 563, 569,
571, 577, 601, 607, 613, 617, 619, 641, 643, 653, 659, 661,
677, 691, 701, 709, 733, 739, 743, 751, 757, 761, 773, 787,
809, 811, 821, 823, 827, 829, 839, 853, 857, 863, 877, 881,
883, 907, 911, 929, 937, 941, 947, 953, 967, 977, 983, 997 ...

This paper appears to have open access:
http://tinyurl.com/mtvsn5
See Section 5.1 for Cino's record holder, p = 28279,
with 21 integral points.

David

[Non-text portions of this message have been removed]
• ... How does it compare with the brute force of issquare for the largest integer point on x^3 - y^2 = 22189 I wonder? The result is already published, but I
Message 6 of 22 , Jun 18, 2009
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cino hilliard <hillcino368@...> wrote:

> I contacted the Pari group and got a workaround
> for eliptic curves from Karim.

How does it compare with the brute force of
"issquare" for the largest integer point on
x^3 - y^2 = 22189
I wonder?

The result is already published, but I thought
that it might be an interesting test case.
Here is an "issquare" timing:

c=0;gettime;
{for(x=1,10^9,if(issquare(x^3-22189),
print(x);c=c+1;if(c==2,break)))}
print(ceil(gettime/10^3)" seconds")

29585
140292677
98 seconds

David
• ... _________________________________________________________________ Découvrez Windows Live Spaces et créez votre site Web perso en quelques clics !
Message 7 of 22 , Jun 19, 2009
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---
>
>
> the collatz function to it as in f(n) = 3n+1/2 if n is odd
>
> and f(n) = n/2 n is even.
>
> Consider the number 33
>
>
>
> 33, 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
>
>
>
> The odd numbers are 33, 25 ;19,29 ,11 ,17 ,13 ,5 and so on.
>
>
>
> As u can see 6 numbre prime : 19 ;29; 11 ;17 ; 13; 5 .
>
>
>
> I believe that the collatz function can be used as a prime generating function though i am not very sure in this.
>
>
>
> i have found n = 111012973909 is the largest integer with all odd are prime
>
>
>
> 111012973909, 333038921728, 166519460864, 83259730432, 41629865216, 20814932608, 10407466304, 5203733152, 2601866576, 1300933288, 650466644, 325233322, 162616661, 487849984, 243924992, 121962496, 60981248, 30490624, 15245312, 7622656, 3811328, 1905664, 952832, 476416, 238208, 119104, 59552, 29776, 14888, 7444, 3722, 1861, 5584, 2792, 1396, 698, 349, 1048, 524, 262, 131, 394, 197, 592, 296, 148, 74, 37, 112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
>
>
>
> all odd are prime : 111012973909 ; 162616661 ; 1861 ; 349 ; 131 ; 197 ; 37 ; 7 ;11 ; 17 ; 13 ; ; 5
>
>
>
> rachid
>

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• ... The word largest makes no sense here. The 271-digit prime (5*2^897-1)/3 manifestly contains no composite odd integer in its Collatz sequence and it would
Message 8 of 22 , Jun 19, 2009
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sta staf <sta_staf@...> wrote:

> i have found n = 111012973909 is the largest integer
> with all odd are prime

The word "largest" makes no sense here.
The 271-digit prime (5*2^897-1)/3 manifestly contains no
composite odd integer in its Collatz sequence and it would be
easy to find larger primes by this, or a similar, construction.

David
• yes but also the 271-figit prime (5*2^897-1)/3 makes no sense here you think (5*2^897-1)/3) = prime numbre !!!!!!!!!!! give me n integer 111012973909
Message 9 of 22 , Jun 19, 2009
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yes but also the 271-figit prime (5*2^897-1)/3 makes no sense here

you think (5*2^897-1)/3) = prime numbre !!!!!!!!!!!

give me n integer 111012973909 < n < 2^50 with all odd integer are prime .

becuse we canot test the sequnce :(5*2^897-1)/3

thanks

rachid

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• ... to which I calmly reply, that I do so think. Here is my proof, using APR-CL in Pari-GP: N=(5*2^897-1)/3; if(isprime(N),print(OK)) OK It appears that
Message 10 of 22 , Jun 19, 2009
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sta staf <sta_staf@...> wrote:

> the 271-figit prime (5*2^897-1)/3 makes no sense here
> you think (5*2^897-1)/3) = prime numbre !!!!!!!!!!!

to which I calmly reply, that I do so think.
Here is my proof, using APR-CL in Pari-GP:

N=(5*2^897-1)/3;
if(isprime(N),print(OK))
OK

It appears that coherence is in inverse proportion
to the number of exclamation marks used by the poster.

David (with no exclamation marks)
• but your seqaunce :(5*2^897-1/3) this sequance of coltaze is : all Even numbers . the only odd is the ferst digit : :(5*2^897-1/3) rachid
Message 11 of 22 , Jun 19, 2009
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this sequance of coltaze is : all Even numbers .

the only odd is the ferst digit : :(5*2^897-1/3)

rachid

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• Hi, ... the workaround, diffcubes2(n,p) = { local(x,y,c,a); a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p)))); a; } is useless for large n.
Message 12 of 22 , Jun 19, 2009
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Hi,

>>cino hilliard <hillcino368@...> wrote:
>> I contacted the Pari group and got a workaround
>> for eliptic curves from Karim.

the workaround,

diffcubes2(n,p) =
{
local(x,y,c,a);
a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p))));
a;
}
is useless for large n. Maybe someone can tweek.

(10:25:03) gp > diffcubes2(10000000,22189)
*** Set: the PARI stack overflows !
current stack size: 800000000 (762.939 Mbytes)
[hint] you can increase GP stack with allocatemem()

>How does it compare with the brute force of
>"issquare" for the largest integer point on
>x^3 - y^2 = 22189
>I wonder?

>c=0;gettime;
>{for(x=1,10^9,if(issquare(x^3-22189),
>print(x);c=c+1;if(c==2,break)))}
>print(ceil(gettime/10^3)" seconds")

>29585
>140292677
>98 seconds

elipissq(n) =
{
c=0;
for(x=1,10^9,if(issquare(x^3-n),
print(x);c=c+1;if(c==2,break)));
}

Pari 2.3.4
dell 8200 2.53 ghz xp pro

42 minutes. say what?

Pari 2.4.2
dell 8200 2.53 ghz xp pro

225 sec Go figure. They must have done some stuff on issquare in v2.4.2.

Pari 2.4.2

dell 6x620 3.2 ghz vista ult 64 bit op sys
112 sec close to David's bench.

Sage online calculator

http://www.sagenb.com/home/hillcino368/0/

sage: time EllipticCurve([0,0,0,0,-22189]).integral_points()

[(29585 : 5088706 : 1), (140292677 : 1661699554612 : 1)]
Time: CPU 0.22 s, Wall: 0.87 s
A tough and user friendly program.

I wonder why Pari has not implemented the Mordell algorithm. Maybe 2.4.3?
Perhaps someone here can write a script or show a c program with gmp that
does this. A gcc/gmp would be great.

Dénouement

A child playing in a room with toys could solve the riddle of the difference between
two squares by arranging toy blocks and counting. Later he could puzzle with numbers
by squaring, subtracting, noticing patterns and conjecturing. For any positive numbers
a,b, a^2 - b^2 = (a-b)*(a+b). Later, he finds out how to symbolically multiply
(a-b)*(a+b) to get a^2-b^2 proving his prior conjectures correct.

So a^2 - b^2 =(a-b)*(a+b) could be the third most fundamental process in arithmetic
the average being second and counting being first. It is that easy of a riddle.

Then later, the child goes into another room with more blocks. He starts making pyramids,
cubes, and squares. Suddenly, with a flash of insight, he about cube - square = number
or a^3 - b^2 = n? It is not that easy of a riddle.

I doubt my Mathematica I bought in 80's and Maple in the 90's does this as Sage.

[Rant]

It is disheartning that students can get these for \$125 and seniors only get 1/2 off
of retail (\$2000) for Mathematica.

[End Rant]

Cheers and Roebuck,

Cino

[Non-text portions of this message have been removed]
• ... As indeed I had expected :-) Sometimes brute force is best, as in my prior hack: c=0;gettime; {for(x=1,10^9,if(issquare(x^3-22189),
Message 13 of 22 , Jun 19, 2009
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cino hilliard <hillcino368@...> wrote:

> the workaround,
> diffcubes2(n,p) =
> {
> local(x,y,c,a);
> a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p))));
> a;
> }
> is useless for large n.

As indeed I had expected :-)

Sometimes brute force is best, as in my prior hack:

c=0;gettime;
{for(x=1,10^9,if(issquare(x^3-22189),
print(x);c=c+1;if(c==2,break)))}
print(ceil(gettime/10^3)" seconds")

29585
140292677
98 seconds

probes deeply into the theory of elliptic curves.

Best regards

David
• Hello Rachid, ... It d still satisfy the property you re expecting it to satisfy: All odd terms of the sequence are primes. However, unless I m very much
Message 14 of 22 , Jun 19, 2009
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Hello Rachid,

Even if this was true:
> this sequance of coltaze is : all Even numbers .
> the only odd is the ferst digit : :(5*2^897-1/3)

It'd still satisfy the property you're expecting it to satisfy: "All odd
terms of the sequence are primes."

However, unless I'm very much mistaken, the sequence looks like this:
(5*2^897-1)/3, 5*2^896, 5*2^895, ..., 5*2^2, 5*2^1, 5, 8, 4, 2, 1 and as
far as my memory goes, the term "5" looks like an odd prime to me.

However, as you wanted a small example of this kind (< 2^50), consider
297784399189. Any objections against it?

Peter

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
• ... sta staf wrote, ... Dear Rachid: Until you learn to be more disciplined, I shall no longer reply. Please isolate your error in the above
Message 15 of 22 , Jun 19, 2009
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sta staf <sta_staf@...> wrote,
untidily and incorrectly:

> this sequance of coltaze is : all Even numbers .
> the only odd is the ferst digit : :(5*2^897-1/3)
> rachid

Dear Rachid: Until you learn to be more disciplined,
in the above statement, bearing in mind that 5
is an odd prime. Then consider that you asked
only for a Collatz sequence in which no odd
member is composite, with which request my
simple construction perfectly complies.

Discipline is a stern master:
I shall not allow you to "wriggle"
out of the hole you have carelessly
and untidily dug yourself into.

You may think me stern. But please recall
that mathematics is far sterner than I am.

Stay well,

David
• ... After reading http://www.primepuzzles.net/puzzles/puzz_476.htm Rachid will perhaps not object to my next claim, namely that (2^1322*(5*2^897-1)/3-1)/3 is a
Message 16 of 22 , Jun 20, 2009
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> The 271-digit prime (5*2^897-1)/3 manifestly contains no
> composite odd integer in its Collatz sequence and it would be
> easy to find larger primes by this, or a similar, construction.

Rachid will perhaps not object to my next claim, namely that
(2^1322*(5*2^897-1)/3-1)/3 is a 668-digit prime whose Collatz
sequence contains no composite odd integer.

As both Jens and I have noted, left extensibility is trivial,
given enough computing power. Hence my comment to Rachid:

> The word "largest" makes no sense here.

David
• ** For Your Eyes Only ** ** High Priority ** the agent never got back to me, in spite of three different demands I assume they sold it I looked at another one
Message 17 of 22 , Jun 20, 2009
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** For Your Eyes Only **
** High Priority **

the agent never got back to me, in spite of three different demands
I assume they sold it
I looked at another one so I may decide to move after all.
I trust & hope your vacation be goo

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> The 271-digit prime (5*2^897-1)/3 manifestly contains no
> composite odd integer in its Collatz sequence and it would be
> easy to find larger primes by this, or a similar, construction.

http://www.primepuzzles.net/puzzles/puzz_476.htm )
Rachid will perhaps not object to my next claim, namely that
(2^1322*(5*2^897-1)/3-1)/3 is a 668-digit prime whose Collatz
sequence contains no composite odd integer.

As both Jens and I have noted, left extensibility is trivial,
given enough computing power. Hence my comment to Rachid:

> The word "largest" makes no sense here.

David

Please find our Email Disclaimer here: http://www.ukzn.ac.za/disclaimer/

[Non-text portions of this message have been removed]
• you have the right david there is always another prime can often be extended to the left like your numbre :(2^1322*(5*2^897-1)/3-1)/3) is a Construction
Message 18 of 22 , Jun 20, 2009
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you have the right david there is always another prime can often be extended to the left

is a Construction Sequence . very easy to Making the collatz sequence contains no composite odd integer.

yes ( The word "largest" makes no sense here)

rachid

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• ... En Angleterre, nous avons l habitude de tirer à gauche. Amitiés David
Message 19 of 22 , Jun 20, 2009
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sta staf <sta_staf@...> wrote:

> you have the right david there is always another prime
> can often be extended to the left
> like your numbre : (2^1322*(5*2^897-1)/3-1)/3

En Angleterre, nous avons l'habitude de tirer à gauche.

Amitiés

David
• Hi, I did a brute force in gcc/gmp. http://groups.google.com/group/elliptic-curves/web/gcc-gmp-brute-force-elliptic-curve Output: n start = 22189 n stop =
Message 20 of 22 , Jun 23, 2009
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Hi,

I did a brute force in gcc/gmp.

Output:

n start => 22189
n stop => 22189
x range => 150000000
max dups => 1
n prime => 1
29585 22189 5088706
140292677 22189 1661699554612
count = 2
31.9994

for dell 8200 2.53 ghz

24.9 sec

for dell gx620 3.2 ghz vista ult

Working with a developer of sage I was pointed to ratpoints.

>> http://www.mathe2.uni-bayreuth.de/stoll/programs/index.html, is free
>> (GPL) and only requires gmp.

>> ratpoints looks promising but I can't install it. Using Msys, I keep getting
>> the cannot find -lgmp.

I installed gmp 4.3.1 with msys which was used for the above timings.

I searched long for this. There ain't much out there on it. In fact, google -lgmp
gets a no hit.

Still asking, how do we determine the x range for ratpoints and how does sage
do it?

>ratpoints takes 0.09s to fine all integral points with x<10^8! But
>there is no proof, which Sage provides. Sage takes longer as the
>curve has rank 5.

>I cannot help with Windows problems. gmp is a standard library (Gnu
>MultiPrecision) but I do not know how to use it on Windows.

>Very soon ratpoints will be available within Sage -- we are just testing it now.

Maybe someone in primenumbers can load ratpoints in a windows xp pro of vista
platform.

Anyway, the gcc/gmp brute force is considerably faster than Pari for x range 10^8-10^9.

Cino

Date: Fri, 19 Jun 2009 23:14:46 +0000
Subject: [PrimeNumbers] Re: primes of the form x^3 - y^2

cino hilliard <hillcino368@...> wrote:

> the workaround,
> diffcubes2(n,p) =
> {
> local(x,y,c,a);
> a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p))));
> a;
> }
> is useless for large n.

As indeed I had expected :-)

Sometimes brute force is best, as in my prior hack:

c=0;gettime;
{for(x=1,10^9,if(issquare(x^3-22189),
print(x);c=c+1;if(c==2,break)))}
print(ceil(gettime/10^3)" seconds")

29585
140292677
98 seconds

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• ... Sage is being sagacious, here. It s good that you can access such wisdom without paying big dollars for Magma. Congrats to Cino, for persistence. David
Message 21 of 22 , Jun 23, 2009
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