Sorry, an error occurred while loading the content.

## Factorization of 8 consecutive integers > 10^500

Expand Messages
• Let x=13860*(10^60+720251) N=x^2*(x^2-23)*(x^2-41)*(x^2-64)/55440 then N+k is factorized for k in [0,7]. Proof:
Message 1 of 7 , Aug 9, 2007
Let

x=13860*(10^60+720251)
N=x^2*(x^2-23)*(x^2-41)*(x^2-64)/55440

then N+k is factorized for k in [0,7].

Proof:

Software used:
GMP-ECM
Msieve
OpenPFGW
Pari-GP
Primo

9 August 2007
• ... Congratulations! That was amazingly fast. I wondered whether anybody would even try 8 for years with the limit set at 500 digits. I guess you are also
Message 2 of 7 , Aug 9, 2007
> x=13860*(10^60+720251)
> N=x^2*(x^2-23)*(x^2-41)*(x^2-64)/55440
>
> then N+k is factorized for k in [0,7].

Congratulations!
That was amazingly fast. I wondered whether anybody would
even try 8 for years with the limit set at 500 digits.
I guess you are also trying your ecm luck on N-1 and N+8.
http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated.

--
Jens Kruse Andersen
• ... Mainly thanks to Jarek, for his nice quartic Ansatz, which was easy to adapt, to give only 8 c125 factors, in 3 of the 8 targets, instead of the 11 that
Message 3 of 7 , Aug 9, 2007
--- In primeform@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:

> That was amazingly fast. I wondered whether anybody would
> even try 8 for years with the limit set at 500 digits.

Mainly thanks to Jarek, for his nice quartic Ansatz,
which was easy to adapt, to give only 8 c125 factors,
in 3 of the 8 targets, instead of the 11 that you spotted.

I had a nasty moment, at the very end, with a c115 that
looked as if it might need GGNFS, or MPQS, but finally it
cracked with ECM, which is /embarrassingly/ parallel.

David
• Nice! That was a very quick response to the challenge. S.
Message 4 of 7 , Aug 9, 2007
Nice! That was a very quick response to the challenge.

S.

> --- In primeform@yahoogroups.com, "Jens Kruse Andersen"
> <jens.k.a@...> wrote:
>
>
>>That was amazingly fast. I wondered whether anybody would
>>even try 8 for years with the limit set at 500 digits.
>
>
> Mainly thanks to Jarek, for his nice quartic Ansatz,
> which was easy to adapt, to give only 8 c125 factors,
> in 3 of the 8 targets, instead of the 11 that you spotted.
>
> I had a nasty moment, at the very end, with a c115 that
> looked as if it might need GGNFS, or MPQS, but finally it
> cracked with ECM, which is /embarrassingly/ parallel.
>
> David
• ... Jarek s construction involves this Diophantine problem: Find a pair of positive coprime integers [a,b], with a
Message 5 of 7 , Aug 10, 2007
--- In primeform@yahoogroups.com,
"Sean A. Irvine" <sairvin@...> wrote:

> Nice! That was a very quick response to the challenge.

Jarek's construction involves this Diophantine problem:

Find a pair of positive coprime integers [a,b], with a<b,
such that there exist two pairs of positive integers,
[c,d] and [e,f], with c < d, c < e < f,
2*a^2 + 2*b^2 = c^2 + d^2 = e^2 + f^2
and distinct positive integers
g = |(4*a*b)^2 -(d^2-c^2)^2|
h = |(4*a*b)^2 -(f^2-e^2)^2|
satisfying max(g,h) < 6*gcd(g,h).

There are (at least) 6 solutions:

[ a, b] [ c, d] [ e, f] g/h

[ 1,47] [18,64] [24,62] 4/3
[ 9,32] [ 1,47] [19,43] 4
[12,31] [ 1,47] [23,41] 3
[19,43] [18,64] [46,48] 1/3
[23,24] [19,43] [23,41] 3/4
[23,41] [24,62] [46,48] 1/4

of which Jarek chose the last.

The other 5 solutions may be obtained from Jarek's,
by noting that

[18,64]/2 = [ 9,32]
[24,62]/2 = [12,31]
[46,48]/2 = [23,24]

and then permuting pairs of coprime integers.

Question: Are there any more solutions?
If not, then Jarek's construction is essentially unique.

Here is a test with b running up to 500:

{jarek(a,b)=local(s,t,C,G,c,d,g,h);
s=2*(a^2+b^2);t=sqrtint(s/2);C=[];G=[];
for(c=1,t,if(issquare(s-c^2,&d),g=abs((4*a*b)^2-(d^2-c^2)^2);
if(d>c&&g,G=concat(G,g);C=concat(C,[[c,d]]))));
for(i=2,length(G),h=G[i];for(j=1,i-1,g=G[j];
if(max(g,h)<6*gcd(g,h),print([[a,b],C[j],C[i],g/h]))))}

for(a=1,500,for(b=a+1,500,if(gcd(a,b)==1,jarek(a,b))))

[[1, 47], [18, 64], [24, 62], 4/3]
[[9, 32], [1, 47], [19, 43], 4]
[[12, 31], [1, 47], [23, 41], 3]
[[19, 43], [18, 64], [46, 48], 1/3]
[[23, 24], [19, 43], [23, 41], 3/4]
[[23, 41], [24, 62], [46, 48], 1/4]
Goodbye!

David
• That reminded me... http://tech.groups.yahoo.com/group/primenumbers/message/9901
Message 6 of 7 , Aug 11, 2007
• ... Andrey s suggested method for 11 consecutive factorizations at 200 digits was to find factorizations of x^6-a, in the 5 hardest cases: a=2,3,5,6,7, and
Message 7 of 7 , Aug 11, 2007
--- In primeform@yahoogroups.com, "Andrey Kulsha"
<Andrey_601@...> wrote:

> That reminded me...

Andrey's suggested method for 11 consecutive factorizations
at 200 digits was to find factorizations of x^6-a, in the
5 hardest cases:

a=2,3,5,6,7,

and then tackle the 6 easier cases:

a=-1,0,1,4,8,9,

where there are obvious algebraic factorizations,
to split the burden.

This looks rather tough at 500 digits.

David
Your message has been successfully submitted and would be delivered to recipients shortly.