## Re: Puzzle re factorization

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• ... ronhallam@... wrote, after many lines of working, ... Thus we are unable to solve the puzzle without obtaining the complete
Message 1 of 8 , Nov 25, 2012
"ronhallam@..." <ronhallam@...> wrote,

after many lines of working,
> (4 7 0 1 10 )mod 11

in agreement with my previous one-line derivation:
> print(vecsort(eval(Set(vector(10,q,(q-1/q)%11)))));
> [0, 1, 4, 7, 10]

Thus we are unable to solve the "puzzle" without obtaining
the complete factorization of a 133-digit target.

A small example illustrates this. With
n = 1110187 = 1 mod 11, we may write n = p*q
with q > p > 1 in 7 different ways and obtain
all 5 of the distinct possible values of q-p modulo 11:

{n=1110187;fordiv(n,p,q=n/p;if(q>p&&p>1,print1((q-p)%11" ")));}

4 0 1 4 4 7 10

where 10 happens to be the residue of (q-p) that
would be obtained from the first factorization
produced by Fermat's method, namely n = 1027*1081.

David
• ... 2203353588507632689540389884099234780425798106022737554148425561277092513841358286596173426627853446495163308636257511927689252306411 In this case, n = p*q
Message 2 of 8 , Nov 25, 2012
"ronhallam@" <ronhallam@> wrote:

> Work out the difference mod 6, 7 and 11 on the following number:

2203353588507632689540389884099234780425798106022737554148425561277092513841358286596173426627853446495163308636257511927689252306411

In this case, n = p*q with (q-p) divisible by the primes
2, 3, 5, 7, 11, 4590011 and 186453571036492667910025371580824511.

Puzzle: Use these data to find another prime that divides (q-p).

Comment: This puzzle may be solved, systematically, in less
than 3 CPU-seconds, without appealing to extraneous data.

David
• ... The answer is: 5558358186556723282830357110595686293 And here s the (ugly) code which found it: n= v=[2, 3, 5, 7, 11, 4590011,
Message 3 of 8 , Nov 27, 2012
> "ronhallam@" <ronhallam@> wrote:
>
> > Work out the difference mod 6, 7 and 11 on the following number:
>
> 2203353588507632689540389884099234780425798106022737554148425561277092513841358286596173426627853446495163308636257511927689252306411
>
> In this case, n = p*q with (q-p) divisible by the primes
> 2, 3, 5, 7, 11, 4590011 and 186453571036492667910025371580824511.
>
> Puzzle: Use these data to find another prime that divides (q-p).

And here's the (ugly) code which found it:

n=<the long number>
v=[2, 3, 5, 7, 11, 4590011, 186453571036492667910025371580824511]

vv=[Mod(0,1)];for(k=1,#v,s=sqrt(Mod(n,v[k])); \
vv=concat(apply(x->[chinese(x,s),chinese(x,-s)],vv))); \
m=prod(k=1,#v,v[k]);vv=lift(vv); \
for(k=1,#vv,r=vv[k];s=(n-r^2)/m;apb=lift(Mod(s,m)/r); \
atb=(s-apb*r)/m; if(issquare(apb^2-4*atb), \
d=sqrtint(apb^2-4*atb); print(factor(d))))

Mat([5558358186556723282830357110595686293, 1])
Mat([5558358186556723282830357110595686293, 1])

(the answer is printed twice due to Mod(1,2) being equal to Mod(-1,2); if
you want to get rid of it, start the first loop with k=2 and initialize vv

> Comment: This puzzle may be solved, systematically, in less
> than 3 CPU-seconds, without appealing to extraneous data.

Puzzle solved? Check.
Systematically? Check.
Less than 3 CPU-seconds? Check.
Extraneous data? Unsure.
Elegant: Certainly not!

Peter
• ... Nice work, Peter. Here is my version, using OpenLenstra: {read( OpenLenstra.gp ); http://tech.groups.yahoo.com/group/primeform/message/2492
Message 4 of 8 , Nov 27, 2012
Peter Kosinar <goober@...> wrote:

Nice work, Peter. Here is my version, using OpenLenstra:

\\ http://tech.groups.yahoo.com/group/primeform/message/2492
n=3^240*polcyclo(770,4/3)/(2311*512053081);
v=[2,3,5,7,11,4590011,186453571036492667910025371580824511];
d=prod(k=1,#v,v[k]);v=vector(#v,k,sqrt(Mod(n,v[k])));u=[v[1]];
for(j=2,#v,w=vector(#u,k,chinese(u[k],v[j]));
u=concat(w,vector(#u,k,chinese(u[k],-v[j]))));u=lift(u);
for(k=1,#u,t=Lenstra(n,u[k],d)[3];if(#t,t=factor((t[2]-t[1])/d);
print("Solution: "t[1,1]" in "gettime" ms");break()));}

Solution: 5558358186556723282830357110595686293 in 42 ms

Comment: The complete factorization of the cyclotomic
number was given by Paul Leyland:
http://www.leyland.vispa.com/numth/factorization/anbn/4+3.txt
385 2311.512053081.200512409435077328148952740169875706164009384669091. P83

David
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