- --- In primenumbers@yahoogroups.com,

"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)))));

Thus we are unable to solve the "puzzle" without obtaining

> [0, 1, 4, 7, 10]

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 - --- In primenumbers@yahoogroups.com,

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

The answer is: 5558358186556723282830357110595686293

> "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

to [Mod(1,2)] instead of [Mod(0,1)]).

> Comment: This puzzle may be solved, systematically, in less

Puzzle solved? Check.

> than 3 CPU-seconds, without appealing to extraneous data.

Systematically? Check.

Less than 3 CPU-seconds? Check.

Extraneous data? Unsure.

Elegant: Certainly not!

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

Peter Kosinar <goober@...> wrote:

> The answer is: 5558358186556723282830357110595686293

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

{read("OpenLenstra.gp");

\\ 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