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## Re: Lucas super-pseudoprime puzzle

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• ... The best pari-GP script I can come up with is /quadratic/ in n_max, so will take about 23360000 secs or 9 months at 3.6 GHz to reach n_max=10^7 and
Message 1 of 46 , Nov 2, 2010
>
> Definition: The integer pair (q,n) is a "Lucas super-pseudoprime"
> (LSPS) pair if and only if n is composite, n > q > 0, and
> V(p,q,n) = p mod n, for every integer p.
>
> With n = 11*29*37*43 = 507529, there are precisely 37 LSPS pairs
> (q,n), namely those with q = (11*k + 3)*29*43, for k = 0 to 36.
> Mike Oakes remarked that 507529 is not a Carmichael number.
>
> Puzzle: Find an odd square-free non-Carmichael number
> such that there exist more than 70,000 LSPS pairs (q,n).
>
> Comment: There is a solution with n < 10^7.

The best pari-GP script I can come up with is /quadratic/ in n_max, so will take about 23360000 secs or 9 months at 3.6 GHz to reach n_max=10^7 and (presumably) solve your puzzle.
Here it is:-

\\For odd non-prime n, finds first q s.t. lucasV(p,q,n)=p mod n for all p
\\(Using CRT, q can be reconstructed from its remainders)
v(p,q,n,m)=2*polcoeff(lift(Mod((p+x)/Mod(2,m),x^2+4*q-p^2)^n),0);
n_min=3; n_max=10^5;
default(primelimit,n_max);
gettime;
{
forstep(n=n_min,n_max,2,
if((n%10^5)==1,print("n="n));
f=factor(n);
facs=#f[,1];
n_ok=1;
if(facs>1,
for(ind=1,facs,
if(f[ind,2]>1,n_ok=0;break(1)); \\not squarefree
m=f[ind,1];
count=0;
for(q=1,m,
ok=1;
for(p=1,m,
if(lift(v(p,q,n,m)-p)<>0,ok=0;break(1));
); \\end for p
if(ok,count++;break(1)); \\add code here to remember q, for CRT
); \\end for q
if(count==0,n_ok=0;break(1));
); \\end for ind
if(n_ok,print("n="n" "factor(n)" is ok"));
); \\end if facs>1
); \\end forstep n
tim=gettime;
print(" took "tim" ms");
}
/*
n_max time
10^4 24 secs
10^5 2336 secs
*/

What am I missing??

Mike
• ... http://physics.open.ac.uk/~dbroadhu/cert/dbmo116.out gives my 116, in the format [n, factors, number of solutions] With n
Message 46 of 46 , Nov 9, 2010
"mikeoakes2" <mikeoakes2@...> wrote:

> > My revised count up to 2*10^10 is 116.
> My (original) count up to 2*10^10 was 105.
> So it must have missed 11, i.e. a bigger proportion.