- djbroadhurst wrote:
> --- In primenumbers@yahoogroups.com,

No I didn't, although it might look that way.

> Paul Leyland <paul@...> wrote:

>

>> I think you're using different vocabularies.

>> In the Cunningham project, and 16^137-1 == 2,548-

>> in their nomenclature, algebraic factors are

>> distinguished from Aurifeuillian factors.

>

> The problem was that Jane saw Aurifeullian and algebraic

> as being mutually exclusive adjectives. They are not.

What I was doing was objecting to the following, posted earlier in the

thread

<quote>> Exercise: Find the algebraic factors of 16^137-1.

</quote>

> Comment: The maximum number of digits in any such

> factor is 42, so if your routine produces a larger

> factor it does not know enough algebra.

because, as I showed, there is an alegbraic (composite) factor of

16^137-1 that has 83 (i.e. more than 42) digits.

>

Let me make it quite clear: I agree with you, but the statement that the

> The Cunningham project uses this vocabulary:

>

> http://www.ams.org/publications/online-books/conm22-conm22-chVII.pdf

>> L,M Aurifeuillian algebraic factorization (see III C)

>

> precisely as I did here:

>

> http://tech.groups.yahoo.com/group/primenumbers/message/22518

>> There are two types of algebraic factorization for x^n-1:

>> 1) cyclotomic algebraic factorization

>> 2) Aurifeuillian algebraic factorization

>

> So at least John Brillhart, D. H. Lehmer, J. L. Selfridge,

> Bryant Tuckerman, and S. S. Wagstaff, Jr agree with me,

> even if Jane or Paul may not :-)

maximum number of digits in any algebraic factor of 16^137-1 is 42 is

wrong, because it's 83.

>

Best wishes

> David

--

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

"mikeoakes2" <mikeoakes2@...> wrote:

> > ABC2 if($a>$b,($a^137+$b^137)/($a+$b),0) & ($a^137-$b^137)/($a-$b)

It seems that no-one spotted the Aurifeuillian method for

> > a: from 216 to 262

> > b: from 216 to 262

> > does not give a double hit, as Mike has probably checked.

> Your intuition is spot on, as per usual, good Sir:

> they were pfgw jobs of precisely this form that I ran.

> Exercise 4: Find integers (a,b) with min(a,b) > 46225 and

As previously remarked, the trivial Ansatz

> N(a,b) = (a^137 - b^137)/(a - b) equal to the product of

> two primes, each with less than 330 decimal digits.

> Comment: This may solved, systematically, in less than

> 30 seconds, including the time for primality proofs.

(a,b) = (x^2,y^2) cannot solve Exercise 4.

Solution: Aurifeuillian algebraic factorization results from

(a,b) = (x^2,137*y^2), with min(x,sqrt(137)*y) > 215. Since

max(x,sqrt(137)*y) < 10^(329/136) < 263, we have x in the

range [216,262] and y in the range [19,22], yielding a pair

of PRPs in 0.27 seconds and a proof in 26 seconds:

{g=factor((z^274-137^137)/(z^2-137))[,1]~;

for(x=216,262,for(y=19,22,if(gcd(x,y)==1,

f=abs(y^136*subst(g,z,x/y));

if(ispseudoprime(f[1])&&ispseudoprime(f[2]),F=f;

print1([[x,y],[#Str(F[1]),#Str(F[2])]]" in "gettime" ms. ")))));

if(isprime(F)==[1,1],print("Proof in "round(gettime/10^3)" sec."));}

[[220, 21], [323, 329]] in 270 ms. Proof in 26 sec.

Hence we have the solution N(a,b) = P323*P329,

with (a,b) = (220^2,137*21^2) = (48400,60417).

David