## Re: [PrimeNumbers] Re: Algebraic factoring

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• ... No I didn t, although it might look that way. What I was doing was objecting to the following, posted earlier in the thread ... because,
Message 1 of 45 , Feb 2, 2011
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> 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.

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

What I was doing was objecting to the following, posted earlier in the
<quote>
> Exercise: Find the algebraic factors of 16^137-1.

> 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.
</quote>

because, as I showed, there is an alegbraic (composite) factor of
16^137-1 that has 83 (i.e. more than 42) digits.

>
> 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:
>
>> 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 :-)

Let me make it quite clear: I agree with you, but the statement that the
maximum number of digits in any algebraic factor of 16^137-1 is 42 is
wrong, because it's 83.

>
> David

Best wishes
--
Jane
• ... It seems that no-one spotted the Aurifeuillian method for ... As previously remarked, the trivial Ansatz (a,b) = (x^2,y^2) cannot solve Exercise 4.
Message 45 of 45 , Feb 7, 2011
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"mikeoakes2" <mikeoakes2@...> wrote:

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

It seems that no-one spotted the Aurifeuillian method for

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

As previously remarked, the trivial Ansatz
(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
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