22579Re: Algebraic factoring
- Feb 7, 2011--- In email@example.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 andAs 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:
print1([[x,y],[#Str(F),#Str(F)]]" 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).
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