Loading ...
Sorry, an error occurred while loading the content.

Re: [PrimeNumbers] Re: Algebraic factoring

Expand Messages
  • Jane Sullivan
    ... Surely the largest algebraic factor of 16^137-1 is 4^137+1 which has 83 digits? There are twelve prime factors of 16^137-1 (=2^548-1), which are 3 5 1097
    Message 1 of 45 , Feb 1, 2011
    View Source
    • 0 Attachment
      djbroadhurst wrote:
      > --- In primenumbers@yahoogroups.com,
      > "j_chrtn" <j_chrtn@...> wrote:
      >
      >> "djbroadhurst" <d.broadhurst@> wrote:
      >>> Exercise: Find the algebraic factors of 16^137-1.
      >>
      >> Same game with 61^73-60^73, 140^71-139^71 and 138^127-137^127.
      >
      > That looks to me like a very different ball game.
      > How might L�on-Fran�ois-Antoine help us to find
      > /algebraic/ factors in the cases given by Jean-Louis?
      >
      > Hints for the exercise:
      > The 6 non-unit /algebraic/ factors of
      > 16^137-1 = f1*f2*f3*f4*f5*f6
      > may be found in one's head; no need for a computer.
      > The largest, f6, has 42 digits, yet may be written in 7 characters.
      > The next-to-largest, f5, may be written in 12 characters.
      >

      Surely the largest algebraic factor of 16^137-1 is 4^137+1 which has 83
      digits?

      There are twelve prime factors of 16^137-1 (=2^548-1), which are
      3 5 1097 15619 189061 168434085820849 32127963626435681
      105498212027592977 32032215596496435569 5439042183600204290159
      206875670104957744917147613 921525707911840587390617330886362701

      > David

      --
      Jane
    • djbroadhurst
      ... 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
      View Source
      • 0 Attachment
        --- In primenumbers@yahoogroups.com,
        "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
      Your message has been successfully submitted and would be delivered to recipients shortly.