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Re: Algebraic factoring

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  • 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 1 of 45 , Feb 7, 2011
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      --- 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
    • 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
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        --- 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
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