## Re: Algebraic factoring

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• Hi David, ... I got this bit and managed to get the 4 algebraic factors. f3*f2 = 2^137+2^69+1 f4*f1 = 2^137+1 f5 = 2^137-2^69+1 f6 = 2^137-1 My question is how
Message 1 of 45 , Feb 1, 2011
Hi David,
>
>
>
> Jack Brennen <jfb@> wrote:
>
> > (2^137-2^69+1)
> > (2-1)
> > (2^136+2^135+2^134+...+2+1)
> > (2+1)
> > (2^136-2^135+2^134-...-2+1)
> > (2^137+2^69+1)
>
> Err, Jack, I think you missed the algebraic factor 2^2+1 ?
>
> > 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.
>
> Solution:
> f1 = 2^2-1
> f2 = 2^2+1
> f3 = (2^137+2^69+1)/(2^2+1)
> f4 = (2^137+1)/(2+1)
> f5 = 2^137-2^69+1
> f6 = 2^137-1
>
> The algebra that Jane missed is
> 4^(2*k-1) + 1 = (2^(2*k-1) - 2^k + 1)*(2^(2*k-1) + 2^k + 1)
> and if that ain't algebra, I don't know what is :-)
I got this bit and managed to get the 4 algebraic factors.
f3*f2 = 2^137+2^69+1
f4*f1 = 2^137+1
f5 = 2^137-2^69+1
f6 = 2^137-1
My question is how does one know algebraically (or at least without trial dividing) that 2^2+1 divides 2^137+2^69+1 and 2^2-1 divides 2^137+1?
cheers
Ken

> David
>
• ... 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
"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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