## Re: Algebraic factoring

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• ... Thankyou ... I certainly didn t mean to imply this. I was sure you had done it algebraically. It just wasn t apparent to me how. As I said I found 4
Message 1 of 45 , Feb 2, 2011
>
>
>
> Jack Brennen <jfb@> wrote:
>
> > Actually, the original number 16^137-1 is equal to 2^548-1,
> > correct? x^548-1 is easily seen to be divisible by x^4-1,
> > which is divisible by x^2+1.
> > So 2^548-1 is divisible by 2^2+1.
Thankyou

> Thanks Jack for explaining to my good friend Ken,
> by computing (shame!) the integer 2^2+1 and
> then actually using trial division (crime!)
I certainly didn't mean to imply this.
I was sure you had done it algebraically.
It just wasn't apparent to me how.
As I said I found 4 factors but then got stuck.
I desired learning how to derive the additional factors (for future use, originality is not my strong point but I willingly "stand on the shoulders of giants").
cheers
Ken
> by whatever integer 2^2+1 turned out to be.
> I never did any such heinous thing, honest, Guv.
>
> 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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