- Hi David,
--- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

>

>

>

> --- In primenumbers@yahoogroups.com,

> 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

> - --- In primenumbers@yahoogroups.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 and

As 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:

{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