- 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.

On 2/2/2011 3:08 PM, kraDen wrote:

> Hi David,

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

>>

>>

>>

>> --- In primenumbers@yahoogroups.com,

>> "kraDen"<kradenken@> wrote:

>>

>>> My question is how does one know algebraically

>>> (or at least without trial dividing) that

>>> 2^2+1 divides 2^137+2^69+1

>>

>> We want to show that y = 2^2 + 1 divides N = 2^137 + 2^69 + 1.

>> Let u = 2^68. Then

>> N - y = 2*u^2 + 2*u - 2^2 = 2*(u-1)*(u+2).

>> But y divides u - 1 = (y-1)^34 - 1. Hence y divides N - y.

>> Hence y divides N and we are done.

> Thankyou.

> However this seems to be proving it is after already knowing it is (hopefully that statement is understandable).

> However this still doesn't explain to to me how you (algebraically) knew that 2^2+1 was a divisor.

> Instead it looks like you have achieved the factor by trial division and then proved it algebraically.

>

> cheers

> Ken

>

>

>> Note that we do not need to compute the value of 2^2+1.

>> I believe that I can do that in my head, but fortunately

>> it is not necessary for this algebraic exercise :-)

>>

>> David

>>

>

>

>

>

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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)

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