## Re: [PrimeNumbers] Re: Algebraic factoring

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• 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
Message 1 of 45 , Feb 2, 2011
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,
>>
>>
>>
>>
>>> 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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• ... 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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