## Re: [PrimeNumbers] Re: Algebraic factoring

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• 16^137-1 == 16*u^8-1, if we choose u == 16^17 == 2^68. 16*u^8-1 == (2*u^2-2*u+1)(2*u^2-1)(2*u^2+1)(2*u^2+2*u+1). Substitute back u == 2^68, we get:
Message 1 of 45 , Feb 1, 2011
16^137-1 == 16*u^8-1, if we choose u == 16^17 == 2^68.

16*u^8-1 == (2*u^2-2*u+1)(2*u^2-1)(2*u^2+1)(2*u^2+2*u+1).

Substitute back u == 2^68, we get:

(2^137-2^69+1)(2^137-1)(2^137+1)(2^137+2^69+1)

The two middle terms factor algebraically and we end up with
the complete algebraic factorization of 16^137-1 into 6 factors,
one of which is unfortunately the trivial unit factor (1), so
of no use for numerical factorization:

(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)

On 2/1/2011 10:10 AM, Jane Sullivan wrote:
>>
>>>> Surely the largest algebraic factor of 16^137-1 is
>>>> 4^137+1 which has 83 digits
>>>
>>> Not so, Jane. According to Léon-François-Antoine
>>> the largest algebraic factor has 42 digits.
>>> Try googling him; he is rather useful in this case :-)
>
> In that case, I'm afraid Léon-François-Antoine is wrong.
> (4^137+1)(4^137-1) = 4^274-1 = 4^(2x137)-1 = (4^2)^137-1 = 16^137-1.
> If you say that is not an algebraic factorization, then I'm sorry,
> you're wrong. It is a simple difference of two squares factorization
> (a^2-b^2) = (a+b)(a-b).
>
> We may be talking at cross purposes here: I think you are looking at
> Aurifeuillian factorizations, but there are algebraic factors that are
> not Aurifeuillian.
>
>>
>> Here is a really big hint; try googling:
>>
>> "Léon-François-Antoine" -Fleury
>>
>> /exactly/ as written above.
>
> Yes I googled that, but I can't find anything relating to 137. Maybe I
> gave up too easily.
>
>>
>> David
>
> Best wishes
• ... 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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