## Re: Algebraic factoring

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• ... There is also an algebraic (composite) factor of 16^137-1 that has 165 (i.e. more than 83) digits, namely (16^137-1)/(2^2-1). In your first posting on this
Message 1 of 45 , Feb 2, 2011
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"Jane Sullivan" <janesullivan@...> wrote:

> there is an algebraic (composite) factor of
> 16^137-1 that has 83 (i.e. more than 42) digits.

There is also an algebraic (composite) factor of
16^137-1 that has 165 (i.e. more than 83) digits,
namely (16^137-1)/(2^2-1).

In your first posting on this subject
you courteously acknowledged my hints
(which I had hoped would resolve any ambiguity):

> 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.
> The largest, f6, has 42 digits, yet may be written in 7 characters.
> The next-to-largest, f5, may be written in 12 characters.

In that sense, the answer is 42.

In a more general sense, there are 62 non-unit proper
algebraic divisors ranging from 2^2-1 to (16^137-1)/(2^2-1),
with the following digit-counts:

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;
F = [f1,f2,f3,f4,f5,f6];
V = vector(165);
for(k=1,62,V[#Str(prod(j=1,6,F[j]^binary(64+k)[j+1]))]++);
for(i=1,165,if(V[i],print(i"-digit algebraic divisors: "V[i])));

1-digit algebraic divisors: 2
2-digit algebraic divisors: 1
41-digit algebraic divisors: 2
42-digit algebraic divisors: 12
43-digit algebraic divisors: 2
82-digit algebraic divisors: 4
83-digit algebraic divisors: 16
84-digit algebraic divisors: 4
123-digit algebraic divisors: 2
124-digit algebraic divisors: 12
125-digit algebraic divisors: 2
164-digit algebraic divisors: 1
165-digit algebraic divisors: 2

Best regards

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
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"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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