- View Sourcedjbroadhurst wrote:
> --- In primenumbers@yahoogroups.com,

Surely the largest algebraic factor of 16^137-1 is 4^137+1 which has 83

> "j_chrtn" <j_chrtn@...> wrote:

>

>> "djbroadhurst" <d.broadhurst@> wrote:

>>> Exercise: Find the algebraic factors of 16^137-1.

>>

>> Same game with 61^73-60^73, 140^71-139^71 and 138^127-137^127.

>

> That looks to me like a very different ball game.

> How might L�on-Fran�ois-Antoine help us to find

> /algebraic/ factors in the cases given by Jean-Louis?

>

> 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.

>

digits?

There are twelve prime factors of 16^137-1 (=2^548-1), which are

3 5 1097 15619 189061 168434085820849 32127963626435681

105498212027592977 32032215596496435569 5439042183600204290159

206875670104957744917147613 921525707911840587390617330886362701

> David

--

Jane - View Source--- 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