20220829, 06:41  #12  
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May 2003
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GMPECM 7.0.4 [configured with GMP 6.2.1, enableasmredc] [ECM] Input number is 9614319269...4094423041 (82177 digits) Using B1=1000000, B2=914058610, polynomial Dickson(3), sigma=0:4416831086328320100 Step 1 took 27856254ms Step 2 took 7836917ms ********** Factor found in step 2: 2749553154323378809339903 Found prime factor of 25 digits: 2749553154323378809339903 Composite cofactor 3496...9247 has 82153 digits The above has been lightly edited. Note it was found on the first curve tried. 

20220829, 07:40  #13  
May 2003
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I don't think I'll bother with the 160kdigit next term, it's probably time to look more closely at the algebra. 

20220829, 09:16  #14  
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? factor(2749553154323378809339902) %1 = [ 2 1] [ 3 1] [ 232357 1] [ 423601 1] [4655840911681 1] ? factor(2749553154323378809339904) %2 = [ 2 18] [ 313 1] [ 113083 1] [296333355629 1] 

20220829, 14:08  #15 
May 2003
255_{10} Posts 
It's nice when things drop out just at the ideal moment. For instance, I was just about to give up on this number:
Run 100 out of 100: Using B1=10315491, B2=1031549035133066160, polynomial Dickson(12), sigma=1:1661695324 Step 1 took 12909ms Step 2 took 6496ms ********** Factor found in step 2: 125960894984050328038716298487435392001 Found prime factor of 39 digits: 125960894984050328038716298487435392001 Prime cofactor 260850953670702126641628144218597843056427843281352119936811457220827426196135100642146995717121 has 96 digits Bosh! All the known factors in the numerators are now up at http://fatphil.org/maths/sqrt_cf/sqrt5.html I'll need to read up on the denominators a bit more before deciding if there's anything more to discover. 
20220829, 14:48  #16  
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You should have switched to GNFS for the C135 IMO. Last fiddled with by xilman on 20220829 at 14:49 

20220829, 16:11  #17 
May 2003
3×5×17 Posts 
Now I've worked out the algebraic structure, these are just a^n+b^n and a^nb^n sequences. I don't think there's any need to hunt for factors any more, at least on the numerator side, but I'll continue to fill out a decent sized table of factors on the denominator side, mostly for schitzengiggles. A deeper hunt for PRPs on the denominator side should be done still.

20220829, 17:18  #18 
"Daniel Jackson"
May 2011
14285714285714285714
3·251 Posts 
I just checked, and your 82177digit number is Lucas(393216)/2. Therefore, it's divisible by Lucas(131072)
Last fiddled with by Stargate38 on 20220829 at 17:20 Reason: forgot to mention which number 
20220829, 17:59  #19  
May 2003
3·5·17 Posts 
Quote:
Congrats, you've solved the first half of the conjecture! Edit: how did you work out it was halfalucas? Last fiddled with by fatphil on 20220829 at 18:00 

20220829, 18:29  #20 
"Daniel Jackson"
May 2011
14285714285714285714
1361_{8} Posts 
I copypasted the full number into FactorDB. After hitting "Factorize", I then clicked on the link under where it says "number". Since the number was already on the DB, it simplified it to the shortest known formula.
Last fiddled with by Stargate38 on 20220829 at 18:30 
20220829, 19:07  #21 
Feb 2017
Nowhere
1100001001110_{2} Posts 
The fundamental unit > 1 of the ring of algebraic integers in Q(sqrt(5)) is the "golden ratio" .
We have Thus, the Lucas and Fibonacci numbers satisfy . Both L_{n} and F_{n} are even when n is divisible by 3. The numerators and denominators of the SCF for sqrt(5) (solutions to x^{2}  5y^{2} = ยฑ1) are thus x = (1/2)L_{3k} and y = (1/2)F_{3k}. Now L_{3k} is divisible by L_{d} for every divisor of 3k whose cofactor is odd, and F_{3k} is divisible by F_{d} for every divisor d of 3k. You have to be a little careful in translating this to results about (1/2)L_{3k} and (1/2)F_{3k}. 
20220829, 19:08  #22  
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