And another large pair, [url]http://factorization.ath.cx/index.php?id=1100000000310446320[/url], ((10^6161*61-169)/9-1)/2 should enable a proof that [url]http://factorization.ath.cx/index.php?id=1100000000291798445[/url], (10^6161*61-169)/9 is prime. And factordb definitely show them both as PRP now!

Chris

And an even larger pair, proving [url]http://factorization.ath.cx/index.php?id=1100000000414241880[/url], ((2^21222*177+1)/81546809+1)/2118 should enable a proof that [url]http://factorization.ath.cx/index.php?query=%282%5E21222*177%2B1%29%2F81546809[/url], (2^21222*177+1)/81546809 is prime.

Chris

After adding algebraic factors to 10^7346*3-48 I found that [url]http://factorization.ath.cx/index.php?id=1100000000260558239[/url], (10^3671*25-1)/21029372841 is PRP so proving it would prove [url]http://factorization.ath.cx/index.php?id=1100000000291841399[/url], 10^7346*3-49, is prime.

Chris

Another one, with two smaller PRPs, [url]http://factorization.ath.cx/index.php?id=1100000000767426548[/url] (2300 digits) and [url]http://factorization.ath.cx/index.php?id=1100000000767426553[/url] (370 digits. Since [url]http://factorization.ath.cx/index.php?id=1100000000439186756[/url], (9832^2143-1)/9831, is 8553 digits you probably need to prove both to enable a N-1 proof (N-1 is 5% factored now).

Chris

I'm doing well, another large algebraic PRP.

Proving [url]http://factorization.ath.cx/index.php?id=1100000000767998993[/url] (3173 digits) prime will enable a N-1 proof for [url]http://factorization.ath.cx/index.php?id=1100000000322074068[/url], (2^28799*21+1)/43 (8670 digts).

Chris

[QUOTE=chris2be8;398694]I'm doing well, another large algebraic PRP.

Proving [url]http://factorization.ath.cx/index.php?id=1100000000767998993[/url] (3173 digits) prime will enable a N-1 proof for [url]http://factorization.ath.cx/index.php?id=1100000000322074068[/url], (2^28799*21+1)/43 (8670 digts).

Chris[/QUOTE]

Using my own Ruby program the 3173 is 2-PRP but does not pass "the main test". I checked with pari's ispseudoprime() and it says it is composite. :smile:

Factordb certainly said it was PRP when I made that post. Still PRP does stand for *probably* prime. Thanks for checking anyway.

And here's another. Proving [URL]http://factorization.ath.cx/index.php?id=1100000000768813028[/URL] (4040 digits) will enable a N-1 proof that [URL]http://factorization.ath.cx/index.php?id=1100000000502087527[/URL], (9199^2267-1)/9198 (8982 digits) is prime.

Chris (now triple checking factordb shows them as PRP).

PS. In both cases the PRP appeared after I added algebraic factors to the large PRP. So factordb hadn't had long to check if it was prime.

[QUOTE=chris2be8;398776]Factordb certainly said it was PRP when I made that post. Still PRP does stand for *probably* prime. Thanks for checking anyway.

And here's another. Proving [URL]http://factorization.ath.cx/index.php?id=1100000000768813028[/URL] (4040 digits) will enable a N-1 proof that [URL]http://factorization.ath.cx/index.php?id=1100000000502087527[/URL], (9199^2267-1)/9198 (8982 digits) is prime.

Chris (now triple checking factordb shows them as PRP).

PS. In both cases the PRP appeared after I added algebraic factors to the large PRP. So factordb hadn't had long to check if it was prime.[/QUOTE]

I did say "PRP", but after I posted, someone changed it to "C". :smile:

[url]http://factorization.ath.cx/index.php?id=1100000000768976722[/url] proved by N+1
[url]http://factorization.ath.cx/index.php?id=1100000000768977696[/url] proved by N-1

[QUOTE=Batalov;396779][URL="http://factordb.com/index.php?id=1100000000762116112"]Phi(5,-205*2^771)[/URL]
Note that N-1 is divisible by 205*2^771 and by 205*2^771-1 which is prime.

Similar proofs are possible for Phi[SUB]7[/SUB](n) and Phi[SUB]15[/SUB](n) (with CHG and some extra factoring).
Maybe some [URL="http://factordb.com/index.php?query=%28%282%5E96*173%29%5E11%2B1%29%2F%282%5E96*173%2B1%29"]Phi[SUB]11[/SUB](n)[/URL]...?[/QUOTE]


What is the formula for algebraic factors of N-1? I found that [url]http://factorization.ath.cx/index.php?id=1100000000256709216&open=prime[/url], (13^11105+1)/(13^2221+1)/26521-1, is divisible by 13^2220-1, are there any other significant factors?

Chris

[QUOTE=chris2be8;399212]What is the formula for algebraic factors of N-1? I found that [url]http://factorization.ath.cx/index.php?id=1100000000256709216&open=prime[/url], (13^11105+1)/(13^2221+1)/26521-1, is divisible by 13^2220-1, are there any other significant factors?

Chris[/QUOTE]
N = Phi[SUB]5[/SUB](x) = x[SUP]4[/SUP] + x[SUP]3[/SUP] + x[SUP]2[/SUP] + x + 1,
so N-1 = x(x+1)(x[SUP]2[/SUP]+1).
If x is k*b[SUP]n[/SUP] and x+1 = k*b[SUP]n[/SUP]+1 is prime, then you get a 50% factored N-1;
if x+1 is a semiprime and x can be >33.3% factored, then you have a >33.3% factored N-1.

What you found is a N = Phi[SUB]5[/SUB](-x) = Phi[SUB]10[/SUB](x).
To prove it, you need a >33% factorization factorization of N-1 = x(x-1)(x[SUP]2[/SUP]+1), but you have a denominator and for most numbers of this form you will not find that factorization.
_______________________________________

On a similar topic, I got interested in primitive factors of Fibonacci and Lucas numbers, and their N-1 and N+1 have very interesting algebraic properties. Have a look at [URL="http://factordb.com/index.php?query=I%28137439%29%2FI%2845813%29%2F17"]this[/URL] or [URL="http://factordb.com/index.php?query=I%28166737%29%2FI%2855579%29%2F2"]this[/URL] (these are primU(137439) and primU(166737)). You will get the idea from seeing N-1 (which needed some help):
[URL="http://factordb.com/index.php?query=I%28137439%29%2FI%2845813%29%2F17-1"]primU(137439)-1[/URL] has nontrivial gcd's with
I3 I4 I5 I6 I8 I9 I10 I12 I15 I18 I23 I24 I30 I45 I46 I48 I69 I83 I90 I92 I138 I166 I184 I249 I276 I332 I498 I509 I552 I664 I996 I1018
I1527 I1909 I1992 I2545 I3054 I3818 I4581 I5090 I5727 I7635 I7636 I9162 I11454 I15270 I15272 I22905 I22908 I45810 I45816
and this set completes its algebraic factorization 100%. However, individual terms (the larger ones) are not factored too much, so the total factorization is too short.
Similar with primU(166737)-1, it has gcd's with
I3 I4 I5 I6 I7 I10 I14 I20 I28 I35 I59 I70 I118 I140 I157 I177 I314 I354 I397 I471 I794 I942 I1588 I1985
I2779 I3970 I5558 I7940 I9263 I11116 I13895 I18526 I27789 I27790 I55578 I55580
(the even-valued Fibonacci's here are standing in for the lucas half-parts: I(2n)=I(n)*L(n); I simply run the gcd script using fibonacci(), the gcd() takes care of the rest).