 mersenneforum.org Odd perfect related road blocks II
 Register FAQ Search Today's Posts Mark Forums Read  2009-05-04, 16:16 #1 jchein1   May 2005 748 Posts Odd perfect related road blocks II Alex, fivemack, ATH, Andi47, Andi HB & all other factoring specialists, Let M be a smallest product of odd primes such that square free of ( gcd ( f(x) ,g(x)) ) | M for all integer x > 1 , where f(x) and g(x) are two given relatively prime integral polynomials. For example, if f(x) = (x2 + x + 1)3 - 33 & g(x) = x6 + x3 + 1 = 0, then M = 3 * 19 * 163 if f(x) = ((x4 + x3 + x2 + x + 1)5 - 55 ) /(x4 + x3 + x2 + x + 1 - 5 ) & g(x) = x20 + x15 + x10 + x5 + 1, then M = 5 * 151 * 701 * 2551 * 24251 * 34651 * 144853351 * 659575601 * 1271785993801. By applying successive (my) extended Eculid algorithms, I have encountered if 1) f(x) = (x2 + x + 1)3 – 33 & g(x) = (x162 + x81 + 1)3 – 33 , then Code: M = a * 598644007623271861103907634394401713031280184082045961436816127153183697504366019739512180572430920157281903457051081235141165004401529332661832140639096711933077 (c162, 2900 curves done), 2) f(x) = ((x2 + x + 1)9 - 39 ) / ((x2 + x + 1)3 - 33 ) & g(x) = x162 + x81 + 1 then Code: M = b * 142255953422080949010042135770532701808988159571292214113059412169543088033374620681986196789531788565741609436257609136785819948534868895759387087438788760412860551 (c165, 2500 curves done ), where a and b are products of known primes, c162 and c165 are two critical roadblocks I am unable to complete by myself. Note that: these M‘s are part of my “pushing “ algorithms concerning several ongoing papers (first one is “On the divisibility of systems of cyclotomic polynomials of degrees 3 and 5”) . Alex may be familiar with c162 and c165 at certain degrees. I sincerely hope it will get completed this time, otherwise one of my lemmas will get killed. Can someone please kindly look into it or give a try before I finally give up. I also found c1.x + c0 = 0 (mod c162), where Code: c1 = 69267298939868734352823569242638987830151794956582757323663257304713979808535979493103174703607420534575191258096981359057081634837910521017237222985183216764829376 c0 = 90396638571991941300717867224156466586094273057103180468944748956569143785197229469169728666741827825163688170928805513389627480981342184801410418423802620496945754 c1.x + c0 = 0 (mod c165), where Code: c1 = 76753267437136028631737826218354419681592252499734800494051831532232913505065057592910545783967666343514329739056408237901354351164460723435804959401745218854773 c0 = 209851117037629231974563829915727038069629516190909815572098319849496046563710374917531353670135870468192288676994164729691076296875290909453103109496164911219198 Is this helpful? I am looking for one factor(only) from each of the following. Some of them may have posted before *, although these composites did not affect the proof, they caused me a lot of unnecessary work, please help and thank you in advance. 1) σ(59345478426821800746377014559617^4) = 3538361.c121 # 2) σ(99995282631947^10) = 23.c139 3) σ(62060021^18) = c141 * =2859153813495302135105360393 * 65219432427202213218611042380245134951516556220905814475152737256300473009279433432471898046319778085284772450023 (mklasson, ecm) 4) σ(347568611538691^10) = 23.617.683.692539.c133 5) σ(10177286401^12) =15731.c116 * 6) σ(6294091^22) = c150 * 7) σ(5229043^22) = 967.194443.c140 * 8) σ(846041103974872866961^6) =1709.c123 = 1709 . 5155723754893919994283900206097487201 . 41621437657162669816374890034224257842979191889517024692601717156897218627524115006963 (fivemack, ecm) 9) σ(934415109937^12) = 2549.c141 = 2549 . 2636398433195487889353395293207 . 65932167085234904931409415588486535060018564833464183835891750325639105962353333551790922471214875173575350927 (mklasson, ecm) 10) σ(P82^2) = 61.433.4651.156817.c150 *, where P82= 3404253904642598840161913302701587626837449819596812423232352 24442734524 2057474631 11) σ(797161^28) = 59.31727.c159 12) σ(172545754771028210096747645881^6) = 631.c173 13) σ(172827552198815888791^6) = 113.c120 = 113 . 3335162523320119257180081115662029635417481 . 70710259042622884083135682057512365547361929153437894550723481191426825243649 (fivemack, snfs) 14) σ(177635683940025046467781066894531^4) = 5.431.c126 * - most wanted & - Alex resolved half a dozen and wblipp got one last year. # - I completed at least several hundreds or even thousand curves per each composite a year ago but lost the record. Best regards Joseph Last fiddled with by fivemack on 2009-05-05 at 00:30   2009-05-04, 17:02 #2 akruppa   "Nancy" Aug 2002 Alexandria 2,467 Posts Joseph, I made some typographical changes to fix the exponents and to avoid horizontal scrolling. Can you please check that I didn't introduce any errors? Alex   2009-05-04, 17:08   #3
10metreh

Nov 2008

2·33·43 Posts Quote:
 Originally Posted by jchein1 1) σ(59345478426821800746377014559617^4) = 3538361.c121 #
You seem to have missed a p11: 68280416627. It leaves a c110.

Quote:
 Originally Posted by jchein1 5) σ(10177286401^12) =15731.c116 * 13) σ(172827552198815888791^6) = 113.c120
These are both difficulty ~120 via a sextic. I suggest you factor them yourself. They should be easy.

Last fiddled with by 10metreh on 2009-05-04 at 17:14   2009-05-04, 17:22   #4
akruppa

"Nancy"
Aug 2002
Alexandria

2,467 Posts Quote:
 Originally Posted by 10metreh These are both difficulty ~120 via a sextic. I suggest you factor them yourself. They should be easy.
You sound like you know a lot about factoring and stuff! Can you please explain how you make a sextic for the second number?

Curiously yours,
Alex   2009-05-04, 17:25   #5
mklasson

Feb 2004

2×3×43 Posts Quote:
 Originally Posted by 10metreh You seem to have missed a p11: 68280416627. It leaves a c110.
You seem to have miscalculated. That p11 is not a factor.   2009-05-04, 17:26   #6
10metreh

Nov 2008

2×33×43 Posts Quote:
 Originally Posted by akruppa You sound like you know a lot about factoring and stuff! Can you please explain how you make a sextic for the second number? Curiously yours, Alex
Sorry if I was making a mistake, but I was simply diving out the factor x - 1 from x^7 - 1 to get the polynomial x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 with root 172827552198815888791.

(Yes, you are more experienced than me, so you are giving me advice if you correct me.)

And about the p11 issue: I'm not sure. I probably had a typo in the input.

Edit: found the typo. I missed out a digit in the number Last fiddled with by 10metreh on 2009-05-04 at 17:34   2009-05-04, 17:41 #7 mklasson   Feb 2004 2×3×43 Posts p31=2636398433195487889353395293207 | sigma(934415109937^12) / 2549 with cofactor p110.   2009-05-04, 17:48 #8 akruppa   "Nancy" Aug 2002 Alexandria 2,467 Posts Ok, I understand that you want to factor the resultant of f(x) and g(x), for example Code: ? f1(x) = (x^2+x+1)^3 - 3^3 ? g1(x) = x^6 + x^3 + 1 ? factorint(polresultant(f1(x), g1(x))) %15 = [3 6] [19 3] [163 1] ? f2(x) = (((x^5-1)/(x-1))^5-5^5) / (x^4 + x^3 + x^2 + x + 1 - 5) ? g2(x) = (x^25-1)/(x^5-1) ? factorint(polresultant(f2(x), g2(x))) %16 = [5 16] [151 1] [701 1] [2551 1] [24251 1] [34651 1] [144853351 1] [659575601 1] [1271785993801 1] The next two polynomials have a common factor (is this intended?) Code: ? f3(x) = (x^2+x+1)^3-3^3 ? g3(x) = (x^162+x^81+1)^3-3^3 ? gcd(f3(x), g3(x)) %23 = x - 1 ? polresultant(f3(x)/(x-1),g3(x)/(x-1))  The next two are coprime Code: ? f4(x)=((x^2+x+1)^9-3^9)/((x^2+x+1)^3-3^3) ? g4(x)=x^162 + x^81 + 1 ? polresultant(f4(x),g4(x))  I wonder if we can use the fact that the number to factor divides the resultant of the two polynomials to make SNFS polynomials. Then the factorization would be quite simple, if ECM should fail. If not, both could still be factored with GNFS, although with much more effort. Alex   2009-05-04, 17:52 #9 mklasson   Feb 2004 2·3·43 Posts sigma(62060021^18): ********** Factor found in step 2: 2859153813495302135105360393 Found probable prime factor of 28 digits: 2859153813495302135105360393 Probable prime cofactor 65219432427202213218611042380245134951516556220905814475152737256300473009279433432471898046319778085284772450023 has 113 digits   2009-05-04, 17:52   #10
akruppa

"Nancy"
Aug 2002
Alexandria

9A316 Posts Quote:
 Originally Posted by 10metreh Sorry if I was making a mistake, but I was simply diving out the factor x - 1 from x^7 - 1 to get the polynomial x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 with root 172827552198815888791. (Yes, you are more experienced than me, so you are giving me advice if you correct me.)
Sorry, I made a mistake, too: I meant the first number, where you start with the 13th cyclotomic polynomial. My apologies.

I thought your telling Joseph off ("I suggest you factor them yourself.") wasn't called for.

Alex   2009-05-04, 18:04   #11
10metreh

Nov 2008

232210 Posts Quote:
 Originally Posted by akruppa Sorry, I made a mistake, too: I meant the first number, where you start with the 13th cyclotomic polynomial. My apologies. I thought your telling Joseph off ("I suggest you factor them yourself.") wasn't called for. Alex
Actually I wasn't meaning to tell him off. I was just informing him that there were easy numbers in that group. (I did not say "I order you to factor them yourself".)

Anyway: I managed to get the polynomials x^6 + x^5 - 5x^4 - 4x^3 + 6x^2 + 3x - 1 and -10177286401x + 103577158487979532802 using phi (yes, I know that is your program). I presume the method used is similar to the one mentioned on the wiki for x^13k-1.

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