... Thankyou ... I certainly didn't mean to imply this. I was sure you had done it algebraically. It just wasn't apparent to me how. As I said I found 4...
22534
Mark
marku606
Feb 3, 2011 4:43 am
... [snip] Other counterexamples are 7,19, 23,43,47, ..... unless I somehow misunderstood your conjecture. But with such easy counterexamples I don't...
22535
Dimiter Skordev
dskordev
Feb 3, 2011 12:50 pm
I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A...
22536
maximilian_hasler
maximilian_h...
Feb 3, 2011 1:27 pm
Where do you disagree with the following reasoning I mailed you privately: Let M = max F and P the set of prime numbers. The set P+F is the set of all numbers...
22537
Jack Brennen
jbrennen
Feb 3, 2011 3:46 pm
A concrete example, perhaps... Let the set F be {3,5}. All you need to do is find a prime number p such that 2p-3 and 2p-5 are not prime. Use the chinese...
22538
Dimiter Skordev
dskordev
Feb 3, 2011 3:50 pm
Thank you for your answer. It makes your statement rather plausible, but, anyway, the existence of a number of the form 2p with prime p in some of the holes of...
22539
Jack Brennen
jbrennen
Feb 3, 2011 4:36 pm
My previous message gave the outline of a proof of your original statement which doesn't explicitly use the concept of prime gaps, but rather Dirichlet's...
22540
Dimiter Skordev
dskordev
Feb 3, 2011 5:10 pm
Dear Jack, Your example is nice. But why you do not look for a prime number p of the simpler form 77X+19 ? Best regards, Dimiter...
22541
Jack Brennen
jbrennen
Feb 3, 2011 5:38 pm
Okay, yeah, that would work. :) If p is of the form 77x+19, then 2p is of the form 154x+38, and 2p-3 is always divisible by 7 and 2p-5 is always divisible by...
22542
djbroadhurst
Feb 3, 2011 6:56 pm
... Why bother using 7 and 11 in the first place? We can use one prime in F = {3,5} to wipe out the other: p = 15*x + 19. Then 3 divides 38-5 and 5 divides...
22543
djbroadhurst
Feb 3, 2011 8:26 pm
... More generally, if F = {q,r} contains two distinct odd primes, then any one of the infinity of primes of the form p = q*r*x + (q+r)/2 solves the problem,...
22544
Kermit Rose
kermit1941
Feb 3, 2011 9:13 pm
I followed the links given by David, and taking the hints given, constructed this identity: m**2 A**4 - m * (m-2) A**2 B**2 + B**4 = ( m * A**2 - m * A * B +...
22545
djbroadhurst
Feb 4, 2011 5:24 am
... That 4th degree equation is useful (in the Aurifeullian sense) only at m=2. As I explained, the general Aurifeullian identity for Phi(n,x) requires (among...
22546
djbroadhurst
Feb 4, 2011 10:18 am
... Jean-Louis gave a cyclotomic number with no algebraic help, to indicate his valour in factorization the hard way. Exercise 3: Prove that N = (46225^137 -...
22547
Dimiter Skordev
dskordev
Feb 4, 2011 3:50 pm
On January 8, 2011, I posted a message with the subject "Primality with respect to multiplication modulo n". It was a query about proving or refuting a certain...
22548
mikeoakes2
Feb 4, 2011 3:57 pm
... 1. Any prime factor of N is of the form 2*k*137+1, and pfgw with the -f{2*137} option quickly finds these 4 small factors: 1097 2741 23291 5679199 2. Note...
22549
Jaroslaw Wroblewski
jarek372000
Feb 4, 2011 5:34 pm
Below are new record 17 & 18 prime tuplets (and new 17 & 18 Simultaneous Primes records): Prime 18-tuplet 601884606346328759496455407 + d, d = 0, 4, 10, 12,...
22550
djbroadhurst
Feb 4, 2011 6:24 pm
... Congrats to Mike for spotting the factorization. (Also thanks for suppressing many lines of algebraic output :-) ... Indeed. Here is the whole thing, in a...
22551
djbroadhurst
Feb 4, 2011 6:37 pm
... Remark: The same thing works neatly for (2500^137 - 137^137)/(2500 - 137) {f=factor((x^274-137^137)/(x^2-137))[,1]; ...
22552
djbroadhurst
Feb 4, 2011 6:52 pm
... No problem, Dimeter. ... Thanks. Where you write ... which avoids having to use your detailed checks, value by value. ... and then we have only to write...
22553
Jens Kruse Andersen
jkand71
Feb 4, 2011 7:17 pm
... Congratulations! http://users.cybercity.dk/~dsl522332/math/simultprime.htm is updated. -- Jens Kruse Andersen...
22554
djbroadhurst
Feb 5, 2011 1:32 am
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...
22555
mikeoakes2
Feb 5, 2011 12:32 pm
... I misread the rules, and left a different job (based on Exercise 3) running for 1 hour. It found that, with b=137 and a=c^2, for a<=10^8 there is exactly...
22556
djbroadhurst
Feb 5, 2011 1:34 pm
... The rules were intended to be /very/ helpful, allowing the unique Aurifeuillian solution to be found very quickly, by an entirely systematic method. (A...
22557
mikeoakes2
Feb 5, 2011 3:07 pm
... (Yes I know, that's why I was careful to claim it as the solution for a different problem.) With regard to your above gloss on the rules, I wonder how much...
22558
Paul Leyland
xilmanuk
Feb 5, 2011 3:54 pm
... Note that in the case of factoring 2^{4n+2}+1 the Aurifeuillian factors can be found very quickly by non-algebraic methods. In particular, Fermat's...
22559
djbroadhurst
Feb 5, 2011 4:34 pm
... Indeed. Fermat is perfectly tuned to that historical case. But now try it with an exercise in Q(sqrt(137)): how long might it take Fermat to factorize ...
22560
djbroadhurst
Feb 5, 2011 5:12 pm
... That won't work for Exercise 4, since pfgw -f{137}, applied to ABC2 if($a>$b,($a^137+$b^137)/($a+$b),0) & ($a^137-$b^137)/($a-$b) a: from 216 to 262 b:...
22561
mikeoakes2
Feb 5, 2011 5:49 pm
... Your intuition is spot on, as per usual, good Sir: they were pfgw jobs of precisely this form that I ran. Mike...
22562
James Merickel
merk7777777
Feb 5, 2011 6:38 pm
Hi. This is my first post here. I'm inquiring as to whether somebody might want to put a fair amount of power behind a search I'm doing. I have the first...
