User talk:Ninho

Welcome to my little math world !
Hullo ! Not much to see here I'm afraid. I'll think about making this page nicer... meanwhile, here's my sketch of a proof of a result in elementary number theory which I set myself to prove as an exercise in neuronal activity. The demonstration isn't meant to be published or peer-reviewed, I being but an isolated amateur and not actively at that; on the flip side, it is simple enough that it can be read, understood and checked without undue effort by someone interested in the field. I hope I didn't leave a gaping flaw, one I would have to blush about... Amateurs are amateurs after all, just expect them to be amateurish at times... As this page is in progress, so please make a note to come back later and see if it's got any better; your comments at the bottom of this page will be welcome...

Note : I make no claim that the below amusing proposition is new - if you know it's been asserted before, please leave a note in the comments section.

I'm copy/pasting from the sketch of a proof which I prepared to send to the maintainer of the site : primepuzzles.net (apparently his mail form does not work). Therefore the argument was made concise by omitting or collapsing "trivial" steps. You've been warned.

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Proposition : There is no Carmichael number having exactly three prime factors, two of which are twin primes.
To prove this we shall assume such a number exists and proceed ad absurdum.

Let p, p+2, r be three positive odd primes such that their product is a Carmichael number. Note that we do not assume r>p or otherwise; we'll soon find out however.

Korselt's well-known relations of divisibility give :
 * p-1 | (p+2)r-1
 * p+1 | p.r-1
 * r-1 | p(p+2)-1

Of those we won't even use the first; the second condition yields :
 * p+1 | r+1  hence p<r as announced, so 2 < p < p+2 < r

Now let k denote the implied factor in the third of Korselt's relations. We must have 2 ≤ k < p (well known, or easily derived from the above inequality.)

Rewriting Korselt's third relation as an equation and adding 2k to each side we get :

k(r+1) = (p+1)² +2k -2

and since p+1 divides r+1 so it must divide the positive quantity 2k-2. The latter being less than 2p, the ratio (2k-2)/(p+1) can only be unity, in other words : 2k = p+3.

Bringing things together, Korselt's third becomes : r+1 = 2.(p+1)(p+2)/(p+3) , but this is impossible since the right hand side cannot be integral whenever p>1.

The proposition follows ab absurdo. QED.

Does it extend to more than three factors ?
Having shown no Carmichael number of three prime factors (C3) can be divisible by twin primes, one may feel entitled to ask whether this impossibility extends to Carmichael numbers having 4, 5 or more prime factors. It turns out not to be the case, as a list of factored out Carmichael numbers (such as that one - external link -) will show upon inspection.

Among C4 : 41041 = 7 x 11 x 13 x 41, the smallest 4-prime-factor Carmichael number, already has a pair of twins among his factors, and so has the next one :  62745 = 3 x 5 x 47 x 89.

The smallest C5 is even more twin-friendly, with not one but two pairs of twin prime factors : 825265 = 5 x 7 x 17 x 19 x 73.

Among C6 : 413631505 sports twin factors 5 and 7. (This is the second smallest C6).

The second smallest C7, 6295936465, also has factors 5 and 7.

The first C8 is 232250619601 and has twin factors 11 and 13...

As we now see, C3 numbers stand apart from the general Carmichael numbers as far as twin factors go. It seems a significantly large proportion of the numbers in C4, C5 and higher classes come with twin factors, at least among (relatively) small members. Does this trend carry on as numbers grow, or are we seing a "small number effect" ? Is there an infinite quantity of numbers in C4, respectively C5 and higher Cn, having twin factors ? Could it be that those collections were finite for every Cn, their union still being infinite ? Big questions ! Let's not forget nobody even knows with certainty whether any individual classes Cn are infinite (only their union, viz the set of all Carmichael numbers was proven infinite), and of course to this day nobody even knows whether the set of all twin primes is itself infinite. More homework for mathematicians, though I guess any of the above questions are way tougher than the subject of the above "proposition" !

Back to the C3s
This section is being written. It is hardly a sketch now.

Let pxqxr be the general C3, now assuming p < q < r. We are interested in what values d= q-p can take. Of course d must be even. Above we showed d > 2, the next question then arises, can  d equal 4 ?

... ... It can't ! ! Demonstration similar to but more complicated than the case of twins above, will be here.... Need to check and simplify as possible... ... ...

Next value 6 is possible, as exemplified by Hardy-Ramanujan's taxi-cab number 1729=7×13×19 !

... ... more text ... ...

In general, for any fixed value of d, there are only a finite number of C3s. Proof goes along the same line as the particular cases above, needs some quiet time to check!

More careful analysis might exhibit an explicit upper bound for p (or for p×q×r) given 'd'''.

I suspect similar results hold for r-q ...

Visitors' comments and notes
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Ninho (talk) 12:13, 3 December 2008 (UTC) Ninho

Your account will be renamed
Hello,

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