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 numbers theory which I set myself to prove as an exercise in neuronal activity. The demo hasn't been published or peer-reviewed, i'm but an isolated amateur and not very active at that; on the flip side, it is simple enough that it can be read, understood and checked without undue effort by an interested person. 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...
Note : I have no idea if the amusing result below is new, or even if you'll agree to find it interesting.
Final note : 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). Hence the PGP armour and signature ! Hence the sketchy style also.
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Proposition : There is no Carmichael number of 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.
The well-known conditions of Korselt specialise thus :
- 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 quantity 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 when p>1.
The proposition follows ab absurdo. QED.
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