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From: David C. Ullrich on 11 Jun 2010 10:23 On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" <lab2k1(a)gmail.c0m> wrote: >"TCL" <tlim1(a)cox.net> wrote... >> On Jun 10, 3:56 pm, riderofgiraffes wrote: >>> >>> This isn't new, this isn't deep, and it isn't a mystery, >>> but I sometimes use this to get kids interested in proof. >>> >>> For prime p>3, p^2-1 is a multiple of 24. >>> >>> It's not hard to prove, but it's a good one to get kids >>> thinking about how we can know something is always true >>> without just checking every number. >> >> Here is an improvement: >> >> For any odd number n not divisible by 3, n^2-1 is a multiple of 24. > >Right, of course. However, whether that's an improvement or not >(pedagogically speaking) largely depends on the target audience and >intended point. The original statement leaves room for "relevancy of >the hypothesis" followups like "now that you've proved it, where was >the primeness of p used in the proof? ...or, if it wasn't, then what >could be a more general condition on p for the conclusion to hold". Or: The original is actually _harder_ to prove, even though it's a weaker statement. >Liviu > > >
From: alainverghote on 11 Jun 2010 14:11 On 11 juin, 16:23, David C. Ullrich <ullr...(a)math.okstate.edu> wrote: > On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" <lab...(a)gmail.c0m> wrote: > >"TCL" <tl...(a)cox.net> wrote... > >> On Jun 10, 3:56 pm, riderofgiraffes wrote: > > >>> This isn't new, this isn't deep, and it isn't a mystery, > >>> but I sometimes use this to get kids interested in proof. > > >>> For prime p>3, p^2-1 is a multiple of 24. > > >>> It's not hard to prove, but it's a good one to get kids > >>> thinking about how we can know something is always true > >>> without just checking every number. > > >> Here is an improvement: > > >> For any odd number n not divisible by 3, n^2-1 is a multiple of 24. > > >Right, of course. However, whether that's an improvement or not > >(pedagogically speaking) largely depends on the target audience and > >intended point. The original statement leaves room for "relevancy of > >the hypothesis" followups like "now that you've proved it, where was > >the primeness of p used in the proof? ...or, if it wasn't, then what > >could be a more general condition on p for the conclusion to hold". > > Or: The original is actually _harder_ to prove, even though > it's a weaker statement. > > > > >Liviu- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents -- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Yes, David Since p^2-1 = 24 for p=5 We may say any prime number is equal to 6q+1 or 6q+5 , q>0 That is p^2-1 = 12q(3q+1) or 12(q+1)(3q+2) ... Alain
From: Chip Eastham on 11 Jun 2010 14:24 On Jun 10, 7:15 pm, TCL <tl...(a)cox.net> wrote: > On Jun 10, 3:56 pm, riderofgiraffes <mathforum.org...(a)solipsys.co.uk> > wrote: > > > This isn't new, this isn't deep, and it isn't a mystery, > > but I sometimes use this to get kids interested in proof. > > > For prime p>3, p^2-1 is a multiple of 24. > > > It's not hard to prove, but it's a good one to get kids > > thinking about how we can know something is always true > > without just checking every number. > > > Just an idle thought. I'll now return you to your usual > > program of deep maths and troll baiting. > > Here is an improvement: > > For any odd number n not divisible by 3, n^2-1 is a multiple of 24. > > -TCL It also gets us to a converse, for integer n: n^2 - 1 is a multiple of 6 if and only if n is not divisible by 2 or 3. --c
From: Bill Dubuque on 11 Jun 2010 21:46 Chip Eastham <hardmath(a)gmail.com> wrote: > On Jun 10, 7:15 pm, TCL <tl...(a)cox.net> wrote: >> On Jun 10, 3:56 pm, riderofgiraffes <mathforum.org...(a)solipsys.co.uk> >> wrote: >> >>> This isn't new, this isn't deep, and it isn't a mystery, >>> but I sometimes use this to get kids interested in proof. >> >>> For prime p>3, p^2-1 is a multiple of 24. >> >>> It's not hard to prove, but it's a good one to get kids >>> thinking about how we can know something is always true >>> without just checking every number. >> >>> Just an idle thought. I'll now return you to your usual >>> program of deep maths and troll baiting. >> >> Here is an improvement: >> >> For any odd number n not divisible by 3, n^2-1 is a multiple of 24. >> >> -TCL > > It also gets us to a converse, for integer n: > > n^2 - 1 is a multiple of 6 if and only if n is > not divisible by 2 or 3. Below are some more general results from some of my old posts: KY <wkfkh...(a)yahoo.co.jp> wrote: > > Mod[n^9 - n^3, 9] = 0 HINT Trivial if 3|n, else 3|n^6-1 by ELT = Euler's little theorem [1], i.e. (a,n) = 1 => a^E(n) = 1 (mod n) for E(n) = Euler_phi(n) More generally see my post [4] on the Carmichael lambda function (esp. Theorem 2) - which I've appended below for convenience: LEMMA m squarefree => n^(phi(m)+1) = n (mod m) This has a very simple direct proof, e.g. see my sci.math post [1] on Apr 24, 2003. However, with only a little more effort one may prove the following generalizations of the Euler/Fermat little Theorems to arbitrary moduli. THEOREM 1 For natural numbers a,e,n with e,n>1 n|a^e-a for all a <=> n squarefree & prime p|n => p-1|e-1 PROOF (<=) Since a squarefree natural divides another iff all its prime factors do, we need only show p|a^e-a for each prime p|n, or, that a != 0 => a^(e-1) = 1 mod p, which, since p-1|e-1, follows from a != 0 => a^(p-1) = 1 mod p, by FlT (Fermat's little Theorem). (=>) Given that n|a^e-a for all naturals a we must show (1) n is squarefree and (2) p|n => p-1|e-1. (1) If n isn't squarefree it has a nontrivial divisor aa hence aa|n|a^e-a => aa|a =><= (via e>1 => aa|a^e) (2) Let a be a generator of the multiplicative group of Z/p. Thus a has order p-1. Now p|n|a(a^(e-1)-1) but not p|a, so a^(e-1) = 1 mod p, thus e-1 must be divisible by p-1, the order of a mod p. QED Notice that the least such e is e = 1 + lcm{p_i - 1}, i.e. 1 + L(n), where L = Carmichael lambda defined below. i.e. a^(L(n)+1) = a (mod n) for all a, squarefree n. Case e = n is Korselt's criterion for Carmichael numbers [2]. Similarly we prove an analogous criterion for n|a^e-1 by using the cyclicity of the group Z/p^k* (vs. Z/p* above). for odd p, and using Z/2^k* = C(2) * C(2^(k-2)) for k>2, i.e. with notation: p^k||n means p^k|n but not p^(k+1)|n THEOREM 2 For natural numbers a,e,n with e,n>1 n|a^e-1, all a coprime n <=> prime p^k||n => E(p^k)|e where E(p^k) = phi(p^k) for odd primes p, or p=2, k<3 and E(2^k) = 2^(k-2) for k>2 The latter exception is due to the reduced universal expt in the noncyclic group Z/2^k* = C(2) * C(2^(k-2)) for k>2. Notice that the least such e is L(n) := lcm{E(p_i^k_i)}. Function L(n) is known as the Carmichael lambda function, or the universal exponent of the multiplicative group Z/n*. --Bill Dubuque [1] http://google.com/groups?selm=y8zhe8n5383.fsf%40nestle.ai.mit.edu [2] http://google.com/groups?selm=y8zmzpsrxr2.fsf%40nestle.csail.mit.edu [3] http://en.wikipedia.org/wiki/Euler's_totient_function [4] http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist;task=show_msg;msg=1000.0001.0001.0001.0001.0003
From: Liviu on 12 Jun 2010 01:40
"David C. Ullrich" <ullrich(a)math.okstate.edu> wrote... > On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" wrote: >>"TCL" <tlim1(a)cox.net> wrote... >>> On Jun 10, 3:56 pm, riderofgiraffes wrote: >>>> >>>> This isn't new, this isn't deep, and it isn't a mystery, >>>> but I sometimes use this to get kids interested in proof. >>>> >>>> For prime p>3, p^2-1 is a multiple of 24. >>>> >>>> It's not hard to prove, but it's a good one to get kids >>>> thinking about how we can know something is always true >>>> without just checking every number. >>> >>> Here is an improvement: >>> >>> For any odd number n not divisible by 3, n^2-1 is a multiple of 24. >> >>Right, of course. However, whether that's an improvement or not >>(pedagogically speaking) largely depends on the target audience and >>intended point. The original statement leaves room for "relevancy of >>the hypothesis" followups like "now that you've proved it, where was >>the primeness of p used in the proof? ...or, if it wasn't, then what >>could be a more general condition on p for the conclusion to hold". > > Or: The original is actually _harder_ to prove, even though > it's a weaker statement. True. Which is why every now and then the IMO has a problem stated in terms of the current year like "let n=2010 and prove that...". Leaving it to the imagination to figure out what exactly it is about that particular number which actually matters in the given context. Liviu |