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From: alainverghote on 12 Jun 2010 07:00 On 12 juin, 07:40, "Liviu" <lab...(a)gmail.c0m> wrote: > "David C. Ullrich" <ullr...(a)math.okstate.edu> wrote... > > > > > > > On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" 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. > > 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- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Good morning, We've got particular solutions based upon simple polynomial equalities, here From A^3- B^3 = (A-B)^3 +3AB(A-B) and n^3-n =(n-1)n(n+1)<=>0 mod.3 Or (n^3)^3-n^3 = (n^3-n)^3+n^4*[3(n^3-n)] <=> 0 mod.9 Alain
From: Rob Johnson on 12 Jun 2010 08:13 In article <d9f37225-9ec7-4d19-b45b-2e1dbb9fadba(a)a30g2000yqn.googlegroups.com>, "alainverghote(a)gmail.com" <alainverghote(a)gmail.com> wrote: >On 12 juin, 07:40, "Liviu" <lab...(a)gmail.c0m> wrote: >> "David C. Ullrich" <ullr...(a)math.okstate.edu> wrote... >> > On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" 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. >> >> 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- Masquer le texte des messages precedents - >> >> - Afficher le texte des messages precedents - > >Good morning, > >We've got particular solutions based upon >simple polynomial equalities, here >From A^3- B^3 = (A-B)^3 +3AB(A-B) >and n^3-n =(n-1)n(n+1)<=>0 mod.3 > >Or (n^3)^3-n^3 = (n^3-n)^3+n^4*[3(n^3-n)] <=> 0 mod.9 Another way to look at this: If x^2 = 1 mod 9, then (x-1)(x+1) = 0 mod 9. (x+1) - (x-1) = 2 mod 9, therefore, both cannot be 0 mod 3, so either x = +1 mod 9 or x = -1 mod 9. Since phi(9) = 6, x^6 = 1 mod 9 for all x not divisible by 3. This means that if x is not divisible by 3, x^3 must be either +1 or -1 mod 9. If x is divisiblw by 3, then x^3 is 0 mod 9. Thus, x^3 is one of -1, 0, or +1 mod 9. Therefore, x^9 - x^3 = (x^3 - 1) x^3 (x^3 + 1) = 0 mod 9 Actually, if 3 divides x then 9 divides x^2, so we have x^8 - x^2 = (x^3 - 1) x^2 (x^3 + 1) = 0 mod 9 Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font
From: alainverghote on 12 Jun 2010 08:53 On 12 juin, 14:13, r...(a)trash.whim.org (Rob Johnson) wrote: > In article <d9f37225-9ec7-4d19-b45b-2e1dbb9fa...(a)a30g2000yqn.googlegroups..com>, > > > > > > "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> wrote: > >On 12 juin, 07:40, "Liviu" <lab...(a)gmail.c0m> wrote: > >> "David C. Ullrich" <ullr...(a)math.okstate.edu> wrote... > >> > On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" 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. > > >> 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- Masquer le texte des messages precedents - > > >> - Afficher le texte des messages precedents - > > >Good morning, > > >We've got particular solutions based upon > >simple polynomial equalities, here > >From A^3- B^3 = (A-B)^3 +3AB(A-B) > >and n^3-n =(n-1)n(n+1)<=>0 mod.3 > > >Or (n^3)^3-n^3 = (n^3-n)^3+n^4*[3(n^3-n)] <=> 0 mod.9 > > Another way to look at this: > > If x^2 = 1 mod 9, then (x-1)(x+1) = 0 mod 9. (x+1) - (x-1) = 2 > mod 9, therefore, both cannot be 0 mod 3, so either x = +1 mod 9 or > x = -1 mod 9. > > Since phi(9) = 6, x^6 = 1 mod 9 for all x not divisible by 3. This > means that if x is not divisible by 3, x^3 must be either +1 or -1 > mod 9. If x is divisiblw by 3, then x^3 is 0 mod 9. Thus, x^3 is > one of -1, 0, or +1 mod 9. Therefore, > > x^9 - x^3 = (x^3 - 1) x^3 (x^3 + 1) = 0 mod 9 > > Actually, if 3 divides x then 9 divides x^2, so we have > > x^8 - x^2 = (x^3 - 1) x^2 (x^3 + 1) = 0 mod 9 > > Rob Johnson <r...(a)trash.whim.org> > take out the trash before replying > to view any ASCII art, display article in a monospaced font- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Good afternoon Rob, OK, we may also look at moduli: n mod.9 0 1 2 3 4 5 6 7 8 n^3 0 1 8 0 1 8 0 1 8 Thence n->n^3 {0,1,8} -> {0,1,8} Idem n^3->n^9 Any comments? Alain
From: Bill Dubuque on 12 Jun 2010 13:06 rob(a)trash.whim.org (Rob Johnson) wrote: > In article <d9f37225-9ec7-4d19-b45b-2e1dbb9fadba(a)a30g2000yqn.googlegroups.com>, > "alainverghote(a)gmail.com" <alainverghote(a)gmail.com> wrote: >>On 12 juin, 07:40, "Liviu" <lab...(a)gmail.c0m> wrote: >>> "David C. Ullrich" <ullr...(a)math.okstate.edu> wrote... >>>> On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" 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. >>> >>> 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. You should've said that you got the problem below from my other post in this thread, so others didn't try to reinvent the wheel (as below). >> We've got particular solutions based upon simple polynomial equalities, >> here. From A^3 - B^3 = (A-B)^3 + 3AB(A-B) >> and n^3-n = (n-1)n(n+1) <=> 0 mod.3 >> >> Or (n^3)^3-n^3 = (n^3-n)^3+n^4*[3(n^3-n)] <=> 0 mod.9 That's more work than applying ElT (Euler-phi little Theorem) as I did in my simple one-line proof in my other post here, viz. trivially 9|n^9-n^3 if 3|n, else 9|n^6-1|n^9-n^3 by ElT i.e. (a,n) = 1 => a^E(n) = 1 (mod n) for E(n) = Euler_phi(n) Generally such elementary problems are all special cases of the generalized Euler-Fermat-Carmichael theorems that I discussed in my other post here (originally posted to Ask an Algebraist [4]). > Another way to look at this: > > If x^2 = 1 mod 9, then (x-1)(x+1) = 0 mod 9. (x+1)-(x-1) = 2 mod 9, > therefore, both cannot be 0 mod 3, so x = +-1 mod 9. > Since phi(9) = 6, x^6 = 1 mod 9 for all x not divisible by 3. This > means that if x is not divisible by 3, x^3 must be either +-1 mod 9. > 3|x => x^3 = 0 mod 9. Thus, x^3 is one of -1,0,+1 mod 9. Therefore, > > x^9 - x^3 = (x^3 - 1) x^3 (x^3 + 1) = 0 mod 9 > > Actually, if 3 divides x then 9 divides x^2, so we have > > x^8 - x^2 = (x^3 - 1) x^2 (x^3 + 1) = 0 mod 9 It's trivial: 3|x => 9|x^2|x^9-x^3 That's a long-winded form of said ElT-based proof. You can find some far-reaching generalizations in my other post here. Btw, the link [4] there to my Ask An Algebraist post is now stale. Instead please use [4] Generalized Euler Fermat Carmichael theorems, September 25, 2009 http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2009;task=show_msg;msg=2599 --Bill Dubuque
From: alainverghote on 13 Jun 2010 04:31
On 12 juin, 19:06, Bill Dubuque <w...(a)nestle.csail.mit.edu> wrote: > r...(a)trash.whim.org (Rob Johnson) wrote: > > In article <d9f37225-9ec7-4d19-b45b-2e1dbb9fa...(a)a30g2000yqn.googlegroups.com>, > > "alainvergh...(a)gmail.com" <alainvergh...(a)gmail.com> wrote: > >>On 12 juin, 07:40, "Liviu" <lab...(a)gmail.c0m> wrote: > >>> "David C. Ullrich" <ullr...(a)math.okstate.edu> wrote... > >>>> On Thu, 10 Jun 2010 23:43:38 -0500, "Liviu" 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. > > >>> 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. > > You should've said that you got the problem below from my other post > in this thread, so others didn't try to reinvent the wheel (as below). > > >> We've got particular solutions based upon simple polynomial equalities, > >> here. From A^3 - B^3 = (A-B)^3 + 3AB(A-B) > >> and n^3-n = (n-1)n(n+1) <=> 0 mod.3 > > >> Or (n^3)^3-n^3 = (n^3-n)^3+n^4*[3(n^3-n)] <=> 0 mod.9 > > That's more work than applying ElT (Euler-phi little Theorem) > as I did in my simple one-line proof in my other post here, viz. > > trivially 9|n^9-n^3 if 3|n, else 9|n^6-1|n^9-n^3 by ElT > > i.e. (a,n) = 1 => a^E(n) = 1 (mod n) for E(n) = Euler_phi(n) > > Generally such elementary problems are all special cases of the > generalized Euler-Fermat-Carmichael theorems that I discussed > in my other post here (originally posted to Ask an Algebraist [4]). > > > Another way to look at this: > > > If x^2 = 1 mod 9, then (x-1)(x+1) = 0 mod 9. (x+1)-(x-1) = 2 mod 9, > > therefore, both cannot be 0 mod 3, so x = +-1 mod 9. > > Since phi(9) = 6, x^6 = 1 mod 9 for all x not divisible by 3. This > > means that if x is not divisible by 3, x^3 must be either +-1 mod 9. > > 3|x => x^3 = 0 mod 9. Thus, x^3 is one of -1,0,+1 mod 9. Therefore, > > > x^9 - x^3 = (x^3 - 1) x^3 (x^3 + 1) = 0 mod 9 > > > Actually, if 3 divides x then 9 divides x^2, so we have > > > x^8 - x^2 = (x^3 - 1) x^2 (x^3 + 1) = 0 mod 9 > > It's trivial: 3|x => 9|x^2|x^9-x^3 > > That's a long-winded form of said ElT-based proof. You can find some > far-reaching generalizations in my other post here. Btw, the link [4] > there to my Ask An Algebraist post is now stale. Instead please use > > [4] Generalized Euler Fermat Carmichael theorems, September 25, 2009http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2009;task=sho... > > --Bill Dubuque- Masquer le texte des messages précédents - > > - Afficher le texte des messages précédents - Bonjour Bill, Since I'm very 'primitive' in math I do not know a word about Euler and Carmichael theorems. For the time being I play with polynomials equalities thence my attack A^3 - B^3 = (A-B)^3 + 3AB(A-B). Is there any extension with moduli to forms (x^n)^p-x^p ? Best regards, Alain |